# CHEM N-12 November In the electrolytic production of Al, what mass of Al can be deposited in 2.00 hours by a current of 1.8 A?

Save this PDF as:

Size: px
Start display at page:

Download "CHEM N-12 November In the electrolytic production of Al, what mass of Al can be deposited in 2.00 hours by a current of 1.8 A?"

## Transcription

1 CHEM N-1 November 014 In the electrolytic production of Al, what mass of Al can be deposited in.00 hours by a current of 1.8 A? The number of moles of electrons passed in.00 hours by a current of 1.8 A is: number of moles of electrons = It / F = (1.8 A)( s) / C mol -1 = 0.13 mol Aluminium is produced from Al O 3 which contains Al electrons are needed to produce each Al so 3 mol of electrons are needed to produce 1 mol of Al. This quantity of electrons will therefore deposit: number of moles of Al = 0.13 / 3 mol = mol As Al has a molar mass of 6.98 g mol -1, this quantity corresponds to: mass of Al = number of moles molar mass = mol 6.98 g mol -1 = 1. g Answer: 1. g What products would you expect at the anode and the cathode on electrolysis of a 1 M aqueous solution of NiI? Explain your answers. At the cathode, there are two possible reduction reactions: Ni + (aq) + e - à Ni(s) H O(l) + 4H + (aq) + 4e - à H (g) + OH - (aq) E o = -0.4 V E o = V Reduction of Ni + (aq) is easier, even without considering an overpotential for water. At the anode, there are two possible oxidation reactions: I - (aq) à I (g) + e - H O(l) à O (g) + 4H + (aq) + 4e - E o = -0.6 V E o = -0.8 V Both reactions will have an overpotential but oxidation of iodine is easier and this will probably occur. Overall, Ni(s) will be produced at the cathode and I (g) will be produced at the anode.

2 CHEM N-13 November 014 An electrochemical cell is consists of 1.0 L half-cells of Fe/Fe + and Cd/Cd + with the following initial concentrations: [Fe + ] = M, [Cd + ] = 0.00 M. What is the initial E cell at 5 C? 8 From the reduction potential table, E o cell (Fe + (aq) + e Fe(s)) = V E o cell (Cd + (aq) + e Cd + (aq)) = V The Fe + /Fe half cell has the more negative reduction potential so it is the half cell that is turned around to act as the oxidation half cell: E cell o (Fe(s) Fe + (aq) + e ) = V In combination with the Cd + /Cd reduction half cell, this gives an overall reaction and cell potential of: Fe(s) + Cd + (aq) Fe + (aq) + Cd(s) E o = ((+0.44) + (-0.40)) V = V For this reaction with [Fe + (aq)] = M and [Cd + (aq)] = 0.00 M: Q = [Fe! aq ] = [Cd! aq ] 0.00 For the e - reaction, the Nernst equation gives the cell potential as: E cell = E - RT nf lnq = (0.04 V) (8.314 J K!1 mol!1 )(98 K) ( C mol!1 ) ln = +0.0 V What is E cell when [Cd + ] reaches 0.15 M? Answer: +0.0 V [Cd + (aq)] has decreased from 0.00 M to 0.15 M: a change of 0.05 M. [Fe + (aq)] will increase by the same amount: [Fe + (aq)] = ( ) M = 0.85 M. Using the Nernst equation again gives: E cell = E - RT nf lnq = (0.04 V) (8.314 J K!1 mol!1 )(98 K) ( C mol!1 ) ln = +0.0 V Answer: +0.0 V ANSWER CONTINUES ON THE NEXT PAGE

3 CHEM N-13 November 014 What is [Cd + ] when E cell reaches V? E cell = (0.04 V) (8.314 J K!1 mol!1 )(98 K) ( C mol!1 ) so: [Fe! aq ] [Cd! aq ] = 7.0 ln [Fe+ aq ] [Cd + aq ] = V If the change from the initial concentrations is x, [Cd + (aq)] = (0.00 x) M and [Fe + (aq)] = ( x) M: x 0.00!x = 7.0 x = M So that [Cd + (aq)] = ( ) M = 0.15 M. Answer: 0.15 M What are the equilibrium concentrations of both ions? Using E = RT lnk = 0.04 V gives K =.5 nf so: [Fe! aq ] [Cd! aq ] =.5 If the change from the initial concentrations is x, [Cd + (aq)] = (0.00 x) M and [Fe + (aq)] = ( x) M: x 0.00!x =.5 x = M So that [Cd + (aq)] = ( ) M = 0.04 M and [Fe + (aq)] = ( ) M = M [Cd + ] = 0.04 M [Fe + ] = M

4 CHEM N-11 November 013 What is the electrochemical potential of the following cell at 5 o C? Fe FeSO 4 (0.010 M) (FeSO 4 (0.100 M) Fe 3 As this is a concentration cell, E o = 0 V. The cell notation corresponds to the M solution being the cathode, where reduction occurs, and the M solution being the anode, where oxidation occurs. The two half cells are: Anode: Fe(s) à Fe + (aq, M) + e - Cathode: Fe + (aq, M) + e - à Fe(s) Overall: Fe + (aq, M) à Fe + (aq, M) The potential is given by the Nernst equation for this two electron reaction: E = E - RT nf lnq = (0 V) (8.314 J K!1 mol!1 )(98 K) mol!1 ln (0.010) (0.100) = V Answer: V Calculate the mass of aluminium which can be produced with the same quantity of electricity that is used to produce 1.00 kg of copper metal. As the molar mass of Cu is g mol -1, 1.00 kg corresponds to: number of moles = mass / molar mass = g / g mol -1 = 15.7 mol. Reduction of Cu + requires mol of electrons. Hence, the number of electrons requires to produce 15.7 mol is: number of moles of electrons = 15.7 mol = 31.5 mol Reduction of a mole of Al 3+ requires 3 mol of electrons. Hence, the number of moles of aluminium produced by 31.5 mol of electrons is: number of moles of aluminium = 31.5 / 3 mol = 10.5 mol As the molar mass of aluminium is 6.98 g mol -1, this corresponds to: mass = number of moles molar mass = 10.5 mol 6.98 g mol -1 = 83 g. Answer: 83 g ANSWER CONTINUES ON THE NEXT PAGE

5 CHEM N-11 November 013 Explain why Na(s) cannot be obtained by the electrolysis of aqueous NaCl solutions. From the table of standard reduction potentials: H O + e H (g) + OH (aq) E = 0.83 V Na + (aq) + e Na(s) E =.71 V Water has a much greater reduction potential than Na + and hence is preferentially reduced, even when the overpotential of water is considered.

6 K = Answer: CHEM N-11 November 01 A galvanic cell consists of a Cr 3+ /Cr half-cell with unknown [Cr 3+ ] and a Ni + /Ni halfcell with [Ni + ] = 1.0 M. The electromotive force of the cell at 5 C was measured to be 0.55 V. What is the concentration of Cr 3+ in the Cr 3+ /Cr half-cell? 6 From the standard reduction potentials, Cr 3+ (aq) + 3e - à Cr(s) Ni + (aq) + e - à Ni(s) E o = V E o = -0.4 V The most negative is reversed to give an overall reaction and cell potential of 3Ni + (aq) + Cr(s) à 3Ni(s) + Cr 3+ (aq) E o = ( ) V = 0.50 V From the Nernst equation for this 6 electron reaction, E = E - RT RT ln[cr3! aq lnq = E - ] nf nf [Ni! aq ] 3 = (0.50 V) (8.314 J K!1 mol!1 )(98 K) mol!1 ln [Cr3! aq ] (1.0) 3 Solving this gives, [Cr 3+ (aq)] = M Answer: M Calculate the equilibrium constant of the reaction at 5 C. The equilibrium constant is related to the standard cell potential through: E o = RT nf lnk Using E o = V, 0.50 V = (8.314 J K!1 mol!1 )(98 K) mol!1 lnk Solving this gives: ANSWER CONTINUES ON THE NEXT PAGE

7 CHEM N-11 November 01 Calculate the standard Gibbs free energy of the reaction at 5 C. The Gibbs free energy change is related to the standard cell potential through: ΔG o = -nfe o = - 6 (96485 mol -1 ) (0.50 V) = -90 kj mol -1 Answer: -90 kj mol -1 Express the overall reaction in the shorthand voltaic cell notation. Cr(s) Cr 3+ (aq) Ni + (aq) Ni(s)

8 CHEM N-1 November 01 A strip of copper and a strip of zinc are embedded in a lemon, and are connected by wires to a voltmeter; a voltage is generated and can be read at the voltmeter. What chemical reactions are occurring that lead to the generation of current? 3 Zn(s) Zn + (aq) + e at the anode H + (aq) + e H (g) at the cathode Assuming there are no losses in the circuit and the conditions are similar to standard, what voltage can be read at the voltmeter? The Zn + / Zn reduction potential is the more negative so is reversed to give E o ox = V. The H + / H reduction potential is E o red = 0.00 V. Overall E o = E o ox + E o red = ( ) V = V

9 CHEM N-13 November 01 A 0.0 ml sample of 0.11 M Fe + in an acid solution was used to titrate 3.5 ml of a KMnO 4 solution of unknown concentration. Write the balanced redox reaction that occurs in solution upon titration, and calculate the molarity of the KMnO 4 solution. 4 From the standard reduction potentials, the two relevant half cells are: MnO 4 - (aq) + 8H + (aq) + 5e - à Mn + (aq) + 4H O(l) Fe + (aq) à Fe 3+ (aq) Giving the overall reaction: 5Fe + (aq) + MnO 4 - (aq) + 8H + (aq) à 5Fe 3+ (aq) + Mn + (aq) + 4H O(l) The number of moles of Fe + used in the titration is: number of moles of Fe + = concentration volume = 0.11 mol L -1 ) (0.000 L) = mol From the balanced equation, the number of moles of MnO 4 - (aq) is therefore: number of moles of MnO 4 - = 1/ mol = mol This amount is present in 3.5 ml so its concentration must be: concentration = number of moles / volume = mol / L = M Answer: M

10 CHEM N-10 November 010 How many minutes would be required to obtain 10.0 g of liquid mercury by passing a constant current of 0.17 A through a solution containing Hg (NO 3 ) (aq)? The number of moles of Hg in 10.0 g is: amount of mercury = mass / molar mass = (10.0 g) / (00.59 g mol -1 ) = mol The redox reaction is the electron process below: Hg + (aq) + e - Hg Hence, for each mole of Hg, 1 mol of electrons is required so mol of electrons are required. The number of moles of electrons passed by a current I in a time t is given by: number of moles of electrons = It / F The time required to pass mol of electrons using I = 0.17 A is therefore: t = ( mol) (96485 C mol -1 ) / (0.17 A) = 8000 s = 470 mins Answer: 470 mins

11 CHEM N-11 November 010 Calculate ΔG for the following reaction: 3Cu(s) + Cr 3+ (aq) 3Cu + (aq) + Cr(s) 3 The two electrode potentials are: Cu(s) à Cu + (aq) + e - Cr 3+ (aq) +3e - à Cr(s) E ox = V E red = V The overall cell potential is therefore: E = E ox + E red = (-0.53 V) + (-0.74 V) = -1.7 V Using ΔG = -nfe for this 3 electron transfer reaction: ΔG = -( C mol -1 ) (-1.7 V) = +368 kj mol -1 Answer: +368 kj mol -1 Is the reaction spontaneous under standard conditions? Give a reason for your answer. No. ΔG > 0 and it must be negative for a spontaneous reaction. Equivalently, E < 1 and it must be positive for a spontaneous reaction.

12 CHEM N-15 November 009 The standard reduction potential of phosphorous acid to hypophosphorous acid is V, with the following half-reaction: H 3 PO 3 (aq) + H + (aq) + e H 3 PO (aq) + H O(l) What would the reduction potential be for this half reaction at a temperature of 5 C in an aqueous solution with ph of.3 and concentrations of [H 3 PO 3 (aq)] = 0.37 M and [H 3 PO (aq)] = M? 3 As ph = -log 10 [H + (aq)], [H + (aq)] = M. With [H 3 PO 3 (aq)] = 0.37 M and [H 3 PO (aq)] = M, the reaction quotient, Q, is given by: Q = H 3 PO aq = (0.0005) H 3 PO aq [H! aq ] 0.37 (10!.3 ) = 7 The reduction potential for this electron reduction is given by the Nernst equation: E = E - RT (8.314 (5!73) lnq = ( ln(7)) V = V nf Answer: V A number of bacteria can reduce the nitrate ion in the presence of sulfur. A simplified unbalanced redox reaction can be written as: S(s) + NO 3 (aq) SO (g) + NO(g) Balance this redox equation for acidic conditions. 3S(s) + 4H + (aq) + 4NO 3 (aq) 3SO (g) + 4NO(g) + H O(l)

13 CHEM N-16 November 009 What is the value of the equilibrium constant for the following reaction at 98 K? Fe 3+ (aq) + 3Sn(s) Fe(s) + 3Sn + (aq) Relevant electrode potentials can be found on the data page. 3 The relevant reduction potentials are: Fe 3+ (aq) + 3e - à Fe(s) E = V Sn + (aq) + e - à Sn(s) E = V As the Sn + / Sn couple is the more negative, it is reversed giving: E = ( ) V = 0.10 V The equilibrium constant, K, is related to the standard reduction potential using: E = (RT/nF) lnk lnk = nfe / RT = ( ) / ( ) = 3.37 K = e 3.37 = Answer:

14 CHEM N-9 November 008 A galvanic cell is made of a Zn + /Zn half cell with [Zn + ] =.0 M and an Ag + /Ag half cell with [Ag + ] = M. Calculate the electromotive force of the cell at 5 C. 5 The standard reduction reactions and potentials for the two half cells are: Zn + (aq) + e - Zn(s) Ag + (aq) + e - Ag(s) E = V E = V The least positive (Zn + /Zn) couple is reversed giving the overall reaction: Zn(s) + Ag + (aq) Zn + (aq) + Ag(s) E = (+0.76 V) + (0.80) = 1.56 V As non-standard concentrations are used, the cell potential is calculated using the Nernst equation. The reaction involves the transfer of e - so with n = this becomes: E = E RT RT lnq = E nf nf ln [Zn+ aq ] [Ag + aq ] = (+1.56 V) J K 1 mol 1 (98 K) ( C mol 1 ) ln = V Answer: V Calculate the equilibrium constant of the reaction at 5 C. The equilibrium constant is related to the standard cell potential: Hence, E = RT nf lnk lnk = E nf RT = (1.56 V) ( C mol 1 ) J K 1 mol 1 (98 K) = 11.5 K = Answer: K = Calculate the standard Gibbs free energy of the reaction at 5 C. Using ΔG = nfe : ΔG = -( C mol -1 ) (+1.56 V) = 301 kj mol 1 Answer: 301 kj mol 1 ANSWER CONTINUES ON THE NEXT PAGE

15 CHEM N-9 November 008 Indicate whether the reaction is spontaneous or not. Give a reason for your answer. As E > 0, ΔG < 0 and K is very large: the reaction is spontaneous. Express the overall reaction in the shorthand voltaic cell notation. Zn(s) Zn + (aq) (.0 M) Ag + (aq) (0.050 M) Ag(s)

16 CHEM N-11 November 008 A melt of NaCl is electrolysed for 35 minutes with a current of 3.50 A. Calculate the mass of sodium and volume of chlorine at 40 C and 1.00 atm that are formed. 4 The number of moles of electrons delivered by a current, I, of 3.50 A in 35 minutes is: number of moles of electrons = =. = mol The overall electrolysis reaction, NaCl(l) Na(s) + ½Cl (g), corresponds to reduction of Na + and oxidation of Cl - : Na + + e - Na and Cl - ½Cl + e - As one mole of electrons would produce one mole of Na and half a mole of Cl : number of moles of Na = mol number of moles of Cl = ½ mol = mol The mass of Na produced is therefore: mass of Na = number of moles atomic mass = (0.076 mol) (.99 g mol -1 ) = 1.8 g Using the ideal gas law, PV = nrt, the volume of Cl produced is: V =... = 0.98 L

17 CHEM N-13 November 008 A concentration cell containing aqueous solutions of Cu(NO 3 ) and solid copper metal is constructed so that the Cu + ion concentration in the cathode half-cell is 0.66 M. Calculate the concentration of the Cu + ion in the anode half-cell if the cell potential for the concentration cell at 5 C is 0.03 V. The cathode and anode reactions are: Cu + (aq) + e - à Cu(s) Cu(s) à Cu + (aq) + e - (cathode) (anode) The standard electrode potential E = 0 V and the potential can be calculated using the Nernst equation for this electron reaction, n = : E = E RT RT lnq = nf nf ln [Cu! aq ] anode [Cu! aq ] cathode = J K 1 mol 1 (98 K) ( C mol 1 ) This gives [Cu + (aq)] anode = 0.06 M. ln [Cu! aq ] anode 0.66 = V Answer: 0.06 M In acid solution, dichromate ion oxidises iron(ii) to iron(iii) as illustrated in the partial equation: Fe + + Cr O 7 Fe 3+ + Cr 3+ Write a balanced equation for this reaction. 3 The half reactions are: Fe + à Fe 3+ + e - Cr O H + + 6e - à Cr H O where H + has been added to the Cr O 7 - / Cr 3+ couple to give H O. To balance the electrons, the first reaction needs to be multiplied by 6. Hence: 6Fe + + Cr O H + à 6Fe 3+ + Cr H O What would happen to the cell potential if the concentration of Cr 3+ were increased? It would decrease. If [Cr 3+ ] is increased, Le Châtelier s principle predicts that the reaction will shift towards reactants, reducing the cell potential.

18 CHEM N-10 November 007 How many minutes would be required to electroplate 5.0 g of manganese by passing a constant current of 4.8 A through a solution containing MnO 4? 5.0 g of manganese corresponds to number of moles = mass atomicmass 5.0 g = g mol = mol The reduction of MnO - 4 is a 7e - process: MnO - 4 (aq) + 8H + (aq) + 7e - Mn(s) + 4H O(l) Production of mol of Mn(s) requires ( ) = 3.19 mol of electrons. The number of moles of electrons passed by a current I in a time t is given by number of moles of electrons = I t F (4.8 A) t 3.19 mol = C mol so t = s = 1100 minutes Answer: 1100 minutes

19 CHEM N-11 November 007 The solubility product constant of AgCl is K sp = M. Using the relevant electrode potentials found on the data page, calculate the reduction potential at 98 K of a half-cell formed by: (a) an Ag electrode immersed in a saturated solution of AgCl. 6 The standard electrode potential for Ag + (aq) + e - Ag(s) is E = V. This refers to the potential with [Ag + (aq)] = 1 M. For the dissolution of AgCl(s) As [Ag + (aq)] = [Cl - (aq)], Ag + (aq) + Cl - (aq), K sp = [Ag + (aq)][cl - (aq)]. [Ag (aq)] K sp = = M Using the Nernst equation, the cell potential at 98 K (5 C) is, E = E logq n The Ag + (aq) + e - Ag(s) half cell involves one electron and so n = 1. The 1 reaction quotient is. Hence, [Ag (aq)] E = (+0.80) log = +0.5 V Answer: E = +0.5 V (b) an Ag electrode immersed in a 0.5 M solution of KCl containing some AgCl precipitate. [Cl - (aq)] = 0.5 M and as K sp = [Ag + (aq)][cl - (aq)], [Ag + (aq)] = K sp [Cl (aq)] M = 0.5M = M The electrode potential is now, E = (+0.80) log = +0.4 V Answer: E = +0.4 V ANSWER CONTINUES ON THE NEXT PAGE

20 CHEM N-11 November 007 Each of these half-cells is connected to a standard Cu + (1 M)/Cu(s) half-cell. In which half-cell, (a) or (b), will clear evidence of a reaction be seen? Describe the change(s) observed. For the Cu + (1 M)/Cu(s) half cell, the reduction potential is E = V. If the half cell is combined with half cell (a), the former has the least positive cell potential and is reversed: Cu(s) Cu + (aq) + e - Ag + (s) + e - Ag(s) ([Ag + (aq)] = M) Cu(s) + Ag + (aq) Cu + (aq) + Ag(s) E = V E = +0.5 V E = (-0.34)+(+0.5) = V If the half cell is combined with half cell (b), the latter has the least positive cell potential and is reversed: Cu + (aq) + e - Cu(s) E = V Ag(s) Ag + (s) + e - ([Ag + (aq)] = M) E = -0.4 V Cu + (aq) + Ag(s) Cu(s) + Ag + (aq) E = (+0.34)+(-0.4) = V Although both reactions have E > 0 V and so are spontaneous, only the second reaction will give clear evidence of a reaction. The Ag + (aq) ions produced will react with the excess Cl - (aq) present to give a white precipitate of AgCl around the electrode

21 CHEM N-10 November 006 The physiological properties of chromium depend on its oxidation state. Consider the half reaction in which Cr(VI) is reduced to Cr(III). CrO 4 (aq) + 4H O(l) + 3e Cr(OH) 3 (s) + 5OH (aq) E o = 0.13 V Calculate the potential for this half reaction at 5 o C, where ph = 7.40 and [CrO 4 (aq)] = M. 3 As ph + poh = and poh = -log 10 ([OH - (aq)]), at ph = 7.40, poh = = 6.60 = -log 10 ([OH - (aq)]) [OH - (aq)] = The reaction quotient for the half-cell reaction is, Q = [OH (aq)] [CrO 4 5 (aq)] = (10 ) 6 ( ) = Using the Nernst equation for this three electron process, E = E o RT lnq = (-0.13) nf (5+73) ln( ) = V Answer: E = V

22 CHEM N-11 November 006 Consider the following reaction at 98 K. Ni + (aq) + Zn(s) Ni(s) + Zn + (aq) Calculate ΔG o for the cell. (Relevant electrode potentials can be found on the data page.) 5 The half-cell reduction reactions and potentials are: Ni + (aq) + e - Ni(s) Zn + (aq) + e - Zn(s) E 0 = -0.4 V E 0 = V In the reaction above, the Zn is undergoing oxidation so its potential is reversed and the overall cell potential is: 0 E cell = (-0.4) (-0.76) = +0.5 V Using G 0 = -nfe 0 for this two electron reaction: G 0 = -() (96485) (+0.5) = J mol 1 = -100 kj mol 1 Answer: -100 kj mol -1 What is the value of the equilibrium constant for the reaction at 98 K? Using E 0 = RT ln K, nf +0.5 = (8.314) (98) lnk () (96485) so K = Alternatively, using G 0 = -RTlnK, = -(8.314) (98) lnk so K = Express the overall reaction in voltaic cell notation. Answer: In the reaction, Zn is being oxidized and hence is the anode. Ni + is being reduced and so Ni is the cathode. In the standard cell notation, the anode is written on the left and the cathode on the right: Zn(s) Zn + (aq) Ni + (aq) Ni(s) ANSWER CONTINUES ON THE NEXT PAGE

23 CHEM N-11 November 006 Using a current of.00 A, how long (in minutes) will it take to plate out all of the silver from 0.50 L of a M Ag + (aq) solution? The number of moles of Ag + (aq) in a 0.50 L of a M solution is, number of moles = volume concentration = = mol The reduction of Ag + (aq) is a one electron process, Ag + (aq) + e - Ag(s), so this number of moles of electrons are required. As the number of moles of electrons delivered by a current I in a time t is, number of moles of electrons = It F =.00 t = t = 137 s =.9 minutes

24 CHEM N-1 November 006 Calculate the standard free-energy change for the following reaction at 98 K. Au(s) + 3Mg + (1.0 M) Au 3+ (1.0 M) + 3Mg(s) The half-cell reduction reactions and potentials are: Au 3+ (aq) + 3e - Au(s) Mg + (aq) + e - Mg(s) E 0 = V E 0 = -.36 V In the reaction above, the Au is undergoing oxidation so its potential is reversed and the overall cell potential is: 0 E cell = (-.36) (+1.50) = V Using G 0 = -nfe 0 for this six electron reaction: G 0 = -(6) (96485) (-3.86) = J mol 1 = kj mol 1 Answer: kj mol -1

25 CHEM N-8 November 005 Calculate the standard free-energy change for the following reaction at 98 K. Au(s) + 3Ca + (1.0 M) Au 3+ (1.0 M) + 3Ca(s) The reduction half cell reactions and E 0 values are: Au 3+ (aq) + 3e - Au(s) Ca + (aq) + e - Ca(s) E 0 = V E 0 = -.87 V In the reaction, Au is being oxidized and so the overall cell potential is: E 0 = ((-.87) (+1.50)) V = V The reaction involves 6 electrons so, using ΔG 0 = -nfe 0 : G 0 = - (6) (96485 C mol -1 ) (-4.37 V) = J mol -1 = kj mol -1 Answer: kj mol -1 Complete and balance the following equation for the reaction between iron(ii) ions and permanganate ions in an acidic solution. Fe + + MnO 4 Fe 3+ + Mn + In acid, the relevant half cells are: Fe + (aq) Fe 3+ (aq) + e - MnO 4 - (aq) + 8H + (aq) + 5e - Mn + (aq) + 4H O(l) Giving an overall reaction: 5Fe + (aq) + MnO 4 - (aq) + 8H + (aq) 5Fe 3+ (aq) + Mn + (aq) + 4H O(l) ANSWER CONTINUES ON THE NEXT PAGE

26 CHEM N-8 November 005 What is the value of the equilibrium constant for the following reaction at 98 K? Fe 3+ (aq) + Sn(s) Sn + (aq) + Fe + (aq) The reduction half cell reactions and E 0 values are: Fe 3+ (aq) + e - Fe + (aq) Sn + (aq) + e - Sn(s) E 0 = V E 0 = V In the reaction, Sn is being oxidized and so the overall cell potential is: E 0 = ((+0.77) (-0.14)) V = V The reaction involves electrons so, using E 0 = RT nf lnk: lnk = E 0 nf 1 RT = (+0.91 V) 96485Cmol J K mol 98K = 70.9 K = e 70.9 = Answer:

27 CHEM N-8 November 004 Consider the following balanced redox reaction. In(s) + 3MnO (s) + 1H + (aq) In 3+ (aq) + 3Mn + (aq) + 6H O If E o = V, what would be the measured potential of this cell at 98 K at the following concentrations? [H + (aq)] = 0.5 M; [In 3+ (aq)] = 0.0 M; [Mn + (aq)] = 0.4 M The reaction quotient Q is given by: Q = [In (aq)] [Mn ] 1 [H (aq)] 3 3 The cell reaction involves a 6e - process (In In(III) and 3Mn(IV) 3Mn(II)). The cell potential can be calculated using the Nernst equation, E = E o RT lnq nf (8.314 J K mol ) (98 K) (0.0) (0.4) = (1.568 V) ln V (6) (96485Cmol ) (0.5) Answer: 1.5 V What is the value of the equilibrium constant for the following reaction at 98 K? Cu + (aq) + Zn(s) Zn + (aq) + Cu(s) Relevant electrode potentials can be found on the data page. The standard reduction potentials for Cu + (aq) / Cu(s) and Zn + (aq) / Zn(s) are V and V respectively. The Zn + (aq) / Zn potential is the less positive and is reversed. The cell potential is therefore: E o cell = ((0.34) (-0.76)) V = 1.10 V The equilibrium constant for this e - RT = ln K : nf process can be calculated using o E cell 1 () (96485Cmol ) lnk = (1.10) = (8.314 JK mol ) (98K) K = e 85.7 = Answer:

28 CHEM J-8 June 003 Consider the following half-reactions and their standard reduction potentials. ClO 3 + 1H e Cl + 6H O E = 1.47 V S O 8 + e SO 4 E =.01 V Give the overall cell reaction. 3 The half cell with the least positive reduction potential is reversed: Cl + 6H O + 5S O 8 - à ClO H SO 4 - Calculate ΔG and hence the value of K c for the cell reaction at 98 K. As the ClO - 3 / Cl half cell is reversed, the standard cell potential is: E o = ((.01) + (-1.47) V = 0.54 V Using ΔG o = -nfe o, the free energy change for this 10e - process is: ΔG o = -(10) (96485 C mol -1 ) (0.54 V) = J mol -1 = -51 kj mol -1 Using ΔG o = -RTlnK c, the equilibrium constant is given by: kj mol -1 = -(8.314 J K -1 mol K) lnk c so K c = Alternatively, K c can be obtained directly from E o using the Nernst equation: E o RT = ln Kc = 0.54 V nf 1 1 (8.314JK mol 98K) 0.54 V = ln K 1 c so K c = ( Cmol ) The reaction essentially proceeds to completion. ΔG = -51 kj mol -1 K c =

### K + Cl - Metal M. Zinc 1.0 M M(NO

Redox and Electrochemistry This section should be fresh in your minds because we just did this section in the text. Closely related to electrochemistry is redox chemistry. Count on at least one question

### Ch 20 Electrochemistry: the study of the relationships between electricity and chemical reactions.

Ch 20 Electrochemistry: the study of the relationships between electricity and chemical reactions. In electrochemical reactions, electrons are transferred from one species to another. Learning goals and

### Chem 1721 Brief Notes: Chapter 19

Chem 1721 Brief Notes: Chapter 19 Chapter 19: Electrochemistry Consider the same redox reaction set up 2 different ways: Cu metal in a solution of AgNO 3 Cu Cu salt bridge electrically conducting wire

### BALANCING REDOX EQUATIONS EXERCISE

+5 6 1. Ag + NO3 Ag 1+ + NO 0 +5 2 +1 +2 2 4 +4 +2 2 2. N2H4 + H2O2 N2 + H2O 2 +1 +1 1 0 +1 2 +6 6 +4 4 3. CO + Fe2O3 FeO + CO2 +2 2 +3 2 +2 2 +4 2 +5 6 +4 4 +4 4 4. NO3 + CO CO2 + NO2 +5 2 +2 2 +4 2 +4

### 1332 CHAPTER 18 Sample Questions

1332 CHAPTER 18 Sample Questions Couple E 0 Couple E 0 Br 2 (l) + 2e 2Br (aq) +1.06 V AuCl 4 + 3e Au + 4Cl +1.00 V Ag + + e Ag +0.80 V Hg 2+ 2 + 2e 2 Hg +0.79 V Fe 3+ (aq) + e Fe 2+ (aq) +0.77 V Cu 2+

### Sample Exercise 20.1 Identifying Oxidizing and Reducing Agents

Sample Exercise 20.1 Identifying Oxidizing and Reducing Agents The nickel-cadmium (nicad) battery, a rechargeable dry cell used in battery-operated devices, uses the following redox reaction to generate

### Review: Balancing Redox Reactions. Review: Balancing Redox Reactions

Review: Balancing Redox Reactions Determine which species is oxidized and which species is reduced Oxidation corresponds to an increase in the oxidation number of an element Reduction corresponds to a

### 2. Write the chemical formula(s) of the product(s) and balance the following spontaneous reactions.

1. Using the Activity Series on the Useful Information pages of the exam write the chemical formula(s) of the product(s) and balance the following reactions. Identify all products phases as either (g)as,

### Chapter 13: Electrochemistry. Electrochemistry. The study of the interchange of chemical and electrical energy.

Chapter 13: Electrochemistry Redox Reactions Galvanic Cells Cell Potentials Cell Potentials and Equilbrium Batteries Electrolysis Electrolysis and Stoichiometry Corrosion Prevention Electrochemistry The

### Name AP CHEM / / Collected Essays Chapter 17 Answers

Name AP CHEM / / Collected Essays Chapter 17 Answers 1980 - #2 M(s) + Cu 2+ (aq) M 2+ (aq) + Cu(s) For the reaction above, E = 0.740 volt at 25 C. (a) Determine the standard electrode potential for the

### Electrochemistry. Chapter 17 Electrochemistry GCC CHM152. Ox # examples. Redox: LEO the lion goes GER. Oxidation Numbers (Chapter 4).

Chapter 17 Electrochemistry GCC CHM152 Electrochemistry Electrochemistry is the study of batteries and the conversion between chemical and electrical energy. Based on redox (oxidation-reduction) reactions

### Potassium ion charge would be +1, so oxidation number is +1. Chloride ion charge would be 1, so each chlorine has an ox # of -1

Chapter 18-1 1. Assign oxidation numbers to each atom in: Ni Nickel ion charge would be +2, so oxidation number is +2 Chloride ion charge would be 1, so each chlorine has an ox # of -1 Mg 2 Ti 4 Magnesium

### Chapter 20 Electrochemistry

Chapter 20 Electrochemistry Electrochemistry deals with the relationships between electricity and chemical reactions. Oxidation-reduction (redox) reactions were introduced in Chapter 4 Can be simple displacement

### Two Types of Chemical Rxns. Oxidation/Reduction. Two Types of Chemical Rxns. Two Types of Chemical Rxns. Review of Oxidation Numbers

Two Types of Chemical Rxns Oxidation/Reduction Chapter 20 1. Exchange of Ions no change in charge/oxidation numbers Acid/Base Rxns NaOH + HCl Two Types of Chemical Rxns Precipitation Rxns Pb(NO 3 ) 2 (aq)

### Useful charge on one mole of electrons: 9.64 x 10 4 coulombs/mol e - = F F is the Faraday constant

Electrochemistry II: Cell voltage and Gibbs Free energy Reading: Moore chapter 19, sections 15.6-15.12 Questions for Review and Thought: 36, 40, 42, 44, 50, 54, 60, 64, 70 Key Concepts and Skills: definition

### Electrochemistry Voltaic Cells

Electrochemistry Voltaic Cells Many chemical reactions can be classified as oxidation-reduction or redox reactions. In these reactions one species loses electrons or is oxidized while another species gains

Electrochemistry - ANSWERS 1. Using a table of standard electrode potentials, predict if the following reactions will occur spontaneously as written. a) Al 3+ + Ni Ni 2+ + Al Al 3+ + 3e - Al E = -1.68

### ELECTROCHEMISTRY. these are systems involving oxidation or reduction there are several types METALS IN CONTACT WITH SOLUTIONS OF THEIR IONS

1 ELECTROCHEMISTRY REDOX Reduction gain of electrons Cu + 2e > Cu(s) Oxidation removal of electrons Zn(s) > Zn + 2e HALF CELLS these are systems involving oxidation or reduction there are several types

### ELECTROCHEMISTRY. these are systems involving oxidation or reduction there are several types METALS IN CONTACT WITH SOLUTIONS OF THEIR IONS

1 ELECTROCHEMISTRY REDOX Reduction gain of electrons Cu 2+ + 2e > Cu (s) Oxidation removal of electrons Zn (s) > Zn 2+ + 2e HALF CELLS these are systems involving oxidation or reduction there are several

### Similarities and Differences Galvanic and Electrolytic Cell:

Electrolytic Cells Voltaic cells are driven by a spontaneous chemical reaction that produces an electric current through an outside circuit. These cells are important because they are the basis for the

### Galvanic cell and Nernst equation

Galvanic cell and Nernst equation Galvanic cell Some times called Voltaic cell Spontaneous reaction redox reaction is used to provide a voltage and an electron flow through some electrical circuit When

### Chemistry 122 Mines, Spring 2014

Chemistry 122 Mines, Spring 2014 Answer Key, Problem Set 9 1. 18.44(c) (Also indicate the sign on each electrode, and show the flow of ions in the salt bridge.); 2. 18.46 (do this for all cells in 18.44

### CHAPTER 21 ELECTROCHEMISTRY

Chapter 21: Electrochemistry Page 1 CHAPTER 21 ELECTROCHEMISTRY 21-1. Consider an electrochemical cell formed from a Cu(s) electrode submerged in an aqueous Cu(NO 3 ) 2 solution and a Cd(s) electrode submerged

### 5.111 Principles of Chemical Science

MIT OpenCourseWare http://ocw.mit.edu 5.111 Principles of Chemical Science Fall 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. Page 1 of 10 pages

### CHM1 Review Exam 12. Topics REDOX

CHM1 Review Exam 12 Topics REDOX REDOX Reactions Oxidation Reduction Oxidizing agent Reducing agent Galvanic (Voltaic) Cells Anode Cathode Salt bridge Electrolyte Half-reactions Voltage o Positive voltages

### Experiment 9 Electrochemistry I Galvanic Cell

9-1 Experiment 9 Electrochemistry I Galvanic Cell Introduction: Chemical reactions involving the transfer of electrons from one reactant to another are called oxidation-reduction reactions or redox reactions.

### Electrochemistry. Chapter 18 Electrochemistry and Its Applications. Redox Reactions. Redox Reactions. Redox Reactions

John W. Moore Conrad L. Stanitski Peter C. Jurs http://academic.cengage.com/chemistry/moore Chapter 18 Electrochemistry and Its Applications Stephen C. Foster Mississippi State University Electrochemistry

### CELL POTENTIAL, E. Terms Used for Galvanic Cells. Uses of E o Values CELL POTENTIAL, E. Galvanic Cell. Organize halfreactions

Electrons move from anode to cathode in the wire. Anions & cations move thru the salt bridge. Terms Used for Galvanic Cells Galvanic Cell We can calculate the potential of a Galvanic cell using one of

### Preliminary Concepts. Preliminary Concepts. Class 8.3 Oxidation/Reduction Reactions and Electrochemistry I. Friday, October 15 Chem 462 T.

Class 8.3 Oxidation/Reduction Reactions and Electrochemistry I Friday, October 15 Chem 462 T. Hughbanks Preliminary Concepts Electrochemistry: the electrical generation of, or electrical exploitation of

### Name Electrochemical Cells Practice Exam Date:

Name Electrochemical Cells Practice Exam Date: 1. Which energy change occurs in an operating voltaic cell? 1) chemical to electrical 2) electrical to chemical 3) chemical to nuclear 4) nuclear to chemical

### MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

Chemistry 1C-Dr. Larson Chapter 20 Review Questions MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1) is reduced in the following reaction: Cr2O7

### Redox and Electrochemistry

Name: Thursday, May 08, 2008 Redox and Electrochemistry 1. A diagram of a chemical cell and an equation are shown below. When the switch is closed, electrons will flow from 1. the Pb(s) to the Cu(s) 2+

### Electrochemistry Worksheet

Electrochemistry Worksheet 1. Assign oxidation numbers to each atom in the following: a. P 4 O 6 b. BiO 3 c. N 2 H 4 d. Mg(BrO 4 ) 2 e. MnSO 4 f. Mn(SO 4 ) 2 2. For each of the reactions below identify

### Discovering Electrochemical Cells

Discovering Electrochemical Cells Part I Electrolytic Cells Many important industrial processes PGCC CHM 102 Cell Construction e e power conductive medium What chemical species would be present in a vessel

### Redox Reactions and Electrochemistry

Redox Reactions and Electrochemistry Problem Set Chapter 5: 21-26, Chapter 21: 15-17, 32, 34, 43, 53, 72, 74 Oxidation/Reduction & Electrochemistry Oxidation a reaction in which a substance gains oxygen

### CHAPTER 13: Electrochemistry and Cell Voltage

CHAPTER 13: Electrochemistry and Cell Voltage In this chapter: More about redox reactions Cells, standard states, voltages, half-cell potentials Relationship between G and voltage and electrical work Equilibrium

### ELECTROCHEMICAL CELLS

1 ELECTROCHEMICAL CELLS Allessandra Volta (1745-1827) invented the electric cell in 1800 A single cell is also called a voltaic cell, galvanic cell or electrochemical cell. Volta joined several cells together

### Chapter 20. MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

Chapter 20 MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1) The gain of electrons by an element is called. A) oxidation B) reduction C) sublimation

### Summer 2003 CHEMISTRY 115 EXAM 3(A)

Summer 2003 CHEMISTRY 115 EXAM 3(A) 1. In which of the following solutions would you expect AgCl to have the lowest solubility? A. 0.02 M BaCl 2 B. pure water C. 0.02 M NaCl D. 0.02 M KCl 2. Calculate

### 38. Consider the following reaction that occurs in a breathalyzer: 2Cr 2 O 7

Electrochemistry Multiple Choice January 1999 37. Consider the following redox reaction: 2MnO 4 - + 3ClO 3 - + H 2 O 3ClO 4 - + 2MnO 2 + 2OH - The reducing agent is A. H 2 O B. ClO 3 - C. MnO 2 D. MnO

### Exam 3 Review Problems: Useful Equations: G o = RTlnK G o = H o T S o G o = nfe o ln K = ne o /0.0257 E cell = E o cell RT/nF lnq

Exam 3 Review Problems: Useful Equations: G o = RTlnK G o = H o T S o G o = nfe o ln K = ne o /0.0257 E cell = E o cell RT/nF lnq 1. Assuming the following reaction proceeds in the forward direction, 2

### Figure 1. A voltaic cell Cu,Cu 2+ Ag +, Ag. gas is, by convention, assigned a reduction potential of 0.00 V.

Voltaic Cells Introduction In this lab you will first prepare a set of simple standard half-cells and then measure the voltage between the half-cells with a voltmeter. From this data you will be able to

### Chapter 6 Oxidation-Reduction Reactions. Section 6.1 2. Which one of the statements below is true concerning an oxidation-reduction reaction?

Chapter 6 Oxidation-Reduction Reactions 1. Oxidation is defined as a. gain of a proton b. loss of a proton c. gain of an electron! d. loss of an electron e. capture of an electron by a neutron 2. Which

### Electrochemical cells

Electrochemical cells Introduction: Sudha Madhugiri D.Chem. Collin College Department of Chemistry Have you used a battery before for some purpose? I bet you have. The type of chemistry that is used in

### AP Chemistry CHAPTER 20- Electrochemistry 20.1 Oxidation States

AP Chemistry CHAPTER 20- Electrochemistry 20.1 Oxidation States Chemical reactions in which the oxidation state of a substance changes are called oxidation-reduction reactions (redox reactions). Oxidation

### Galvanic Cells. SCH4U7 Ms. Lorenowicz. Tuesday, December 6, 2011

Galvanic Cells SCH4U7 Ms. Lorenowicz 1 Electrochemistry Concepts 1.Redox reactions involve the transfer of electrons from one reactant to another 2.Electric current is a flow of electrons in a circuit

### Electrochemistry. Pre-Lab Assignment. Purpose. Background. Experiment 12

Experiment 12 Electrochemistry Pre-Lab Assignment Before coming to lab: Read the lab thoroughly. Answer the pre-lab questions that appear at the end of this lab exercise. The questions should be answered

### 4. Using the data from Handout 5, what is the standard enthalpy of formation of BaO (s)? What does this mean?

HOMEWORK 3A 1. In each of the following pairs, tell which has the higher entropy. (a) One mole of liquid water or one mole of water vapor (b) One mole of dry ice or one mole of carbon dioxide at 1 atm

### Chem 1101. 2 cases from last 2 lectures. Highlights of last lecture. Electrochemistry. Electrochemistry. A/Prof Sébastien Perrier

Chem 1101 A/Prof Sébastien Perrier Room: 351 Phone: 9351-3366 Email: s.perrier@chem.usyd.edu.au Unless otherwise stated, all images in this file have been reproduced from: Blackman, Bottle, Schmid, Mocerino

### 5.111 Principles of Chemical Science

MIT OpenCourseWare http://ocw.mit.edu 5.111 Principles of Chemical Science Fall 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. 26.1 5.111 Lecture

### AP Chemistry 2004 Scoring Guidelines Form B

AP Chemistry 2004 Scoring Guidelines Form B The materials included in these files are intended for noncommercial use by AP teachers for course and exam preparation; permission for any other use must be

### Chapter 11. Electrochemistry Oxidation and Reduction Reactions. Oxidation-Reduction Reactions. Oxidation-Reduction Reactions

Oxidation-Reduction Reactions Chapter 11 Electrochemistry Oxidation and Reduction Reactions An oxidation and reduction reaction occurs in both aqueous solutions and in reactions where substances are burned

### hwkc20_a.mcd S.E. Van Bramer 4/9/2003 Solutions to: Electrochemistry Homework Problem Set S.E. Van Bramer 1/11/97

Solutions to: Electrochemistry Homework Problem Set S.E. Van Bramer 1/11/97 1.Balance the following chemical equations. Assume the reactions occur in acidic solutions (add H 1+ and/or H 2 as neseccary).

### Building Electrochemical Cells

Cautions Heavy metals, such as lead, and solutions of heavy metals may be toxic and an irritant. Purpose To determine the cell potential (E cell ) for various voltaic cells and compare the data with the

### AP* Chemistry ELECTROCHEMISTRY

Terms to Know: AP* Chemistry ELECTROCHEMISTRY the study of the interchange of chemical and electrical energy OIL RIG oxidation is loss, reduction is gain (of electrons) Oxidation the loss of electrons,

### Oxidation-Reduction Reactions

CHAPTER 19 REVIEW Oxidation-Reduction Reactions SECTION 1 SHORT ANSWER Answer the following questions in the space provided. 1. All the following equations involve redox reactions except (a) CaO H 2 O

### Electrochemistry. (Referring to the loss/gain of electrons)

1. What occurs during a redox reaction? Electrochemistry Electrons are transferred. 2. What does the acronym OIL RIG stand for? Oxidation Is Loss Reduction Is Gain (Referring to the loss/gain of electrons)

### HOW TO PREDICT WHETHER A REDOX REACTION WILL BE SPONTANEOUS:

Chemistry 12 UNIT 5 OXIDATION AND REDUCTION PACKAGE #2 HOW TO PREDICT WHETHER A REDOX REACTION WILL BE SPONTANEOUS: Looking at the table in the data booklet on page 8, INCREASING TENDENCY TO REDUCE = INCREASING

### 3. Which of the following describes a conjugate acid-base pair for the following equilibrium? CN - (aq) + CH 3 NH 3 + (aq) H 2 CO 3 (aq) + H 2 O (l)

Acids, Bases & Redox 1 Practice Problems for Assignment 8 1. A substance which produces OH ions in solution is a definition for which of the following? (a) an Arrhenius acid (b) an Arrhenius base (c) a

### Price per cell Price per hour of service Brand A \$8.7/4=\$2.175 \$2.175/3=\$0.725 Brand B \$11.6/2=\$5.8 \$5.8/9=\$0.644

21 st Century Chemistry Structured Question in Topic 5 Chemical Cells and Electrolysis Unit 18-21 1. (a) Both zinc-carbon cell and silver oxide cell are primary cells. (i) Why are they classified as primary

### CHEM1612 2014-N-2 November 2014

CHEM1612 2014-N-2 November 2014 Explain the following terms or concepts. Le Châtelier s principle 1 Used to predict the effect of a change in the conditions on a reaction at equilibrium, this principle

### CHEM1909 2006-N-2 November 2006

CHEM1909 006-N- November 006 High-purity benzoic acid, C 6 H 5 COOH, (H comb = 37 kj mol 1 ) is used to calibrate a bomb calorimeter that has a 1.000 L capacity. A 1.000 g sample of C 6 H 5 COOH is placed

### Electrochemical Half Cells and Reactions

Suggested reading: Chang text pages 81 89 Cautions Heavy metals, such as lead, and solutions of heavy metals may be toxic and an irritant. Purpose To determine the cell potential (E cell ) for various

### Lab 10: RedOx Reactions

Lab 10: RedOx Reactions Laboratory Goals In this laboratory, you will: develop a basic understanding of what electrochemical cells are develop familiarity with a few different examples of redox reactions

### The Electrical Control of Chemical Reactions E3-1

Experiment 3 The Electrical Control of Chemical Reactions E3-1 E3-2 The Task In this experiment you will explore the processes of oxidation and reduction, in which electrons flow between materials, and

### M A S S A C H U S E T T S I N S T I T U T E O F T E C H N O L O G Y. 3.014 Materials Laboratory Fall 2006

D E P A R T M E N T O F M A T E R I A L S S C I E N C E A N D E N G I N E E R I N G M A S S A C H U S E T T S I N S T I T U T E O F T E C H N O L O G Y 3.014 Materials Laboratory Fall 2006 LABORATORY 3:

### MULTIPLE CHOICE. Choose the one alternative that best complet es the statement or answers the question.

MULTIPLE CHOICE. Choose the one alternative that best complet es the statement or answers the question. 1) The first law of thermodynamics can be given as. A) for any spontaneous process, the entropy of

### CHEM 103 Spring 2006 Final Exam 5 June 2006

Name CHEM 103 Spring 2006 Final Exam 5 June 2006 Multiple Choice (5 points each) Write the letter of the choice that best completes the statement or answers the question in the blank provided 1. An aqueous

### Determining Equivalent Weight by Copper Electrolysis

Purpose The purpose of this experiment is to determine the equivalent mass of copper based on change in the mass of a copper electrode and the volume of hydrogen gas generated during an electrolysis reaction.

### AP Chemistry 2009 Free-Response Questions Form B

AP Chemistry 009 Free-Response Questions Form B The College Board The College Board is a not-for-profit membership association whose mission is to connect students to college success and opportunity. Founded

### Copyright 2012 Nelson Education Ltd. Chapter 10: Electrochemical Cells 10-2

Chapter 10 Review, pages 678 683 Knowledge 1. (b) 2. (b) 3. (d) 4. (c) 5. (a) 6. (d) 7. False. In a galvanic cell, electrons travel from the anode to the cathode via the external circuit. 8. True 9. True

### Chapter 8 - Chemical Equations and Reactions

Chapter 8 - Chemical Equations and Reactions 8-1 Describing Chemical Reactions I. Introduction A. Reactants 1. Original substances entering into a chemical rxn B. Products 1. The resulting substances from

### CH 223 Chapter Thirteen Concept Guide

CH 223 Chapter Thirteen Concept Guide 1. Writing Equilibrium Constant Expressions Write the equilibrium constant (K c ) expressions for each of the following reactions: (a) Cu(OH) 2 (s) (b) Cu(NH 3 ) 4

### AP Chemistry 2010 Free-Response Questions

AP Chemistry 010 Free-Response Questions The College Board The College Board is a not-for-profit membership association whose mission is to connect students to college success and opportunity. Founded

### 12. REDOX EQUILIBRIA

12. REDOX EQUILIBRIA The electrochemical series (reference table) 12.1. Redox reactions 12.2. Standard electrode potentials 12.3. Calculations involving electrochemical cells 12.4. Using Eʅ values to predict

### AP Chemistry 2010 Free-Response Questions Form B

AP Chemistry 010 Free-Response Questions Form B The College Board The College Board is a not-for-profit membership association whose mission is to connect students to college success and opportunity. Founded

### o Electrons are written in half reactions but not in net ionic equations. Why? Well, let s see.

REDOX REACTION EQUATIONS AND APPLICATIONS Overview of Redox Reactions: o Change in Oxidation State: Loses Electrons = Oxidized (Oxidation number increases) Gains Electrons = Reduced (Oxidation Number Reduced)

### Chapter 11. Homework #3. Electrochemistry. Chapter 4 73. a) Oxidation ½ Reaction Fe + HCl HFeCl 4 Fe + 4HCl HFeCl 4 Fe + 4HCl HFeCl 4 + 3H +

hapter 4 73. a) Oxidation ½ Reaction Fe + HlHFel 4 Fe + 4HlHFel 4 Fe + 4HlHFel 4 + 3H + Homework #3 hapter 11 Electrochemistry Fe + 4HlHFel 4 + 3H + + 3e - Reduction ½ Reaction H H + H H + + e- H Balanced

### Galvanic Cells and the Nernst Equation

Exercise 7 Page 1 Illinois Central College CHEMISTRY 132 Laboratory Section: Galvanic Cells and the Nernst Equation Name: Equipment Voltage probe wires 0.1 M solutions of Pb(NO 3, Fe(NO 3 ) 3, and KNO

### Solution a homogeneous mixture = A solvent + solute(s) Aqueous solution water is the solvent

Solution a homogeneous mixture = A solvent + solute(s) Aqueous solution water is the solvent Water a polar solvent: dissolves most ionic compounds as well as many molecular compounds Aqueous solution:

### Chapter 20. Thermodynamics p. 811 842. Spontaneity. What have we learned about spontaneity during this course?

Chapter 20 p. 811 842 Spontaneous process: Ex. Nonspontaneous process: Ex. Spontaneity What have we learned about spontaneity during this course? 1) Q vs. K? 2) So.. Spontaneous process occurs when a system

### To determine relative oxidizing and reducing strengths of a series of metals and ions.

Redox Reactions PURPOSE To determine relative oxidizing and reducing strengths of a series of metals and ions. GOALS 1 To explore the relative oxidizing and reducing strengths of different metals. 2 To

### Instructions Answer all questions in the spaces provided. Do all rough work in this book. Cross through any work you do not want to be marked.

GCSE CHEMISTRY Higher Tier Chemistry 1H H Specimen 2018 Time allowed: 1 hour 45 minutes Materials For this paper you must have: a ruler a calculator the periodic table (enclosed). Instructions Answer all

### Multiple Choice Questions NCERT

CHAPTER 3 Metals and Non-metals Multiple Choice Questions 1. Which of the following property is generally not shown by metals? (a) Electrical conduction (b) Sonorous in nature (c) Dullness (d) Ductility

### Appendix D. Reaction Stoichiometry D.1 INTRODUCTION

Appendix D Reaction Stoichiometry D.1 INTRODUCTION In Appendix A, the stoichiometry of elements and compounds was presented. There, the relationships among grams, moles and number of atoms and molecules

### Answers: Given: No. [COCl 2 ] = K c [CO][Cl 2 ], but there are many possible values for [CO]=[Cl 2 ]

Chemical Equilibrium What are the concentrations of reactants and products at equilibrium? How do changes in pressure, volume, temperature, concentration and the use of catalysts affect the equilibrium

### PART I: EQUILIBRIUM. 3. Explain why the tetraamminecopper(ii) solution appears blue.

Part A: Formation of a Complex Ion PART I: EQUILIBRIUM Anhydrous copper(ii) sulfate is white, whereas hydrated copper(ii) sulfate is blue. The structure of the hydrated compound is more accurately represented

### Electrochemistry Revised 04/29/15

INTRODUCTION TO ELECTROCHEMISTRY: CURRENT, VOLTAGE, BATTERIES, & THE NERNST EQUATION Experiment partially adapted from J. Chem. Educ., 2008, 85 (8), p 1116 Introduction Electrochemical cell In this experiment,

Additional Lecture: TITRATION BASICS 1 Definition and Applications Titration is the incremental addition of a reagent solution (called titrant) to the analyte until the reaction is complete Common applications:

### CHEM1001 Example Multiple Choice Questions

HM00 xample Multiple hoice Questions The following multiple choice questions are provided to illustrate the type of questions used in this section of the paper and to provide you with extra practice. It

### CHEMISTRY ATAR COURSE DATA BOOKLET

CHEMISTRY ATAR COURSE DATA BOOKLET 2016 Copyright School Curriculum and Standards Authority, 2016 This document apart from any third party copyright material contained in it may be freely copied, or communicated

### Worksheet 25 - Oxidation/Reduction Reactions

Worksheet 25 Oxidation/Reduction Reactions Oxidation number rules: Elements have an oxidation number of 0 Group I and II In addition to the elemental oxidation state of 0, Group I has an oxidation state

### General Chemistry II Chapter 20

1 General Chemistry II Chapter 0 Ionic Equilibria: Principle There are many compounds that appear to be insoluble in aqueous solution (nonelectrolytes). That is, when we add a certain compound to water

### Determining Equivalent Weight by Copper Electrolysis

Purpose To determine the equivalent mass of copper based on change in the mass of a copper electrode and the volume of hydrogen gas generated during an electrolysis experiment. The volume of hydrogen gas

### Aqueous Solutions. Water is the dissolving medium, or solvent. Some Properties of Water. A Solute. Types of Chemical Reactions.

Aqueous Solutions and Solution Stoichiometry Water is the dissolving medium, or solvent. Some Properties of Water Water is bent or V-shaped. The O-H bonds are covalent. Water is a polar molecule. Hydration

### AP Chemistry 2008 Free-Response Questions

AP Chemistry 008 Free-Response Questions The College Board: Connecting Students to College Success The College Board is a not-for-profit membership association whose mission is to connect students to college

### Question Bank Electrolysis

Question Bank Electrolysis 1. (a) What do you understand by the terms (i) electrolytes (ii) non-electrolytes? (b) Arrange electrolytes and non-electrolytes from the following substances (i) sugar solution

### Decomposition. Composition

Decomposition 1. Solid ammonium carbonate is heated. 2. Solid calcium carbonate is heated. 3. Solid calcium sulfite is heated in a vacuum. Composition 1. Barium oxide is added to distilled water. 2. Phosphorus