CHEM 102 Instructional Objectives Prof. Upali Siriwardane, Chemistry Program, Louisiana Tech University, Ruston, LA 71272 CHAPTER 19 19. Electrochemistry 19.1 Redox Reactions 19.2 Using Half-Reactions to Understand Redox Reactions 19.3 Electrochemical Cells 19.4 Electrochemical Cells and Voltage 19.5 Using Standard Cell Potentials 19.6 E o and Gibbs Free Energy 19.7 Effect of Concentration on Cell Potential 19.8 Neuron Cells 19.9 Common Batteries 19.10 Fuel Cells 19.11 Electrolysis-Forcing Reactant-Favored Reactions to Occur 19.12 Counting Electrons 19.13 Corrosion-Product-Favored Reactions Objectives are as follows: Basic Skills Students should be able to: 1. Identify the oxidizing and reducing agents in a redox reaction (Section 19. 1). 2. Write equations for the oxidation and reduction half-reactions, and use them to balance the net equation (Section 19.2). 3. Identify and describe the functions of the parts of an electrochemical cell; describe the direction of electron flow outside the cell and the direction of the ion flow inside the cell (Section 19-3). 4. Describe how standard reduction potentials are defined and use them to predict whether a reaction will be product-favored as written (Sections 19.4 and 19.5). Calculate Gº from the value of Eº for a redox reaction (Section19.6). 5. Explain how product-favored electrochemical reactions can be used to do useful work, and list the requirements for using such reactions in rechargeable batteries (Section 19.6). 6. Explain how the Nernst equation relates concentrations of redox reactants to Ecell (Section 19.7). 7. Use the Nernst equation to calculate the potentials of cells that are not at standard conditions (Section 19.7). 8. Explain the source of the equilibrium potential across the membrane of a neuron cell (Section 19.8). 9. Describe the chemistry of dry cell battery, the mercury battery, the lead-acid storage battery (Section 19.9). 10. Describe how a fuel cell works, and indicate how it is different from a battery (Section 19. 10). Use standard reduction potentials to predict the products of electrolysis of an aqueous salt solution (Section 19.11).
11. Use standard reduction potentials to predict the products of electrolysis of an aqueous salt solution (Section 19.11) 12. Calculate the quantity of products formed at an electrode during an electrolysis reaction, given the current passing through the cell and the time during which the current flows (Section 19.12) 13. Explain how electroplating works (Section 19.12) 14. Describe what corrosion is and how it can be prevented by cathodic protection (Section 19.13)
CHEM 102 CLASS PROBLEMS Prof. Upali Siriwardane, Chemistry Program, Louisiana Tech University, Ruston, LA 71272 CHAPTER 19 Chapter 19. Electrochemistry Electrochemistry brings together, as a technique, several topics we have already studied: redox reactions, equilibrium, and thermodynamics. Electrochemistry influences many areas of our lives from batteries (car, flashlights, calculators, watches, and electric car) to electroplating, e.g., surgical instruments. Electrochemistry represents the inter-conversion of chemical energy and electrical energy. Electrochemistry involves redox reactions because the electrical energy (flow of electrons) has at its origin the oxidation (loss of electrons) and reduction (gain of electrons) of species. Thermodynamic considerations applies to redox reactions. 1. Can we harness the energy released when a spontaneous reaction occurs? 2. Can we cause a non-spontaneous reaction to occur by supplying energy? You have already seen one way to accomplish 2., by supplying thermal energy (raising the temperature) if S (universe) > 0 or G (system) < 0 G = H - T S If S > 0, then -T S < 0, and we can make a non-spontaneous reaction spontaneous by raising the temperature so that -T S term dominates. Is there a convenient and general way capable of driving reactions in their nonspontaneous directions? How about supplying electrical energy? The answer is yes. Redox Reactions Oxidation = removal of one or more electrons Reduction = addition of one or more electrons Oxidizing agent = substance that is doing the oxidation (taking the electrons) Reducing agent = substance that is doing the reduction (giving the electrons) If something is oxidized, its oxidation state (Ox. State) is increased. If something is reduced, its oxidation state decreases. Balancing Redox Reactions We must keep track of the exchange of electrons among species in a redox reaction taking place in a basic solution. Consider: MnO 4 - (aq) + C 2 O 4 2- (aq) MnO 2 (s) + CO 3 2- (aq)
First step is to assign oxidation numbers and separate half reactions. MnO 4 - (aq) + C 2 O 4 2- (aq) MnO 2 (s) + CO 3 2- (aq) Mn = + 7 C = +3 Mn = + 4 C = +4 Write the skeleton half-reactions. MnO 4 - (aq) MnO 2 (s) and C 2 O 4 2- (aq) CO 3 2- (aq) (oxidation state changes: Mn + 7 to + 4 and C +3 to +4) Balance the atoms which change in oxidation state. MnO 4 - (aq) MnO 2 (s) C 2 O 4 2- (aq) 2 CO 3 2- (aq) Add or remove electrons, as needed, to balance oxidation state changes. 3e - + MnO 4 - (aq) MnO 2 (s) C 2 O 4 2- (aq) 2CO 3 2- (aq) + 2e - Balance the reduction and oxidation half reactions by balancing the electrons gained and lost. 6e - + 2MnO 4 - (aq) 2MnO 2 (s) 3 C 2 O 4 2- (aq) 6CO 3 2- (aq) + 6 e - Now add the half reactions (electrons should not appear; they should cancel!) 2 MnO 4 - (aq) + 3 C 2 O 4 2- (aq) 2 MnO 2 (s) + 6 CO 3 2- (aq) Balance atoms that do not change in oxidation state. Now balance O by adding H 2 O as needed, then balance H by adding H + as needed. 2 H 2 O + 2 MnO 4 - (aq) + 3 C 2 O 4 2- (aq) 2 MnO 2 (s) + 6 CO 3 2- (aq) 2 H 2 O + 2 MnO 4 - (aq) + 3 C 2 O 4 2- (aq) 2 MnO 2 (s) + 6 CO 3 2- (aq) + 4 H + This is the FINAL BALANCED REDOX EQUATION if the reaction was done in acidic or neutral solution When reaction is done in basic solution, add as many OH - to both sides as it takes to neutralize the acid (H + + OH - H 2 O). 2H 2 O + 2MnO 4 - (aq) + 3C 2 O 4 2- (aq) 2MnO 2 (s) + 6 CO 3 2- (aq) + 4 H + Adding 4 OH - to each side gives:
4OH - + 2H 2 O + 2MnO 4 - (aq) + 3C 2 O 4 2- (aq) 2MnO 2 (s) + 6CO 3 2- (aq) + 4 H 2 O and cancel some waters This is the FINAL BALANCED REDOX EQUATION in basic solution 4 OH - (aq) + 2 MnO 4 - (aq) + 3 C 2 O 4 2- (aq) 2 MnO 2 (s) + 6 CO 3 2- (aq) + 2 H 2 O(l) Do Cr 2 O 2-7 (aq) + Cl - (aq) -----> Cr 3+ (aq) + Cl 2 (g) (acidic solution) Cr 2 O 2-7 (aq) + Cl - (aq) -----> Cr 3+ (aq) + Cl 2 (g) (acidic solution) ON Cr = +6 Cl = -1 Cr = +3 Cl= 0 O = -2 Reduction Half Reaction(RHR): 14H + 2- + Cr 2 O 7 + 6e- ----> 2Cr 3+ + 7H 2 O H= +1 Cr = +6 Cr=+3 H= +1, O= - 2 Oxidation Half Reaction(OHR) : 2Cl - ----> Cl 2 + 2e- Cl = -1 Cl= 0 In order to balance redox reactions the electrons coming from oxidation half-reaction has to be equal to electrons going into reduction half-reaction. Since these half reactions do not have same number of electrons we have to multiply these equation by a factor as in linear algebra before canceling an unknown. Reduction Half Reaction(RHR):( 14H + + Cr 2 O 7 2- + 6e- ----> 2Cr 3+ + 7H 2 O) x 1 Oxidation Half Reaction(OHR): ( 2Cl - ----> Cl 2 + 2e- ) x 3 14H + 2- + Cr 2 O 7 + 6e- ----> 2Cr 3+ + 7H 2 O 6Cl - ----> 3Cl 2 +6e- Net ionic equation 14H + 2- + Cr 2 O 7 + 6Cl - ----> 2Cr 3+ + 3Cl 2 + 7H 2 O You should realize at this point what you came up with is a net ionic equation ( discussed in chapter 4). In order to obtain molecular equation, first you have to get the complete ionic equation. 14H + 2- + Cr 2 O 7 + 6Cl - ----> 2Cr 3+ + 3Cl 2 + 7H 2 O Adding spectator ions: 2K + + 8Cl - ----> 2K + + 8Cl - Complete ionic equation: (14H + + 14Cl - ) + ( 2K + 2- + Cr 2 O 7 ) ----> (2Cr 3+ + 6Cl - ) + (2K + + 2Cl - ) + 3Cl 2 + 7H 2 O Molecular equation: 14HCl + K 2 Cr 2 O 7 ----> 2CrCl 3 + 2 KCl + 3Cl 2 + 7H 2 O Sum of the stoichiometric coefficients: 14 + 1 + 2 +2 + 3+ 7 = 29
Using Half-Reactions to Understand Redox Reactions Consider adding zinc metal to a solution of acid (e.g. HCl) - bubbles of gas formed (hydrogen gas). The equation is Zn(s) + 2 H + (aq) Zn 2+ (aq) + H 2 (g) For the reaction as written left-to-right: Zn(s) is the reducing agent, giving electrons (e - ) to H + (aq) H + (aq) is the oxidizing agent, taking e - from Zn (s) Zn(s) is being oxidized (it is losing e - ) (Ox. State Zn 0 increases to Zn 2+ ) H + (aq) is being reduced (it is gaining e - ) (Ox. State H 1+ decreases to H 0 ) For the reverse reaction right-to-left H 2 (g) is the reducing agent, Zn 2+ (aq) is the oxidizing agent H 2 (g) is getting oxidized, Zn 2+ (aq) is getting reduced. Note: The oxidizing agent is the substance being reduced. The reducing agent is the substance being oxidized. Electrochemical Cells Voltaic Cell 1. A voltaic or galvanic cell uses a spontaneous reaction ( G < 0) to generate electrical energy. Thus reactants have higher chemical potential energy (free energy) than products. All batteries contain voltaic cells. (System does work on the surroundings.) Electrolytic Cell 2. An electrolytic cell uses electrical energy to drive a non-spontaneous reaction ( G > 0). Thus, products have higher free energy than reactants, e.g., metal plating. (Surroundings to work on the system.) Voltaic Cells (Zn bar in Cu 2+ solution) - a spontaneous reaction (the resulting Zn/Cu Voltaic cell) We find that the concentration of the zinc solution increases over time, concentration of copper solution decreases over time. We find mass of zinc rod decreases. Copper metal is deposited on the copper rod. Electrons flow through the external circuit, e.g., can light a bulb, run a heater, etc. Left compartment Zn (s) Zn 2+ + 2 e - Right compartment Cu 2+ + 2 e - Cu (s)
Salt bridge serves to keep the solution neutral; without the salt bridge the Zn 2+ solution would become positively charged and the Cu 2+ solution would become negatively charged, and the reaction would stop. reduction occurs at the cathode (= red cat) Notation for a Voltaic Cell Zn (s) + Cu 2+ (aq) Zn 2+ (aq) + Cu (s) Zn (s) Zn 2+ (aq) Cu 2+ (aq) Cu (s) anode cathode salt bridge 1) vertical bars ( ) indicate a phase boundary 2) electrodes placed at far left and right. If electrode not involved in half-reaction, put it at end and show reagents in order they appear in half-reaction. e.g., 2 I - (aq) I 2 (s) + 2 e - (anode i.e. source of e - ) MnO 4 - (aq) + 8 H + (aq) + 5 e - Mn 2+ (aq) + 4 H 2 O (l) (cathode i.e. where e - go and are used in a reduction) Electrodes are graphite, platinum, etc.: not involved in reactions. graphite I - (aq) I 2 (s) MnO 4 - (aq), H + (aq), Mn 2+ (aq) graphite
or Pt (s) I - (aq) I 2 (s) MnO 4 - (aq), H + (aq), Mn 2+ (aq) Pt (s) e.g., Cr bar dipping in Cr(NO 3 ) 3 solution on one side of cell, Ag bar in AgNO 3 solution on the other side, and connected by salt bridge. The Cr electrode is the negative one. Draw diagram, and give cell reaction and cell notation. Since Cr (s) Cr 3+ (aq) is the negative electrode, it is the anode Ag (s) Ag + (aq) is the positive electrode it is the cathode) Cr (s) Cr 3+ (aq) + 3 e - (anode, source of e -, put on left of cell) Ag + (aq) + e - Ag (s) (cathode, where e - are used, put on right of cell) Cr (s) + 3 Ag + (aq) Cr 3+ (aq) + 3 Ag (s) and cell notation is Cr (s) Cr 3+ (aq) Ag + (aq) Ag (s) Electrochemical Cells and Voltage Using Standard Cell Potentials Measured cell potential (E cell ) depends on concentrations. At standard conditions (1 atm gas, 1 M sol n s), E cell = E cell E cell = standard cell potential E cell for Zn/Cu cell = 1.10 V. Zn(s) Zn 2+ (aq, 1 M) Cu 2+ (aq, 1 M) Cu (s) How do we calculate E cell? We use E half-cell for the two half-reactions Half-Cell Electrode Potentials (E half-cell ) Each half of cell has a half-cell potential that together give total E cell or E cell. By convention, half-cell potentials refer to half-rxns written as reductions i.e. ox + e - red (ox = oxidized form; red = reduced form) and this is the way they are listed in Tables Reversing direction changes sign of half-cell potential. So, for Zn (s) + Cu 2+ (aq) Zn 2+ (aq) + Cu (s)
Zn 2+ (aq) + 2 e - Zn (s) E Zn (listed in Tables) Cu 2+ (aq) + 2 e - Cu (s) E cathode = E Cu (listed like this) In the cell, the Zn half-reaction occurs in the opposite direction (oxidation), so we change the sign: Zn (s) Zn 2+ (aq) + 2 e - E cell is the sum of the two half-cell rxns. E anode = -E Zn E cell = E cathode + (-E anode ) = E Cu + (-E Zn ) Since the anode is always an oxid n, E anode = -E half-rxn always E cell = E cat E anode (NOTE: the minus sign already takes into account the fact that the anode reactions occurs in the opposite direction from that listed in the text, so do not change the sign of numbers you take from App D and plug into this equation!) Determining E half-cell with the Standard H 2 Electrode E cell is related to G. G not known absolutely, only differences ( G). Similarly, halfcell E 's always measured vs. a reference electrode. Reference electrode = standard hydrogen electrode (SHE) SHE = Pt solid dipping in 1 M aq strong acid through which 1 atm H 2 gas is bubbled. This is, by convention, given a value of 0.00 V 2 H + (aq, 1M) + 2 e - H 2 (g, 1 atm) E ref = 0.00 V So, construct voltaic cell with SHE on one side and unknown on other; gives unknown E, since E cell is known (measured with a voltmeter) E cell = E cat - E ano If SHE is cat: E cell = 0.00 - E unk E cell = -E unk If SHE is ano: E cell = E unk - 0.00 E cell = E unk This way, E half-cell for any half-cell can be obtained from tables Example: Zn (s) Zn 2+ (aq) Cu 2+ (aq) Cu (s) E cell = E cat E ano = +0.34 V (- 0.76 V) = 1.10 V
(NOTE: use numbers straight from table. Do not change sign) Measured E cell = 1.10 V. Problem: Voltaic cell: Br 2 (aq) + Zn (s) Zn 2+ (aq) + 2 Br - (aq) E cell = 1.83 V. If E Zn = -0.76 V, what is E Br for the half-cell rxn Br 2 (aq) + 2 e - 2 Br - (aq) E cell = E cat E ano Zn (s) Zn 2+ (aq) + 2 e - ; it is the anode E cell = E Br - E Zn 1.83 V = E Br -(- 0.76 V) = E Br + 0.76 V E Br = 1.83 V - 0.76 V E Br = 1.07 V CHECK: E cell = E cat E ano = 1.07 V - (- 0.76 V) = 1.83 V (Correct!) Problem: Br 2 (aq) + 2 V 3+ (aq) + 2 H 2 O (l) 2 VO 2+ (aq) + 4 H + (aq) + 2 Br - (aq) E cell = 1.39 V. What is E for VO 2+ V 3+? Br 2 (aq) + 2 e - 2 Br - (aq) the cathode E cat = 1.07 V E cell = E cat E ano 1.39 V = 1.07 V E anode E anode = 1.07 1.39 = - 0.32 V E VO 2+ V 3+ = - 0.32 V Relative Strengths of Oxidizing and Reducing Agents E half-cell values represent how easy or difficult it is to add electrons comparing E h.c. gives us strengths of oxidizing/reducing agents. e.g. Cu 2+ (aq) + 2 e - Cu (s) E = + 0.34 V 2 H + (aq) + 2 e - H 2 (g) E = 0.00 V Zn 2+ (aq) + 2 e - Zn (s) E = - 0.76 V The bigger (more +ve, or less -ve) the E, the easier is the reduction The smaller (more ve, or less +ve) the E, the harder is the reduction Conversely, The bigger (more +ve, or less -ve) the E, the harder is the oxidation The smaller (more ve, or less +ve) the E, the easier is the oxidation
reducing strength: Zn > H 2 > Cu oxidizing strength: Cu 2+ > H + > Zn 2+ General Rule: Strength of reducing agent increases as E half- becomes smaller (as you go down Table). Strength of oxidizing agent increases as E halfn becomes bigger. Predicting Direction of Spontaneous Reactions E values are related to G we can use them to predict spontaneity? Cu (s) + Zn 2+ (aq)? Cu 2+ (aq) + Zn (s) weaker weaker stronger stronger red. ag. ox. ag. ox. ag. red. ag. stronger red/ox agents win, and push reaction towards the other side spontaneous direction is right-to-left i.e., e - s grabbed by stronger oxid. agent (Cu 2+ ) converting it to Cu (s). (i.e., electrons flow from the smaller E half-rxn to the larger E half-rxn ) ( Note: Negative numbers are smaller than positive numbers!! ) Example Mn 2+ (aq) + 2 e - Mn (s) E half-rxn = -1.18 V Sn 2+ (aq) + 2 e - Sn (s) E half-rxn = -0.14 V Mn (s) + Sn 2+ spont. (aq) Mn 2+ (aq) + Sn (s) Sample Problem NO - 3 (aq) + 4 H + (aq) + 3 e - NO (g) + 2 H 2 O (l) N 2 (g) + 5 H + (aq) + 4 e - N 2 H + 5 (aq) MnO 2 (s) + 4 H + + 2 e - Mn 2+ (aq) + 2 H 2 O (l) E = 0.96 V E = - 0.23 V E = 1.23 V Reducing strength: N 2 H 5 + > NO > Mn 2+ Oxidizing strength: MnO 2 > NO 3 - > N 2 Spont. Rxns: 1. 3 N 2 H + 5 (aq) + 4 NO - 3 (aq) + H + (aq) N 2 (g) + 4 NO + 8 H 2 O (l) 2. 2 NO (g) + 3 MnO 2 (s) + 4 H + 2 NO - 3 (aq) + 3 Mn 2+ (aq) + 2 H 2 O (l) 3. N 2 H + 5 (aq) + 2 MnO 2 (s) + 3 H + N 2 (g) + 2 Mn 2+ (aq) + 4 H 2 O (l)
E cell (1) = 0.96 V - (-0.23 V) = 1.19 V E cell (2) = 1.23 V - (0.96 V) = 0.27 V E cell (3) = 1.23 V (-0.23 V) = 1.46 V An Application: the Reaction of Metals with Acids Some metals (e.g., Fe) dissolve in acid (e.g., HCl) to give H 2 (g), others (e.g., Cu) do not. Why? Consider: M (s) + 2 H + (aq) M 2+ (aq) + H 2 (g) stronger stronger weaker weaker red. agent ox. agent ox. agent red. agent -- for the M (s) to react to give M 2+ (aq) and H 2 (g), the rxn has to be spontaneous left-toright Half-reactions are: M 2+ (aq) + 2 e - M (s) 2 H + (aq) + 2 e - H 2 (g) For rxn. to occur spontaneously, the following must be true: E M E (= SHE) o H 2 (i) (ii) (iii) voltaic cell using these half-rxns must have E cell > 0 (i.e. G<0) M (s) must be a stronger reducing agent than H 2 (g) H + (aq) must be a stronger oxidizing agent than M 2+ (aq) All these are saying the same thing!! and all require E half for M (=E M ) be smaller (less positive or more negative) than E half for H + (= o o E H ) i.e. E M is the anode, and E 2 H is the 2 cathode. o e.g., E cell = E cat - E ano = EH - E M = 0.00 E M 2 for E cell to be positive, E M must be < 0.00 (i.e. negative) i.e., E M < o E H i.e. E M < 0.00 V 2 E M must be listed below H 2 half-cell (SHE) in Table 21.2 or App D i.e., M (s) must be a stronger reducing agent than H 2 (g) i.e., H + (aq) must be a stronger reducing agent than M 2+ (aq) must be listed above M half-cell Co, Mn, Na, Zn, Mg, Al, etc. dissolve in (react with) acid whereas Cu, Hg, Au, Ag, etc. do not
E o and Gibbs Free Energy Electrochemistry involves redox reactions because the electrical energy (flow of electrons) has at its origin the oxidation (loss of electrons) and reduction (gain of electrons) of species. 1. Can we harness the energy released when a spontaneous reaction occurs? 2. Can we cause a non-spontaneous reaction to occur by supplying energy? You have already seen one way to accomplish 2., by supplying thermal energy (raising the temperature) if S > 0. G = H - T S If S > 0, then -T S < 0, and we can make a non-spontaneous reaction spontaneous by raising the temperature so that -T S term dominates. The relationship between Gibbs fee energy and the E cell is summarized by the expression Where G = n F E cell F = Faraday constant = 96500 J mol/v n = number of mols of electrons transferred θ Coupling this with the expression below (See equilibrium notes) θ G = - RT ln K it is fairly simple to derive the expression θ RT E cell = lnk nf at 298K and substituting for R and F, the expression can be simplified to θ 0.0257 E cell = lnk n Analyzing the possible combinations of K, E o cell and G o leads to the following conclusions. K E o cell G o > 1 Positive Negative Conclusion Spontaneous cell reaction =1 0 0 At equilibrium < 1 Negative Positive The Nernst equation Non-spontaneous cell reaction. Reaction is spontaneous in the reverse direction The Nernst Equation and non-standard conditions of temperature AND concentration
The Nernst equation can be used to calculate the voltage generated by the combination of two half-cells when the conditions are NOT standard. One form of the equation is RT E = E cell - lnq nf Where R is the gas constant (8.314 J/K mol), T the Kelvin temperature, n the number of electrons transferred between the species, F the Faraday constant, E o cell is the voltage generated IF the conditions WERE standard, ln represents the natural logarithm and Q is the reaction quotient. The reaction quotient is defined as the concentrations of the product ions, raised to their stoichiometric powers and multiplied together, divided by the concentrations of the reactant ions raised to their stoichiometric powers and multiplied together. Any pure solids or liquids are omitted from the reaction quotient. This equation allows us to compute the cell voltage at any concentration of reactants and products and at any temperature. Commonly it is only the concentrations of ions that have changed and at 298K the expression can be simplified to, Example Calculation 0.0257 E = E cell - lnq n If the reaction below is carried out using solutions that are 5M Zn 2+ and 0.3M Cu 2+ at 298K, what is the actual cell voltage? Zn (s) + Cu 2+ (aq) Cu (s) + Zn 2+ (aq) Firstly, work out the E o cell assuming standard conditions. Zn (s) Zn 2+ (aq) Cu 2+ (aq) Cu (s) Zn 2+ (aq) + 2e - Zn (s) E o = -0.76V Cu 2+ (aq) + 2e - Cu (s) E o = +0.34V E o cell = E R E L = +0.34 - -0.76 = 1.1V Then calculate Q. Since zinc and copper metals are solids, they are omitted from Q. Q = [Zn 2+ ]/[Cu 2+ ] = 5/0.3 = 16.7 Two electrons are transferred between the zinc and copper, so n=2. Plug everything in E = 1.1V - (0.0257/2) * ln (16.7) = 1.06V Effect of Concentration on Cell Potential Concentration cells can arise when the concentration of one of the species participating in a corrosion reaction varies within the electrolyte. Two examples will be given: Electrolyte concentration Consider a metal bathed in an electrolyte containing its own ions. The basic redox reaction where a metal atom losses an electron and enters the electrolyte as an ion can proceed both forward and backwards, and will eventually reach equilibrium. If a region of the electrolyte (adjacent to the metal) were to exhibit a decreased concentration of metal ions, this region would become anodic to the other portions of the
metal surface. As a result, this portion of the metal would react faster in order to increase the local ion concentration. The net affect is that local reaction rates are modulated in order to homogenize reduction ion concentrations within the electrolyte. Neuron Cells and Neuron Action Neurons operate by electricity through charge conduction is by ions, not electrons. Important ions for neural transmission are potassium, sodium, calcium, and chloride. There is a resting potential, maintained by the cell after firing neurotransmitters, it takes time to re-establish. Firing allows the passage of the signals through the synaptic gap. Transduction (the process whereby a transducer accepts waves in one form and gives back related waves in a different form) involves turning energy into an electrical potential as we described in electrochemistry. Neuron Action Neurons operate by electricity. Charge conduction is by ions, not electrons Important ions - potassium, sodium, calcium, chloride There is a resting potential, maintained by the cell After firing, it takes time to re-establish Sight - stimulus in the visible range of the electromagnetic spectrum causes pigment (rhodopsin derived from vitamin A so eat your carrots!) changes that result in an action potential which is transmitted through the neurons. Hearing - sound vibrations are mediated by the ear drum and middle ear bones that cause special nerve endings to flex and produce an electrical discharge Smell and Taste - chemical reactions with certain foods or aromas produce an electrical potential which transmitted through the neurons. Tactile - pressure, temperature, pain, position also produce an electrical potential which transmitted through the neurons.
Common Batteries Batteries are simply galvanic cells, or stacks of galvanic cells. I want to look at a couple of examples. Today we'll focus on the lead acid battery, one of the most important and common batteries. Lead storage batteries are important and hard to replace because they have a high current density, are tough, durable, handle a wide range of environmental conditions, and are rechargeable. Drycell battery: Zn + MnO 2 + H 2 O Zn (OH) 2 + Mn 2 O 3 Drycells are lightweight, durable batteries, easily manufactured for widespread use. This is ideal for portable flashlights and radios. Unfortunately, this type of battery is not rechargeable; once the chemical reaction is complete (all of the zinc has reacted with the manganese (IV) oxide to form manganese (III) oxide and zinc hydroxide), the battery is "dead". Disposable (nonrechargeable) batteries are a major source of trash and pollution; the problem is further compounded by the use toxic metals, such as nickel and cadmium, in some batteries. Car battery The half reactions in these batteries can be written as: - anode: Pb (s) + HSO 4 (aq) PbSO 4(s) + H + + 2e - - cathode: PbO 2(s) + HSO 4 (aq) + 3 H + + 2e - PbSO 4(s) + 2 H 2 O
The anode of the battery is a lead grid coated in "spongy lead" to provide a high surface area and thus high current capacity, while the cathode is a lead grid coated with spongy lead(iv) oxide. (overhead, figure in text) The rechargeability of this battery arises from the fact that the reactants and reaction products of both half cells are deposited on the electrodes and so are available for both the forward and reverse reactions. Fuel Cells Electrolysis-Forcing Reactant-Favored Reactions to Occur Electrolysis is the process in which electrical energy is used to cause a nonspontaneous REDOX reaction to occur. An electrolytic cell is one that is used to carry out electrolysis. A typical cell is shown below. Anions are attracted towards the anode where they under go a process of oxidation. Electrons flow from the anode to the cathode where cations undergo a process of reduction.
Quantitative aspects of electrolysis The amount of a substance produced in an electrolytic cell can be calculated using Faradays law. 1. Calculate the number of Faradays passed in the electrolysis by using the expression Current (in amps) x Time (in seconds) Number of Faradays = 96500 2. Then use the stoichiometry of the electrode process to determine the mass of product deposited at the electrode. Remembering that a process that produces 1 mol of product by the transfer of 1 electron will require 1 Faraday to produce that 1 mol, a process that produces 1 mol of product by the process of 2 electrons will require 2 Faraday s to produces that 1 mol etc. Counting Electrons Electron counting is done by half reaction method and balancing the electrons in both reactions. Corrosion-Product-Favored Reactions Corrosion reactions are based on principles we described earlier for concentration cells. Perhaps the most common concentration cell affecting engineered structures is that of oxygen gas. When oxygen has access to a moist metal surface, corrosion is promoted. however, it is promoted the most where the oxygen concentration is the least. As a result, sections of a metal that are covered by dirt or scale will often corrode faster, since the flow of oxygen to these sections is restricted. An increased corrosion rate will lead to increased residue, further restricting the oxygen flow to worsen the situation. Pitting often results from this "runaway" reaction. Most metal corrosion occurs via electrochemical reactions at the interface between the metal and an electrolyte solution. A thin film of moisture on a metal surface forms the electrolyte for atmospheric corrosion. Wet concrete is the electrolyte for reinforcing rod corrosion in bridges. Although most corrosion takes place in water, corrosion in nonaqueous systems is known. Electrochemical corrosion involves two half-cell reactions; an oxidation reaction at the anode and a reduction reaction at the cathode. For iron corroding in water with a near neutral ph, these half cell reactions can be represented as: Anode reaction: 2Fe => 2Fe 2+ + 4e - Cathode reaction: O 2 + 2H 2 O + 4e - => 4OH -
There are obviously different anodic and cathodic reactions for different alloys exposed to various environments. These half cell reactions are thought to occur (at least initially) at microscopic anodes and cathodes covering a corroding surface. Macroscopic anodes and cathodes can develop as corrosion damage progresses with time. Schematic representation of electrochemical corrosion process (aqueous corrosion of iron under near neutral ph conditions) Schematic representation of current flow (conventional current direction) in a simple corrosion cell
From the above theory it should be apparent that there are four fundamental components in an electrochemical corrosion cell: An anode. A cathode. A conducting environment for ionic movement (electrolyte). An electrical connection between the anode and cathode for the flow of electron current. If any of the above components is missing or disabled, the electrochemical corrosion process will be stopped. Clearly, these elements are thus fundamentally important for corrosion control. Corrosion normally occurs at a rate determined by an equilibrium between opposing electrochemical reactions. The first is the anodic reaction, in which a metal is oxidized, releasing electrons into the metal. The other is the cathodic reaction, in which a solution species (often O 2 or H + ) is reduced, removing electrons from the metal. When these two reactions are in equilibrium, the flow of electrons from each reaction is balanced, and no net electron flow (electrical current) occurs. The two reactions can take place on one metal or on two dissimilar metals (or metal sites) that are electrically connected. Cathodic protection The required cathodic protection current is supplied by sacrificial anode materials or by an impressed current system. By connecting a metal of higher potential to a metallic structure it is possible to create an electrochemical cell in which the metal with lower potential becomes a cathode and is protected. This technique is called cathodic protection. Anodic Protection In contrast to cathodic protection, anodic protection is relatively new. Edeleanu first demonstrated the feasibility of anodic protection in 1954 and tested it on small-scale stainless steel boilers used for sulfuric acid solutions. This was probably the first industrial application, although other experimental work had been carried out elsewhere. Anodic protection possesses unique advantages. For example, the applied current is usually equal to the corrosion rate of the protected system. Thus, anodic protection not only protects but also offers a direct means for monitoring the corrosion rate of a system. Anodic protection can decrease corrosion rate substantially. The primary advantages of anodic protection are its applicability in extremely corrosive environments and its low current requirements. Anodic protection has been most extensively applied to protect equipment used to store and handle sulfuric acid. Sales of anodically protected heat exchangers used to cool sulfuric acid manufacturing plants have represented one of the more successful ventures for this technology. These heat exchangers are sold complete
with the anodic protection systems installed and have a commercial advantage in that less costly materials can be used.
102 HOMEWORK 7 HOMEWORK FOR CHAPTER 19 1. Oxidation number of an atom in a compound or ion a) is always equal to the number of valance electron. b) is always equal to the number of protons in the nucleus. c) is the number of electrons shared/gained/lost relative to the neutral atom. d) is always equal to charge of the compound or ion 2. The following species with the highest oxidation number for manganese is a. MnSO 4. b. MnO 4 -. c. MnO 2. d. Mn 2 O 3. 3. For Pb(s) + PbO 2 (s) + 2 H 2 SO 4 (aq) -> 2 PbSO 4 (s) + 2 H 2 O(l), the substance that is reduced during the reaction is a. Pb(s). b. PbO 2 (s). c. H 2 SO 4 (aq). d. PbSO 4 (s). 4. Calculate G for the cell reaction [ G = -nfe cell] 2Al 3+ (aq) + 6I - (aq) ---> 2Al(s) + 3I 2 (s). Al 3+ (aq) + 3e - ---> Al(s); E ½ = -1.66 I 2 (s) + 2e - ---> 2I - (aq) ; E ½ = +0.54 V a. 6.5 x 10 5 J b. -6.5 x 10 5 J c. -1.3 x 10 5 J d. 1.3 x 10 5 J e. -4.2 x 10 5 J 5. If all of the following species are in their standard states, which is the strongest oxidizing agent? Al 3+ (aq) + 3e- ----> Al(s) E ½ = -1.66 V 2H + (aq) + 2e- ----> H 2 (g) E ½ = 0.00 V Pb 2+ (aq) + 2e- ----> Pb(s) E ½ = +0.13 V K + (aq) + e- ----> K(s) E ½ = -2.92 V F 2 (s) + 2e- ----> 2F - (aq) E ½ = +2.87 V a. 2H + (aq) b. K + (aq) c. F 2 (s) d. Al 3+ (aq) e. Pb 2+ (aq) 6. In the redox reaction given below the oxidation half reaction is 2 Na + Cl 2 ---- 2NaCl a) Cl 2 + 2e - ---- 2Cl - b) Na ---- Na + + e - c) 2 Na + Cl 2 ---- 2NaCl d) Na + + e - ---- Na e) 2Cl - ---- Cl 2 + 2e - 7. What is the oxidation state of Cr in Cr 2 O 7 2-? a) +7 b) +6 c) +12 d) -1 e) -2 8. When the following redox reaction is balanced what is the stoichiometric coefficient of Cl 2 Cr 2 O 7 2- (aq) + Cl - (aq) ----- Cr 3+ (aq) + Cl 2 (g) (acid solution) a) 2 b) 3 c) 6 d) 9 e) 12 9. The reaction below occurs in BASE. When the equation is balanced, what is the sum of the coefficients? Zn + NO 2+ 2-3 Zn(OH) 4 + NH 3 a) 12 b) 15 c) 19 d) 23 e) 27
10. When the following redox reaction is balanced how many electrons are transferred from oxidation to reduction half reaction Cr 2 O 7 2- (aq) + Cl - (aq) ----- Cr 3+ (aq) + Cl 2 (g) (acid solution) a) 3 b) 6 c) 9 d) 12 e) 14 11. Refer to the cell diagrammed below at 25 o C Zn(s) ------ Zn2++(aq) + 2e- ; E o half-cell = -0.76V Cu ------ Cu 2+ (aq) + 2e - ; E o half-cell = +0.34V What is the value of E o cell a) -1.10 V b) -0.76 V c) 1.10 V d) 0.34 V e) -0.42 V 12. Consider the galvanic cell shown above. Which of the following statements about this cell is FALSE? a) This is not a galvanic cell. b) Electrons flow from the Zn electrode to the Cu electrode. c) Oxidtion occurs at the Cu electrode. d) The cell is not at standard conditions. e) To complete the circuit, cations migrate into the left half-cell and anions migrate into the right half-cell from the salt bridge. 13. Consider the cell shown in the figure above. The relevant reduction potentials are Ag + + e - --- Ag E o = 0.80 V Zn 2+ + 2e - --- Zn E o = -0.76 V Which of the following statements is FALSE? a) The silver electrode is the cathode. b) Increasing the [Zn 2+ ] will increase the cell voltage. c) Electrons in the external circuit will flow from zinc to silver. d) The standard cell potential for this galvanic cell is 1.56 V. e) The zinc electrode is the anode. 14. Given that Q = I x t; t and Q = n x z x F; where: Q, measured in coulombs (C), is the quantity of electricity; I, measured in amps (A), is the current; t, measured in seconds (s), is the time; n is the number of moles of substance produced at the electrode; z is the charge on the ion; and, F is a constant, with a value of 96500 C mol -1 ;. How many seconds would it take to deposit 21.40 g of Ag (at. mass = 107.87) from a solution of AgNO 3 using a current of 10.00 amp? a) 9649 s b) 4825 s c) 3828 s d) 1914 s e) none of these
15. What is the value of the non-standard cell potential, E cell, for the galvanic cell with a Zn 2+ concentration of 0.0100 M and the Ag + concentration of1.25 M? Cell reaction: 2Ag + + Zn(s) ----> 2Ag(s) + Zn 2+ ; E cell = 1.56 V [E cell = E cell - (0.592/n) log Q] a. 15 V b. 1.63 V c. 1.25 x 10-2 V d. 8.00 x 10-3 V e. 6.40 x 10-3 V
102 Sample Test 7 SAMPLE TEST FOR CHAPTER 19. Electrochemistry. 1. The study of the relationships between electron flow and redox reactions is called a. thermodynamics b. kinetics c. electrochemistry d. inorganic chemistry e. nuclear chemistry 1. c. electrochemistry 2. Which is not an absolute indicator of a redox reaction? a. One reactant is a strong oxidizing agent. b. One reactant is a strong reducing agent. c. At least one element changes oxidation number. d. A pure element is present as a reactant or a product. e. Oxygen is present. 3. All of these changes indicate an oxidation half-reaction except a. increase in oxidation number b. loss of electrons c. electrons as products d. reactant acting as a reducing agent e. pure oxygen becoming oxide ion 4. Which one of these changes describes a reduction half-reaction? a. increase in oxidation number b. loss of electrons c. electrons as products d. reactant acting as a reducing agent e. pure oxygen becoming oxide ion 5. All of these changes indicate that a reduction has occurred except a. increase in oxidation number b. gain of electrons c. electrons as reactants d. reactant acting as an oxidizing agent e. pure oxygen becoming oxide ion 6. Zn(s) + H 2 SO 4 (aq) H 2 (g) + ZnSO 4 (aq) In the reaction shown, is the oxidizing agent and is the reducing agent. a. Zn; H + b. H + ; Zn c. H 2 ; Zn 2+ d. Zn 2+ ; H 2 e. H + ; Zn 2+ 7. When iron ore is refined to pure iron using coke (a form of carbon), the oxidizing agent is, and the reducing agent is. a. O 2- ; C b. Fe 3+ ; O 2- c. C; Fe 3+ d. Fe 3+ ; C e. O 2- ; Fe 8. Al(s) + Hg 2+ (aq) Al 3+ (aq) + Hg(l) When the reaction shown above is balanced, the coefficients are, and electrons are transferred. a. 1,3,1,3; 6 b. 2,3,2,3; 12 c. 2,3,2,3; 6 d. 3,2,3,2; 12 e. 3,2,3,2; 6 9. The balanced reaction between aluminum metal and aqueous cadmium(ii) ion to produce cadmium metal and aqueous aluminum ion in acidic solution is written a. 2Al(s) + 3Cd 2+ (aq) 2Al 3+ (aq) + 3Cd(s) b. 3Al(s) + 2Cd 2+ (aq) 3Al 3+ (aq) + 2Cd(s) c. Al(s) + 3Cd 2+ (aq) 2Al 3+ (aq) + Cd(s) d. 6Al(s) + 6Cd 2+ (aq) 6Al 3+ (aq) + 6Cd(s)
e. 2Al(s) + 3Cd 2+ (aq) 3Al 3+ (aq) + 2Cd(s) 10. Which statement is not correct? a. Another term for electrochemical cell is voltaic cell. b. A battery is an example of an electrochemical cell. c. An electrochemical cell uses electrical energy to cause a reactant-favored chemical reaction to occur. d. In an electrochemical cell electrons flow from the anode to the cathode in the external circuit. e. Reduction occurs at the cathode in an electrochemical cell. 11. Refer to Graph above consider an electrochemical cell as shown, with Al in AlCl 3 (aq) and Ag in AgNO 3 (aq), and a salt bridge containing KNO 3 (aq). The unbalanced overall chemical reaction is Al(s) + Ag + (aq) Al 3+ (aq) + Ag(s). Which statement is not correct? a. The anode reaction is Al(s) Al 3+ (aq) + 3e -. b. Electrons flow from the Al to the Ag in the external circuit. c. K+ (aq) flow from the silver solution to the aluminum solution via the salt bridge. d. The cathode reaction is Ag + (aq) + e - Ag(s). e. When the cell reaction is balanced, three moles of electrons are transferred. 12. Refer to Graph above. Consider an electrochemical cell as shown, with Zn in ZnCl 2 (aq) and Cu in Cu(NO 3 ) 2 (aq), and a salt bridge containing KNO 3 (aq). The overall chemical reaction is Zn(s) + Cu 2+ (aq) Zn 2+ (aq) + Cu(s). Which statement is correct? a. Copper is oxidized at the anode. b. Zinc is reduced at the cathode. c. One mole of electrons is transferred in this reaction. d. This is an example of a concentration cell. e. Electrons travel from the Zn to the Cu. 13. In the anode compartment of an electrochemical cell, the electrode is being ; electrons are flowing the anode, and are flowing toward the compartment through the salt bridge. a. oxidized; away from; anions b. oxidized; toward; cations c. oxidized; away from; cations d. reduced; away from; cations e. reduced; toward; anions 14. In the cathode compartment of an electrochemical cell, the electrode is being ; electrons are flowing the cathode, and are flowing toward the compartment through the salt bridge. a. oxidized; away from; anions b. oxidized; toward; cations c. oxidized; away from; cations d. reduced; away from; anions e. reduced; toward; cations
15. Which cell notation represents a battery constructed using zinc and iron, with electrons flowing from zinc to iron? a. Fe 3+ (aq) Fe(s) Zn(s) Zn 2+ (aq) b. Zn(s) Zn 2+ (aq) Fe 3+ (aq) Fe(s) c. Zn(s) Zn 2+ (aq) Fe 3+ (aq) Fe 2+ (aq) d. Fe 3+ (aq) Fe 2+ (aq) Zn(s) Zn 2+ (aq) e. Zn(s) Zn 2+ (aq) Fe(s) Fe 3+ (aq) 16. The unit used to measure electrical charge is the a. volt b. ampere c. faraday d. coulomb e. joule 17. The unit used to measure electromotive force (emf) is the a. volt b. ampere c. faraday d. coulomb e. joule 18. The unit used to measure electrical current is the a. volt b. ampere c. faraday d. coulomb e. joule 19. Which statement about standard conditions for electrochemical cell measurements is not correct? a. The cell voltage of a product-favored reaction is positive. b. Gases are at 1 bar pressure. c. Solids and liquids are in the pure state. d. The reference reaction is the oxidation of H 2 (g). e. Solutes are at 0.1 M concentration. Use this list of half-reactions to answer the following questions. MnO - 4 (aq) + 8H + (aq) + 5e - Mn 2+ (aq) + 4H 2 O(l) 1.51 V Cr 2 O 2-7 (aq) + 6e - 2Cr 3+ (aq) + 7H 2 O(l) 1.33 V Pt 2+ (aq) + 2e - Pt(s) 1.20 V Cu 2+ (aq) + 2e - Cu(s) 0.34 V Pb 2+ (aq)+2e - Pb(s) -0.13 V Al 3+ (aq)+3e - Al(s) -1.66 V 20. Refer to Inst above. The strongest oxidizing agent in the table is a. MnO - 4 (aq) b. H + (aq) c. Mn 2+ (aq) d. Al 3+ (aq) e. Al(s) - 21. Refer to Inst above. The weakest oxidizing agent in the table is a. MnO - 4 (aq) b. H + (aq) c. Mn 2+ (aq) d. Al 3+ (aq) e. Al(s) 22. Refer to Inst above. The strongest reducing agent in the table is a. MnO 4 - (aq) b. H + (aq) c. Mn 2+ (aq) d. Al 3+ (aq) e. Al(s) 23. Refer to Inst above. Which combination of reactants could be used to make an electrochemical cell? a. Al 3+ (aq) and Cu(s) b. Pt(s) and Pb 2+ (aq) c. Cu(s) and MnO - 4 (aq) d. Cr 2 O 2-7 (aq) and MnO - 4 (aq) e. Al 3+ (aq) and Cr 3+ (aq) 24. Refer to Inst above. The potential for the product-favored reaction involving aluminum and platinum is a. 6.92 V b. 2.86 V c. 2.58 V d. 0.46 V e. 0.28 V 25. Refer to Inst 19-1. An electrochemical cell is designed using copper as one electrode and another metal higher than copper on the table above as the other electrode. The cell potential is +0.515. The potential for the unknown half-reaction is, and that electrode is the. a. 0.855 V; cathode b. 0.855 V; anode c. 0.175 V; cathode d. 0.175 V; anode e. 0.340 V; cathode 26. The quantity of charge in coulombs associated with one mole of electrons is C. a. 1.60 10-19 b. 6.022 10 23 c. 0.00 d. 96,500 e. 0.0592
27. The relationship between Gibbs free energy and Eº cell is Gº = a. nfeº cell b. -nfeº cell c. -RT ln K d. 0.0592 log K e. 0.0592 log K n n 28. The value of Eº cell for an aluminum-zinc electrochemical cell is 0.897 V at 25 C. Calculate the value of Gº for this cell. a. -86.6 kj b. -17 3 kj c. -260. kj d. -433 kj e. -519 kj 29. Consider the cell reaction Sn(s) + Cu 2+ (aq) Sn 2+ (aq) + Cu(s). The value Eº cell of is 0.447 V at 25 C. Calculate the value of G and K for this cell. a. -43.1 kj; 1.37 10 43 b. 43.1 kj; 3.55 10 7 c. -86.3 kj; 1.34 10 15 d. 86.3 kj; 2.00 10 86 e. 86.3 kj; 7.92 10-16 30. Calculate the value of Eº cell for a cell consisting of Ag in a 2.0 M solution of Ag + (aq) and Cd in a solution of 2.0 M Cd 2+ (aq). The value of Eº cell for this reaction is 1.26 V. Ag is the less active metal. a. -0.22 V b. 1.21 V c. 1.27 V d. 1.31 V e. 1.48 V 31. The value of Eº cell for an iron-tin cell is 0.30 V. What is the value of Eº cell if the concentration of Fe 2+ (aq) is 0.50 M and the concentration of Sn 2+ (aq) is 2.0 M. Sn is the less active metal. a. 0.32 V b. 0.30 V c. 0.28 V d. 0.26 V e. 0.02 V 32. For a zinc-platinum electrochemical cell, calculate the value of Eº cell when the concentration of Pt 2+ (aq) is 0.050 M and the concentration of Zn 2+ (aq) is 1.1 M. Eº cell = 1.96 V under standard conditions. Pt is the less active metal. a. 1.88 V b. 1.92 V c. 1.96 V d. 2.00 V e. 2.04 V 33. Which statement concerning concentration cells is not correct? a. Both cell compartments use the same half-reaction. b. The value of Q for the system is 1. c. The concentration of metal ions in the cathode compartment must be greater than the concentration of metal ions in the anode compartment. d. The electrons flow from the anode to the cathode. e. The cell becomes useless when the concentrations of the metal ions in both compartments becomes equal. 34. The value of ph can be calculated from a measurement of cell potential using the formula ph = a. 2FEº cell b. -2FEº cell c. -RTln Eº cell d. Eº cell /0.0592 V e. 0.0592 Eº cell 35. A voltmeter was set up with an electrode to measure ph, but the readout gave a value of 0.245 V instead of a reading in ph units. What is the ph of the solution? a. 1.03 b. 4.14 c. 4.31 d. 9.69 e. 9.86 36. The function of the "ion pump" in nerve chemistry is to a. transport Na + and K + against their concentration gradients in order to maintain constant values of their concentrations. b. transport Ca 2+ and Cl - against their concentration gradients in order to maintain constant values of their concentrations. c. allow the passage of Na +, K +, Ca 2+, and Cl - faster than simple diffusion allows. d. prevent the transport of any ions against their concentration gradients. e. regulate the speed of transport of all ions across nerve cell membranes. 37. Which group of descriptions of batteries is not correct?
a. Leclanche cell; primary; Zn/NH + 4 /MnO 2 b. alkaline battery; primary; Zn/MnO 2 c. mercury battery; secondary; Zn/HgO d. lead storage; secondary; Pb/PbO 2 /PbSO 4 e. lithium battery; secondary; Li/CoO 2 38. Which statement about lead storage batteries is not correct? a. In order to provide 12 volts, the battery consists of six cells. b. The cathode reaction involves the conversion of Pb 4+ to Pb 2+. c. The anode reaction involves the conversion of Pb to Pb 2+. d. The value of Eº cell is slightly more than 2 V. e. Nitric acid serves as the electrolyte. 39. Which statement concerning the alkaline fuel cell is not correct? a. At the anode, hydrogen gas is oxidized to water. b. At the cathode, oxygen is reduced to hydroxide ions. c. The actual fuel cell reaction causes no pollution of any kind. d. A limiting factor is the cost of producing hydrogen as the fuel. e. All of the statements are correct. 40. Which process is not an electrolytic reaction? a. Na + (aq) + Cl - (aq) 2Na(l) + Cl 2 (g) b. 2H 2 O(l) 2H 2 (g) + O 2 (g) c. 2Al 2 O 3 (l) 4Al(l) + 3O 2 (g) d. Zn(s) + Cu 2+ (aq) Zn 2+ (aq) + Cu(s) e. 2PbSO 4 (s) + 4H 2 O(l) Pb(s) + PbO 2 (s) + 2HSO - 4 (aq) + 2H 3 O + (aq) 41. An electrolytic reaction is a system in which a. a chemical reaction is used to produce electricity. b. electricity is used to produce a chemical reaction. c. the same element is both oxidized and reduced. d. a reactant-favored reaction is forced to produce electricity by the input of heat or light. e. the reaction conditions are manipulated to change the value of Eº cell to a favorable one. 42. Calculate the mass of cobalt that will be deposited when a current of 2.00 A is passed through a solution of CoSO 4 for 10 hours. a. 44.0 g b. 23.7 g c. 22.0 g d. 0.366 g e. 6.11 10-3 g 43. Which statement concerning the number of electrons involved in electrolysis is not correct? a. Electrons can be "counted" by multiplying the current in amps by the time in seconds. b. The charge associated with one electron is 96,500 coulombs. c. In producing one mole of Cl 2 (g) from Cl - (aq), two moles of electrons are produced. d. When one mole of Fe is produced from Fe 3+ (aq), three moles of electrons are needed. e. Electroplating of one mole of silver from a solution of silver ions requires one mole of electrons. 44. How much copper can be plated from a solution of Cu 2+ (aq) at a current of 0.75 amps in 95 minutes? a. 84 g b. 42 g c. 1.4 g d. 0.70 g e. 0.23 g 45. Calculate the time needed to plate out 175 g of nickel from a Ni 2+ solution when a current of 10.0 A is applied. a. 938 hr b. 32 hr c. 16 hr d. 8.0 hr e. 4.0 hr 46. Unwanted oxidation of a metal exposed to the environment is a. corrosion b. electroplating c. galvanization d. cathodic protection e. electrolysis
47. An example of cathodic protection to prevent corrosion is a. coating the metal with paint. b. allowing a film of metal oxide to form on the metal. c. using a film of oil to cover the metal. d. coating the surface with a more easily oxidized metal. e. preventing contact between the metal surface and air. ANSWERS TO SAMPLE TEST FOR CHAPTER 19. Electrochemistry. 1. c. electrochemistry 2. e. Oxygen is present 3. e. pure oxygen becoming oxide ion 4. e. pure oxygen becoming oxide ion 5. a. increase in oxidation number 6. b. H + ; Zn 7. d. Fe 3+ ; C 8. c. 2,3,2,3; 6 9. a. 2Al(s) + 3Cd 2+ (aq) 2Al 3+ (aq) + 3Cd(s) 10. c. An electrochemical cell uses electrical energy to cause a reactant-favored chemical reaction to occur. 11. c. K+ (aq) flow from the silver solution to the aluminum solution via the salt bridge. 12. e. Electrons travel from the Zn to the Cu. 13. a. oxidized; away from; anions 14. e. reduced; toward; cations 15. b. Zn(s) Zn 2+ (aq) Fe 3+ (aq) Fe(s) 16. d. coulomb 17. a. volt 18. b. ampere 19. e. Solutes are at 0.1 M concentration. 20. a. MnO 4 - (aq) 21. d. Al 3+ (aq) 22. e. Al(s) 23. c. c. Cu(s) and MnO 4 - (aq) 24. b. 2.86 V 25. a. 0.855 V; cathode 26. d. 96,500 27. b. -nfeº cell 28. e. -519 kj 29. c. -86.3 kj; 1.34 10 15 30. c. 1.27 V 31. a. 0.32 V 32. b. 1.92 V 33. b. The value of Q for the system is 1. 34. d. Eº cell /0.0592 V 35. b. 4.14 36. a. transport Na + and K + against their concentration gradients in order to maintain constant values of their concentrations. 37. c.mercury battery; secondary; Zn/HgO 38. d. The value of Eº cell is slightly more than 2 V. 39. d. A limiting factor is the cost of producing hydrogen as the fuel. 40. d. Zn(s) + Cu 2+ (aq) Zn 2+ (aq) + Cu(s) 41. b. electricity is used to produce a chemical reaction. 42. c. 22.0 g 43. b. The charge associated with one electron is 96,500 coulombs. 44. c. 1.4 g
45. c. 1.4 g 46. a. corrosion 47. d. coating the surface with a more easily oxidized metal.