Using standard reduction potentials (Appendix E), calculate the standard emf for each of the following reactions: $$ \begin{array}{l} \text { (a) } \mathrm{Cl}_{2}(g)+2 \mathrm{I}^{-}(a q) \longrightarrow 2 \mathrm{Cl}^{-}(a q)+\mathrm{I}_{2}(s) \\ \text { (b) } \mathrm{Ni}(s)+2 \mathrm{Ce}^{4+}(a q) \longrightarrow \mathrm{Ni}^{2+}(a q)+2 \mathrm{Ce}^{3+}(a q) \\ \text { (c) } \mathrm{Fe}(s)+2 \mathrm{Fe}^{3+}(a q) \longrightarrow 3 \mathrm{Fe}^{2+}(a q) \\ \text { (d) } 2 \mathrm{NO}_{3}^{-}(a q)+8 \mathrm{H}^{+}(a q)+3 \mathrm{Cu}(s) \longrightarrow \\ 2 \mathrm{NO}(g)+4 \mathrm{H}_{2} \mathrm{O}(l)+3 \mathrm{Cu}^{2+}(a q) \end{array} $$

For each given reaction, we need to identify the reduction and oxidation half-reactions. (a) Cl2(g) + 2I^-(aq) -> 2Cl^-(aq) + I2(s) Reduction half-reaction: Cl2(g) + 2e^- -> 2Cl^-(aq) Oxidation half-reaction: 2I^-(aq) -> I2(s) + 2e^- (b) Ni(s)+2Ce^4+(aq) ->Ni^2+(aq)+2Ce^3+(aq) Reduction half-reaction: 2Ce^4+(aq) + 2e^- -> 2Ce^3+(aq) Oxidation half-reaction: Ni(s) -> Ni^2+(aq) + 2e^- (c) Fe(s)+2Fe^3+(aq) -> 3Fe^2+(aq) Reduction half-reaction: 2Fe^3+(aq) + 6e^- -> 2Fe^2+(aq) Oxidation half-reaction: 4Fe(s) -> 4Fe^2+(aq) + 8e^- (d) 2NO3^-(aq) + 8H^+(aq) + 3Cu(s) -> 2NO(g)+4H2O(l)+3Cu^2+(aq) Reduction half-reaction: 2NO3^-(aq) + 12H^+(aq) + 10e^- -> 2NO(g) + 6H2O(l) Oxidation half-reaction: 3Cu(s) -> 3Cu^2+(aq) + 6e^-

Using Appendix E, we look up the standard reduction potentials (E°) for the half-reactions. (a) E°(reduction) = +1.36 V (for Cl2 + 2e^- -> 2Cl^-) E°(oxidation) = +0.54 V (for I2 + 2e^- -> 2I^-) (b) E°(reduction) = -1.61 V (for Ce^4+ + e^- -> Ce^3+) E°(oxidation) = -0.24 V (for Ni^2+ + 2e^- -> Ni) (c) E°(reduction) = +0.77 V (for Fe^3+ + 3e^- -> Fe) E°(oxidation) = -0.44 V (for Fe^2+ + 2e^- -> Fe) (d) E°(reduction) = +0.96 V (for NO3^- + 6H^+ + 3e^- -> NO + 3H2O) E°(oxidation) = +0.16 V (for Cu^2+ + 2e^- -> Cu)

To calculate the standard emf (E°) for each reaction, subtract the standard reduction potential of the oxidation half-reaction from the reduction half-reaction. (a) E°(reaction) = E°(reduction) - E°(oxidation) = 1.36 V - 0.54 V = +0.82 V (b) E°(reaction) = E°(reduction) - E°(oxidation) = -1.61 V - (-0.24 V) = -1.37 V (c) E°(reaction) = E°(reduction) - E°(oxidation) = 0.77 V - (-0.44 V) = +1.21 V (d) E°(reaction) = E°(reduction) - E°(oxidation) = 0.96 V - 0.16 V = +0.80 V So, the standard emfs for the given reactions are: (a) +0.82 V (b) -1.37 V (c) +1.21 V (d) +0.80 V