Chapter 18 Electrochemistry examples solutions PDF

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Example: Balance in basic solution MnO4- (aq) + SO3-2 (aq) +3 e- + 4H+ + MnO4H2O +

SO32-

+2OH- + 2 MnO4- + 2H+ + 3SO322 MnO4- + 2H2O + 3SO32-

2 MnO4- + H2O + 3SO32-

SO42- (aq) + MnO2 (s) +2 H2O

x2

+ 2H+ + 2 e-

x3

MnO2 SO42-

2 MnO2 + 3SO42- + H2O

+2OH-

2 MnO2 + 3SO42- + H2O + 2OH-

2 MnO2 + 3SO42- + 2OH-

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Old Exam Question How many electrons will be involved in the balanced equation involving the oxidation of aluminum metal with Fe(II) to form Fe(0) and Al(III)? A. 2 B. 3 C. 6 D. 1 E. 5

Old Exam Question Stuck in a RV in the desert and facing his untimely demise, a former high school teacher wants to make a spontaneous electrochemical cell (i.e. a battery) using mercury oxide and zinc. The reduction half-cell potentials for the reactions are as follows: HgO (aq) + H2O (l) + 2 e- → Hg(l) + 2OH-(aq) Eo½cell = + 0.10 V ZnO + H2O + 2e- → Zn(s) + 2OH-(aq)

Eo½cell = - 1.25 V

What are the standard cell potential and element at the anode, respectively? A. 1.35 V, zinc B. 1.35 V, mercury C. 1.15 V, zinc D. -1.15 V, zinc E. -1.15 V, mercury

Example: What’s the highest voltage battery we can make? We’ll need the chemical with the most positive cell potential for the cathode, and the one with the most negative half potential for the anode (from Appendix 6): The half reactions are: F2 (g) + 2 e-  2 F-(aq) Li+ (aq) + e-  Li (s)

E° = 2.866 V E ° = - 3.05 V

Calculate E°cell

E°cell = 2.866 V – (-3.05 V) = 5.916 V Note this battery contains the most powerful oxidizing agent (F2) and a very strong reducing agent (Li metal). Maybe not a great idea…

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Old Exam Question What would be the appropriate short-hand notation for a battery in which iron metal was being oxidized to Fe(III) and hydrogen ions were being reduced to H2(s) over Pt? A. Fe(III)(aq)|Fe(s)||H+(aq)|H2(g) B. Fe(s)| Fe(III)(aq)||H+(aq)|H2(g)|Pt C. H+(aq)|H2(g)|Pt|| Fe(III)(aq)|Fe(s) D. H+(aq)|H2(g)|Pt|| Fe(s)|Fe(III)(aq) E. Fe||H2

Example: Determine Ecell for the following: Pt(s) | Fe2+ (0.10 M), Fe3+ (0.20 M)  Ag+ (1.0 M) | Ag(s) Hint: The reaction at the anode is Fe2+  Fe3+ + 1 eFe2+ (0.10 M) + Ag+ (1.0M)  Fe3+ (0.20 M) + Ag(s) E°cell = E°cathode – E°anode = E° Ag+/Ag – E°Fe3+/Fe2+ = 0.800 V – 0.771 V = 0.029 V

Ecell =

Eo

0.0592 V 0.0592 V log ([Fe3+]/[Fe2+][Ag+]) log Q = 0.029 V cell n = 0.029 V -

0.0592 V

log [(0.20)/(0.10)(1.0)]

= 0.029 V - 0.018 V = 0.011 V

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