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 even though I only asked you to sketch the one in part (c)); 3. 18.48; 4. 18.14 (Note that Tro uses the term electrode potential for what I ve called a reduction potential ); 5. NT1; 6. 18.55 & 18.56; 7. NT2; 8. 18.66b (Note: You need to determine n here in order to complete the calculation!); 9. 18.110 (Add Part (b): Without doing a calculation, state whether the cell potential will increase or decrease if a bit of solid NaOH were added to the solution containing the H + (i.e., the electrode in contact with the Pt(s)) [Assume the T was kept the same.]? Explain without using the Nernst Equation (use Le Chatelier type of argument as discussed in the Voltaic Cells experiment reading/handout.); 10. 18.80; 11. NT3; 12. 18.136; 13. 18.89 (NOTE: You must assume that the concentrations of the ions in each half cell are 1 M [e.g., standard state conditions]) ; 14. 18.91; 15. 18.100 (Add (b): Would the number of moles of electrons passed be the same if the only difference in the cell were that Cu 2+ was plated out as Cu(s) (instead of Ag + being plated out as Ag(s)? Would the number of moles of solid produced be the same these two cases? Explain. Why is it so important to write out the half reaction when doing electrolysis problems such as this?); 16. NT4 1. 18.44(c). Sketch a voltaic cell for each redox reaction. Label the anode and cathode and indicate the halfreaction that occurs at each electrode and the species present in each solution. Also indicate the direction of electron flow. (Also indicate the sign on each electrode, and show the flow of ions in the salt bridge.) (c) 2 NO 3 (aq) + 8 H + (aq) + 3 Cu(s) 2 NO(g) + 4 H2 O(l) + 3 Cu 2+ (aq) Answer: anode () Cu(s) e Cu 2+ (aq) salt bridge (NO 3 flows left and K + flows right) NO3 K + NO 3 (aq) H + (aq) (+) cathode Cu(s) Cu 2+ (aq) + 2 e 2 NO 3 (aq) + 4 H + (aq) + 3 e NO(g) + 2 H 2 O(l) Pt(s) NO(g) [bubbling over electrode; source, gas inlet and glassware not shown] Explanation/Reasoning: The problem implied that the reaction in the cell was the reaction indicated by the balanced equation, so no knowledge of potentials was used to determine which reaction (forward or reverse) was the actual cell reaction (it was assumed to be the forward reaction). Half reactions were obtained from Table 18.1 Since Cu(s) Cu 2+ (aq) + 2 e is an oxidation half reaction, the left side (as I have drawn it it could be written with the sides swapped ) must be the anode. Also, since electrons are being produced there, they must leave that side and go to the right side. Since this is a voltaic cell, that means the left side must be negative (since electrons spontaneously flow away from a negative toward a positive. A Pt electrode was used on the right side because there is no conductive solid in that half reaction. Negative ions in the salt bridge always flow in the opposite direction as the (negative) electrons in the wire so that there is current looping through the cell. Positive ions flow in the opposite direction as the negative ones. 2. 18.46. Calculate the standard cell potential for each of the electrochemical cells in Problem 44. (Do this for all cells in 18.44 even though I only asked you to sketch the one in part (c).) (a) Ni 2+ (aq) + Mg(s) Ni(s) + Mg 2+ (aq) (b) 2 H + (aq) + Fe(s) H 2 (g) + Fe 2+ (aq) + (c) 2 NO 3 (aq) + 8 H (aq) + 3 Cu(s) 2 NO(g) + 4 H2 O(l) + 3 Cu 2+ (aq) Strategy: Answers: (a) 2.14 V; (b) 0.45 V; (c) 0.62 V PS91

1) Separate the reaction into two half reactions. 2) Look up the standard reduction potential for the reduction half (in a table). You will use this as is. 3) Recognize that the reverse of the oxidation half reaction should match a reduction half reaction in the Table. The standard oxidation potential is just the opposite of the reduction potential. 4) Sum the standard reduction potential (E red ) and oxidation potential (E ox ) to get E cell. Execution of Strategy: (a) Ni 2+ (aq) + Mg(s) Ni(s) + Mg 2+ (aq) Ni 2+ (aq) + 2 e Ni(s) E red 0.23 V Mg(s) Mg 2+ (aq) + 2 e Ni 2+ (aq) + Mg(s) Ni(s) E ox E red (Mg 2+ ) (2.37) V E cell 0.23 + (+2.37) 2.14 V (b) 2 H + (aq) + Fe(s) H 2 (g) + Fe 2+ (aq) 2 H + (aq) + 2 e H 2 (g) E red 0 V Fe(s) Fe 2+ (aq) + 2 e Ni 2+ (aq) + Mg(s) Ni(s) E ox E red (Fe 2+ ) (0.45) V E cell 0 + (+0.45) 0.45 V (c) 2 NO 3 (aq) + 8 H + (aq) + 3 Cu(s) 2 NO(g) + 4 H2 O(l) + 3 Cu 2+ (aq) NO 3 (aq) + 4 H + (aq) + 3 e NO(g) + 2 H 2 O(l) E red 0.96 V Cu(s) Cu 2+ (aq) + 2 e E ox E red (Cu 2+ ) (+0.34) V 2 NO 3 (aq) + 8 H + (aq) + 3 Cu(s) 2 NO(g) + 4 H2 O(l) + 3 Cu 2+ (aq) E cell 0.96 + (0.34) 0.62 V** ** NOTE: Since voltage is a per electron kind of quantity (energy per electron), reduction (or cell) potentials do not change when the (half) reaction is multiplied through by a coefficient! In other words: 2 NO 3 (aq) + 8 H + (aq) + 6 e 2 NO(g) + 4 H 2 O(l) ALSO has E red 0.96 V (NOT 2(0.96 V)!) This is an important difference from H, S, or G. Because of this, one need not actually bother to multiply the half reactions above if one only wants to get the E cell (as in this problem). However, it should be pointed out that one must do that multiplication in order to find n and thus G (see Problem #8 on this set) 3. 18.48. Consider the voltaic cell (my picture is not as pretty [or accurate] as the text graphic, but the key elements are there): (a) Determine the direction of electron flow and label the anode and the cathode. (b) Write a balanced equation for the overall reaction and calculate E cell. (c) Label each electrode as negative or positive. (d) Indicate the direction of anion and cation flow in the salt bridge. Pb(s) 1 M Pb 2+ (aq) 1 M Cl (aq) salt bridge containing NaNO 3 (aq) Pt(s) Cl 2 (g) [bubbling over electrode; source, gas inlet and glassware not shown] PS92

Answers: (a),(c) anode () Pb(s) e (a) Pb(s) Pb 2+ (aq) + 2 e 1 M 1 M Cl (aq) Cl 2 (g) + 2 e 2 Cl (aq) Pb 2+ (aq) (d) NO 3 flows left and Na + flows right NO3 Na + (+) cathode Pt(s) (a),(c) Cl 2 (g) [bubbling over electrode; source, gas inlet and glassware not shown] (b) Pb(s) + Cl 2 (g) Pb 2+ (aq) + 2 Cl (aq) ; E cell 1.49 V Reasoning: 1) NOTE: You cannot ASSUME that the anode will always be on the left! You really have to do part (b) (or at least consider the potentials) before you can do part (a). 2) To determine the balanced equation, start with the reduction half reactions: a) On the left, either Pb(s) Pb 2+ (aq) + 2 e or Pb 2+ (aq) + 2 e Pb(s) is occurring. b) On the right, either Cl 2 (g) + 2 e 2 Cl (aq) or 2 Cl (aq) Cl 2 (g) + 2 e occurs. 3) Then look at the reduction potentials from Table 18.1: Pb 2+ (aq) + 2 e Pb(s) E red 0.13 V Cl 2 (g) + 2 e 2 Cl (aq) E red +1.36 V 4) From these values, you can conclude that it is more favorable to reduce Cl 2 than Pb 2+. This means that Cl 2 is the one that actually gets reduced in the cell (not Pb 2+ ). That means keep the second half reaction and flip the first one. (In other words, Pb is the other reactant with Cl 2 ). The balanced equation is thus obtained by: Cl 2 (g) + 2 e 2 Cl (aq) E red +1.36 V Pb(s) Pb 2+ (aq) + 2 e E ox (0.13 V) Pb(s) + Cl 2 (g) Pb 2+ (aq) + 2 Cl (aq) ; E cell (1.36 + 0.13) 1.49 V 5) Once you know the half reactions (directionality), you can answer part (a). The electrons must be coming to the side with Cl 2 (since it is getting reduced; electrons are on the left side of the half reaction), and from the side with Pb (it is getting oxidized; electrons are on the right side of the half reaction). 6) This also allows you to assign the charges on the electrodes, as long as you remember that in a voltaic cell, electrons get to go where they want to go that is, they flow from negative to positive (this is because the spontaneous redox reaction is actually occurring in the cell, dictating the direction of electron flow). Since we already determined that the electrons flow to the side with Cl 2, that electrode must be the positive one. 7) Although not asked here, knowing the half reactions also allows you to state which is the anode (where oxidation occurs) and the cathode (where reduction occurs). 8) Negative ions in the salt bridge always flow in the opposite direction as electrons in the wire do (Think of it as a big loop. Negative charges cannot both flow to the same side the whole purpose of the salt bridge is to avoid the buildup of charge!) PS93

4. 18.14 Does a large positive electrode potential (i.e., reduction potential ) indicate a strong oxidizing agent or a strong reducing agent? What about a large negative electrode potential? Answers: A large positive electrode potential indicates a strong oxidizing agent (which would be on the left side of the reduction half reaction). This is because a reduction half reaction involves an oxidizing agent as a reactant (on the left), getting reduced. Since a large positive E of any kind means more favorable (than a smaller one), a large positive reduction potential indicates a good oxidizing agent. A large negative reduction potential indicates a strong reducing agent, but on the right side of the reduction half reaction. Technically, I would prefer to say that a large positive oxidation potential indicates a strong reducing agent (with the reducing agent being a reactant, on the left side of the oxidation half reaction). This would be in complete parallel to my response in the prior paragraph. 5. NT1. You are told that metal A is a better reducing agent than metal B. What, if anything, can be said about A + and B +? Explain. Answer: A + will be a poorer oxidizing agent than B +. If A is better at giving (than B), then A + will be poorer at taking (than B + ). The better the giving power of a reducing agent, the poorer is the tendency of the species formed to take electrons back. This is very analogous to the concept that the stronger the acid, the weaker is its conjugate base. 6. 18.55 & 18.56 18.55 Which metal could you use to reduce Mn 2+ ions but not Mg 2+ ions? Answer: Al Reasoning: The standard reduction potential for Mn 2+ is 1.18 V, and for Mg 2+ it is 2.37 V. Thus, in order for a metal to reduce Mn 2+ under standard conditions, the oxidation potential would have to be more positive than 1.18 V (so that E cell (= E red + E ox ) will be positive). That means the reduction potential (of the cation) would need to be more negative than 1.18 V. Similarly, for Mg 2+, look for a reduction potential (for the cation) more positive than 2.37 V (so that the cation would be able to oxidize Mg(s), which means that the metal would not be able to reduce Mg 2+ ). The only cation that has a potential more negative than 1.18 V and more positive than 2.37 V is Al 3+, with a reduction potential of 1.66 V. But be careful to state that it is the metal, Al, that is used to do the reduction. 18.56 Which metal can be oxidized with an Sn 2+ solution but not with an Fe 2+ solution? Answer: Ni and Cd Reasoning: The standard reduction potential for Sn 2+ is 0.14 V, and for Fe 2+ it is 0.45 V. Thus, in order for a metal to be oxidized by Sn 2+ under standard conditions, the oxidation potential would have to be more positive than +0.14 V, which means the reduction potential (of the cation) would need to be more negative than 0.14 V. Similarly, for Fe 2+, look for a reduction potential more negative than 0.45 V. The only cations that has a potential between these values Ni 2+ and Cd 2+, with reduction potentials of 0.23 and 0.40 V. But again, be careful to state that it is the metals, Ni and Cd, that are being oxidized. 7. NT2. Consider only the following species: Na +, Cl, Ag +, Ag, Zn 2+, Zn, Pb (a) Which is the strongest oxidizing agent? (b) Which is the strongest reducing agent? (c) Which species can be oxidized by SO 2 4 in acid at standard state conditions? (d) Which species can be reduced by Al at standard state conditions? Answers: (a) Ag + (b) Zn (c) Zn and Pb (d) Ag + and Zn 2+ NOTE: If you have not included reasoning on your paper, add it now before you look at the reasoning below. Deductions will result without reasoning! PS94

Reasoning: (a) Look for the species on the left of the arrows in Table 18.1 (i.e., on the left side of a reduction half reaction), and see which has the highest (most positive) reduction potential. This is because a higher (more positive) reduction potential means more favorable reduction to occur which means the species being reduced (the oxidizing agent) is a stronger one: Na + (2.71 V) Cl (not on left not considered an oxidizing agent [extremely poor; It would have to become Cl 2!?]) Ag + (+0.80 V) Ag (not on left not considered an oxidizing agent [extremely poor; would have to become Ag!?]) Zn 2+ (0.76 V) Zn (not on left not considered / negligibly poor) Pb (not on left not considered / negligibly poor) (b): Look for the species on the right of the arrows in Table 18.1 (i.e., on the right in a reduction half reaction), and see which has the highest (most positive) oxidation potential (i.e., flip the halfreactions in Table 18.1 and change the sign). This is because a higher (more positive) oxidation potential means more favorable oxidation to occur which means the species being oxidized (the reducing agent) is a stronger one: Na +, Ag +, and Zn 2+ : not considered reducing agents (not on the right side of any reduction half reaction N/A or extremely poor ) Cl (1.36 V) Ag (0.80 V) Zn ((0.76 V) = +0.76 V) Pb ((0.13) = +0.13 V) (c) In order for somebody to be oxidized by SO 2 4, SO 2 4 must accept electrons (i.e., be reduced). Thus consider the half reaction in which SO 2 4 is reduced (i.e., the reduction half reaction) and look at its potential. The standard reduction potential for SO 2 4 is +0.20 V. Thus any reducing agent with an oxidation potential greater than 0.20 V will yield a positive E cell and thus will be able to react with SO 2 4. So look for the species on the right in Table 18.1 that are below where SO 2 4 is (remember that for the oxidation potential, you will need to change the sign!). THIS IS TRICKY! Please take time to fully understand what you are doing and why. Please don t memorize the up here reacts with down there kind of phrases instead of learning why the result is what it is. Shortcuts should come only after you have understood the concept on which the shortcut is based! From part (b), only Zn (+0.76 V) and Pb (+0.13 V) meet the criteria. (d) Being reduced by Al means Al is acting as a reducing agent and is getting oxidized. Thus look at the half reaction in which Al is oxidized (i.e., electrons are being produced). [This oxidation half reaction will be the opposite of the reduction half reaction for Al 3+ in Table 18.1 which has a value of 1.66 V.] The standard oxidation potential for Al is thus +1.66 V (E red (Al 3+ ) = (1.66 V) = +1.66 V). Thus any oxidizing agent with a reduction potential greater than 1.66 V will yield a positive E cell and thus will be able to react with Al. So look for the species on the left in Table 17.1 that are above the line where Al is. From part (a), Ag + (+0.80 V) and Zn 2+ (0.76 V) meet the criteria. Na + s reduction potential is too negative (2.71 V!) 8. 18.66b Use tabulated electrode potentials to calculate G rxn for each reaction at 25C. (b) O 2 (g) + 2 H 2 O(l) + 2 Cu(s) 4 OH (aq) + 2 Cu 2+ (aq) Answer: 2 x 10 4 J or 2 x 10 1 kj PS95

Strategy: 1) Recognize that the appropriate equation to use is: G = nfe 2) As such, one needs to determine E cell and n for the balanced equation. 3) Break the reaction into half reactions. 4) Get the reduction potentials from Table 18.1 and calculate the E cell as in Q2 above. 5) Find n by looking at the half reactions. n is the number of electrons transferred ( lost gained) Execution of Strategy: O 2 (g) + 2 H 2 O(l) + 2 Cu(s) 4 OH (aq) + 2 Cu 2+ (aq) becomes: O 2 (g) + 2 H 2 O(l) + 4 e 4 OH (aq) E red +0.40 V 2 Cu(s) 2 Cu 2+ (aq) + 4 e E ox E red (0.34) V (recall, E red s don t change with coefficient ) E cell (0.40 0.34) 0.06 V and n 4 G nfe (4 mol e )(96485 C/mol e )(+0.06 J/C) 23156 J 2 x 10 4 J 2 x 10 1 kj 9. 18.110 The cell potential of this electrochemical cell depends son the gold concentration in the cathode halfcell: Pt(s) H 2 (g, 1.0 atm) H + (aq, 1.0 M) Au 3+ (aq,? M) Au(s) (a) What is the concentration of Au 3+ in the solution if E cell is 1.22 V? (b) Without doing a calculation, state whether the cell potential will increase or decrease if a bit of solid NaOH were added to the solution containing the H + (i.e., the electrode in contact with the Pt(s))? Explain without using the Nernst Equation (use Le Chatelier type of argument as discussed in the Voltaic Cells experiment reading/handout.) Answers: (a) ~6 x 10 15 M; (b) E cell would increase, because you would decrease the concentration of a product, H + (since OH would react with it to make H 2 O). When the concentration of a product is decreased, the reaction will have a greater tendency to go in the forward direction. This is reflected in a greater (more positive) cell potential. NOTE: You needed to look at the balanced equation for the cell reaction to see if H + was a reactant or a product in order to answer this question! Work / Reasoning: 1) Recognize that the equation to use in this problem is the Nernst Equation: o 0.0592 Ecell Ecell log Q n 2) As such, you ll need to find E cell and n, and determine the balanced chemical equation (in order to get an expression for Q) 3) Finding E cell, n, and the balanced equation: 2 H + + 2 e H 2 ; E red = 0.00 V More negative, less favorable, so flip it. Au 3+ + 3 e Au; E red = +1.50 V Also need to multiply first equation by 3 and second by 2 to get the same number of electrons: 3 H 2 6 H + + 6 e ; E = 0.00 V 2 Au 3+ + 6 e 2 Au; E = +1.50 V 6 [H ] 3 H 2 + 2 Au 3+ 3 Au + 6 H + ; E cell 0.00 + 1.50 1.50 V; Q = 3 3 2 P [Au ] ; n = 6 4) Substituting in (recalling that E cell is given in the problem as 1.22 V) H 2 PS96

6 1 0.0592 V Ecell 1.22 V 1.50 V log 6 1 Au 3 3+ 2 0.0592 V 0.28 V log 6 Au 1 6 3+ 2 6( 0.28 V) 1 1 log 28.37.. log 0.0592 V Au Au 3+ 2 3+ 2 28.37.. 1 3+ 2 1 3+ 1 15 10 Au Au 6.46..x 10 M 3+ 2 28.37.. 28.37.. Au 10 10 6 x 10 15 M 1 x 10 14 (technically, this result doesn t even have 1 SF! But let s not dwell on that 10. 18.80 Consider the concentration cell: salt bridge containing NaNO 3 (aq) Pb(s) Pb(s) 2.5 M Pb 2+ (aq) 5.0 x 10 3 M Pb 2+ (aq) (a) Label the anode and cathode. (b) Indicate the direction of electron flow. (c) Indicate what happens to the concentration of Pb 2+ in each half cell (as time goes by and rxn occurs) Answers: Cathode e salt bridge containing NaNO 3 (aq) Pb(s) Pb(s) Anode [Pb 2+ ] decreases 2.5 M Pb 2+ (aq) 5.0 x 10 3 M Pb 2+ (aq) [Pb 2+ ] increases Reasoning: Pb 2+ will want to increase its concentration in the right side and decrease it in the left side (tendency for high concentration to flow to low concentration ). To make this happen chemically the following reactions should occur: Left side: Pb 2+ + 2 e Pb Right side: Pb Pb 2+ + 2 e PS97

Thus, the left side is the cathode (reduction is occurring), and the right side is the anode (oxidation). It also means that electrons flow from right to left. 11. NT3. True or false questions (correct if false): (a) The chemical reaction that occurs in a voltaic cell is the spontaneous cell reaction, but the chemical reaction that occurs in an electrolysis cell is the nonspontaneous cell reaction (the reverse of the spontaneous cell reaction). TRUE. (b) The anode is the electrode at which reduction occurs, and thus it is always the negative electrode in a voltaic cell (a battery). FALSE Corrected #1: The anode is, by definition, the electrode at which oxidation occurs, and thus it is always the negative electrode in a voltaic cell (a battery). OR Corrected #2: The cathode is, by definition, the electrode at which reduction occurs, and thus it is always the positive electrode in a voltaic cell (a battery). (c) The anode is the electrode at which reduction occurs, and thus it is always the negative electrode in an electrolysis cell. FALSE Corrected #1: The anode is, by definition, the electrode at which oxidation occurs, and thus it is always the positive electrode in an electrolysis cell. OR Corrected #2: The cathode is, by definition, the electrode at which reduction occurs, and thus it is always the negative electrode in an electrolysis cell. Remember that in an electrolysis cell, electrons are forced from the negative lead of the external power source INTO the negative electrode of the cell (i.e., the electrons are forced to go where they would not spontaneously go) and so reduction will occur at the negative electrode (and oxidation at the positive one). What is always true in either kind of electrochemical cell is that the anode is where oxidation occurs and the cathode is where reduction occurs; what changes is whether or not the anode (or cathode) is positive or negative; that depends on the type of cell you have! (d) The standard potential of a cell is constant at a given temperature, and does not change with temperature. FALSE. Corrected: The standard potential of a cell is constant at a given temperature, but it changes as temperature changes. We know from the thermodynamics chapter that G s vary with T (G = H TS ). But E cell changes as G does (G = nfe cell ), so E cell must also vary with T. (e) All productfavored oxidationreduction reactions have a standard cell voltage E cell with a negative sign. FALSE. Corrected: All productfavored oxidationreduction reactions have a standard cell voltage E cell with a positive sign. Reasoning: Productfavored means K > 1. Standard state conditions means Q = 1. So productfavored at standard state conditions means K > Q or reaction is spontaneous (at standard state conditions). But this means G < 0, which means that E cell > 0, which means E cell is positive. PS98

12. 18.136 An electrochemical cell has a positive standard cell potential but a negative cell potential. What is true of Q and K for the cell? (a) K > 1; Q > K (b) K < 1; Q > K (c) K > 1; Q < K; (d) K < 1; Q < K Answer (with explanation: (a). i) A positive E cell means spontaneous at standard state conditions (where Q = 1) and thus the system must have K > 1 (i.e., Q > 1 at equilibrium). ii) A negative E cell means the conditions are such that the reaction not spontaneous in the forward direction (as the equation was originally written) in fact, the reaction is spontaneous in the reverse direction. That means Q must be too big to be at equilibrium (because Q is products over reactants ), which means Q > K 13. 18.89 Consider the electrolytic cell: (a) Label anode and cathode. Write half reactions. (b) Indicate electron flow. (c) Label terminals on battery as positive or negative, and calculate the minimum voltage necessary to drive the reaction. (Assume standard state conditions!) Ni(s) Ni 2+ (aq) Cd 2+ (aq) salt bridge containing NaNO 3 (aq) Cd(s) Strategy (one option anyway; there are others): 1) Recognize that in an electrolysis cell, the reaction that is occurring in the cell is the one that is not spontaneous. That means that in order to proceed with the labeling in this cell (an electrolysis cell), you must first determine what reaction would be spontaneous without the push from the battery driving the cell. In other words, first pretend that the cell is a voltaic cell. (Then you will reverse things later.) 2) Treating the cell as if it were a voltaic cell, calculate the E cell using reduction potentials (from Table 18.1) and write the balanced equation corresponding to this E cell. 3) Recognize that the reaction you have just determined in (2) above must be reversed in order to describe the reaction that will be occurring in the actual cell in this problem (an electrolysis cell, not a voltaic cell). As such, just write the reverse of the balanced equation in (2). 4) Using that equation (broken into half reactions) as your guide, determine the half reactions at each electrode. From there, you can determine which is the cathode and anode (where reduction and oxidation are occurring, respectively), and the direction of electron flow. 5) Recognize that the E cell you calculated in (2) above is the minimum voltage that would need to be applied to drive the reaction. (Technically, slightly higher than that, because if it were at that voltage, the cell would be at equilibrium [no net reaction in either direction].) Why? Let s take an example. If the E cell in (2) were 1.5 V, then the E cell for the reaction in the electrolysis cell is 1.5 V. That means that +1.5 V is needed to be applied to get the cell potential to be 0 (i.e., at equilibrium). Any voltage greater than that will make the reaction occur. Execution of Strategy (and answer, below): PS99

Ni 2+ Ni 2+ + 2 e Ni; E red = 0.23 V Cd 2+ + 2 e Cd; E red = 0.40 V More negative, less favorable, so flip it. + Cd Ni + Cd 2+ ; E cell = +0.17 V for spontaneous reaction (if no external voltage were applied) This is the reaction that would occur without any external voltage pushing in the opposite direction. Thus, in the electrolytic cell noted in this problem, the reaction occurring is represented by the reverse of the above: Ni + Cd 2+ Ni 2+ + Cd; E cell = 0.17 V (for the reaction occurring in the cell) Thus, Ni is getting oxidized to Ni 2+ in this cell (on the left, anode) and Cd 2+ is getting reduced to Cd.(on the right, cathode). Thus, electrons must be flowing toward the right (where reduction occurs). Thus, electrons must be being pushed to the right (electrolytic cell), and so the negative electrode of the battery must be the one on the right side. **The negative sign of E cell here indicates that the reaction won t happen without an external voltage pushing it. The value of 0.17 V means that the external voltage applied would have to be +0.17 V just to get the system to be at equilibrium. Anything higher than that will be needed make the reaction occur as shown. Anode e (+) () e Cathode Ni(s) Cd(s) Ni Ni 2+ + 2e Ni 2+ (aq) Cd 2+ (aq) Cd 2+ + 2e Cd 14. 18.91 Answer: K + + e K; 2 Br Br 2 + 2 e (Don t worry about knowing the state designations) 15. 18.100 (a) 6.8 C 6.8 A ; 1 min 60 sec s 60 s 6.8 C 1 mol e 1 mol Ag plated 107.87 g 72 min x x x x x 32.8.. 33 g Ag plated min s 96485 C 1 mol e 1 mol Ag (b) Would the number of moles of electrons passed be the same if the only difference in the cell were that Cu 2+ was plated out as Cu(s) (instead of Ag + being plated out as Ag(s))? Would the number of moles of solid produced be the same these two cases? Explain. Why is it so important to write out the half reaction when doing electrolysis problems such as this?) Answers: i) Yes, moles of electrons plated is determined by the current and time, not the chemistry. PS910

ii) No. The moles of solid would be half as many, because it takes 2 e to plate out one Cu atom, whereas it only takes 1 e to plate out one Ag. (In other words, the mole ratio in the equation above would be 1 mol Cu / 2 moles e ; i.e., 1/2 ) iii) The half reaction tells you the mole ratio of moles of metal to moles of electrons. That is how you get the proper mole ratio to do the stoichiometry! 16. NT4. (i) Balance the equation under the conditions specified (ii) State the value of n for each balanced redox equation. (a) As 2 O 3 (s) + NO 3 (aq) H3 AsO 4 (aq) + NO(g) (acidic conditions) (b) NO 2 (aq) + Al(s) NH3 (g) + AlO 2 (aq) (basic conditions) Answers: (a) (i) 3 As 2 O 3 (s) + 7 H 2 O + 4 NO 3 (aq) + 4 H + 6 H 3 AsO 4 (aq) + 4 NO(g); (ii) n = 12 (b) (i) NO 2 (aq) 2 Al(s) + H2 O + OH NH 3 (g) + 2 AlO 2 (aq); (ii) n = 6 Work / Procedure: See handout sheet for the procedure followed here ( Half reaction method). I did not assign oxidation numbers. Rather, I let the procedure indicate (at the end) which half was the oxidation half and which the reduction half. (a) Note: State designations for H 2 O(l) and H + (aq) will be omitted in this answer simply for convenience. As 2 O 3 (s) + NO 3 (aq) H3 AsO 4 (aq) + NO(g) (acidic conditions) 1 st Half Reaction: As 2 O 3 (s) H 3 AsO 4 (aq) (Separate out the species containing As from those with N [see below]) As 2 O 3 (s) 2 H 3 AsO 4 (aq) (Balance atoms other than O and H) As 2 O 3 (s) + 5 H 2 O 2 H 3 AsO 4 (aq) (Balance O atoms by adding H 2 O s) As 2 O 3 (s) + 5 H 2 O 2 H 3 AsO 4 (aq) + 4 H + (Balance H atoms by adding H + s) As 2 O 3 (s) + 5 H 2 O 2 H 3 AsO 4 (aq) + 4 H + + 4 e (Balance charge by adding e s) Electrons are being produced, so someone is getting oxidized, and this is the oxidation half reaction 2 nd Half Reaction: NO 3 (aq) NO(g) NO 3 (aq) NO(g) (Balance atoms other than O and H already done!) NO 3 (aq) NO(g) + 2 H2 O (Balance O atoms by adding H 2 O s) NO 3 (aq) + 4 H + NO(g) + 2 H 2 O (Balance H atoms by adding H + s) NO 3 (aq) + 4 H + + 3 e NO(g) + 2 H 2 O (Balance charge by adding e s) Electrons are being used, so somebody is getting reduced, and this is the reduction half reaction Bringing half reactions together: To get the number of electrons to be the same in both reactions (which will be the value of n!), multiply the first by 3 and the second by 4; then add them together: 3 As 2 O 3 (s) + 15 H 2 O 6 H 3 AsO 4 (aq) + 12 H + + 12 e 4 NO 3 (aq) + 16 H + + 12 e 4 NO(g) + 8 H 2 O 3 As 2 O 3 (s) + 15 H 2 O + 4 NO 3 (aq) + 16 H + 6 H 3 AsO 4 (aq) + 12 H + + 4 NO(g) + 8 H 2 O n is the number of electrons transferred per equation unit of reaction. So n = 12 here. PS911

Checking for same species on both sides: Cancelling out H 2 O s and H + s that appear on both sides (the same number of times) yields: 3 As 2 O 3 (s) + 7 H 2 O + 4 NO 3 (aq) + 4 H + 6 H 3 AsO 4 (aq) + 4 NO(g) ; n = 12 Check charge: on left, 0 and on right, also zero. Checks! (b) NO 2 (aq) + Al(s) NH3 (g) + AlO 2 (aq) (basic conditions) Note: All early steps are the same as in part (a), so parenthetical comments will be omitted. 1 st Half Reaction: NO 2 (aq) NH3 (g) NO 2 (aq) NH3 (g) + 2 H 2 O NO 2 (aq) + 7 H + NH 3 (g) + 2 H 2 O NO 2 (aq) + 7 H + + 6 e NH 3 (g) + 2 H 2 O Electrons are being used, so somebody is getting reduced, and this is the reduction half reaction 2 nd Half Reaction: Al(s) AlO 2 (aq) Al(s) + 2 H 2 O AlO 2 (aq) Al(s) + 2 H 2 O AlO 2 (aq) + 4 H + Al(s) + 2 H 2 O AlO 2 (aq) + 4 H + + 3 e Electrons are being produced, so someone is getting oxidized, and this is the oxidation half reaction Bringing half reactions together: To get the number of electrons to be the same in both reactions (which will be the value of n!), multiply the second by 2; then add them together: NO 2 (aq) + 7 H + + 6 e NH 3 (g) + 2 H 2 O 2 Al(s) + 4 H 2 O 2 AlO 2 (aq) + 8H + + 6 e NO 2 (aq) + 7 H + + 2 Al(s) + 4 H 2 O NH 3 (g) + 2 H 2 O + 2 AlO 2 (aq) + 8 H + n is the number of electrons transferred per equation unit of reaction. So n = 6 here. Cancel out species that appear on both sides the same number of times (7 H + s and 2 H 2 O s): NO 2 (aq) 2 Al(s) + 2 H2 O NH 3 (g) + 2 AlO 2 (aq) + H + ; n = 6 (acidic solution) Converting to basic conditions: To convert this to basic conditions, add one OH to both sides for every H +. NO 2 (aq) 2 Al(s) + 2 H2 O + OH NH 3 (g) + 2 AlO 2 (aq) + H + + OH Make each OH + H + become an H 2 O molecule (i.e., you are neutralizing the H + s with OH s): NO 2 (aq) 2 Al(s) + 2 H2 O + OH NH 3 (g) + 2 AlO 2 (aq) + H2 O Finally, cancel out any H 2 O s that appear on both sides (if applicable): NO 2 (aq) 2 Al(s) + H2 O + OH NH 3 (g) + 2 AlO 2 (aq); n = 6 (basic solution) Check charge: Left, 1 + 1 = 2; Right, 2 x 1 = 2. Checks! PS912