18-12 Consider the following oxidation/reduc tion reactions: 2H^++Sn(s)arrow H_(2)(g)+Sn^2+ Ag^++Fc^2+arrow Ag(s)+Fc^3+ Sn^4++H_(2)(g)arrow Sn^2++2H^+ 2Fc^3++Sn^2+arrow 2Fc^2++Sn^4+ Sn^2++Co(s)arrow Sn(s)+Co^2+ (a) Write each net process in terms of two balanced half-reactions. (b) Express cach half-reaction as a reduction. (c) Arrange the half -reactions in (b) in order of de- creasing effectiveness as electron acceptors.

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18-12 Consider the following oxidation/reduc tion reactions: 2H^++Sn(s)arrow H_(2)(g)+Sn^2+ Ag^++Fc^2+arrow Ag(s)+Fc^3+ Sn^4++H_(2)(g)arrow Sn^2++2H^+ 2Fc^3++Sn^2+arrow 2Fc^2++Sn^4+ Sn^2++Co(s)arrow Sn(s)+Co^2+ (a) Write each net process in terms of two balanced half-reactions. (b) Express cach half-reaction as a reduction. (c) Arrange the half -reactions in (b) in order of de- creasing effectiveness as electron acceptors.

Answer 1

(a) To write each net process in terms of two balanced half-reactions, we need to identify the oxidation and reduction reactions in each equation. 1. $2H^{+}+Sn(s)\rightarrow H_{2}(g)+Sn^{2+}$ Oxidation: Sn(s) → Sn^{2+} + 2e^- Reduction: ^+ + 2e^- → H_{2}(g) 2. $Ag^{+}+Fc^{2+}\rightarrow Ag(s)+Fc^{3+}$ Oxidation: Fc^{2+} → Fc^{3+} + e^- Reduction: Ag^+ + e^- → Ag(s) 3. $Sn^{4+}+H_{2}(g)\rightarrow Sn^{2+}+2H^{+}$ Oxidation: H_{2}(g) → 2H^+ + 2e^- Reduction: Sn^{4+} + 2e^- → Sn^{2+} 4. $2Fc^{3+}+Sn^{2+}\rightarrow 2Fc^{2+}+Sn^{4+}$ Oxidation: Sn^{2+} → Sn^{4+} + 2e^- Reduction: 2Fc^{3+} + 2e^- → 2Fc^{2+} 5. $Sn^{2+}+Co(s)\rightarrow Sn(s)+Co^{2+}$ Oxidation: Sn^{2+} → Sn(s) + 2e^- Reduction: Co(s) → Co^{2+} + 2e^- (b) To express each half-reaction as a reduction, we need to reverse the direction of the oxidation reactions. 1. $2H^{+}+Sn(s)\rightarrow H_{2}(g)+Sn^{2+}$ Oxidation: Sn(s) → Sn^{2+} + 2e^- Reduction: 2H^+ + 2e^- → H_{2}(g) 2. $Ag^{+}+Fc^{2+}\rightarrow Ag(s)+Fc^{3+}$ Oxidation: Fc^{2+} → Fc^{3+} + e^- Reduction: Ag^+ + e^- → Ag(s) 3. $Sn^{4+}+H_{2}(g)\rightarrow Sn^{2+}+2H^{+}$ Oxidation: H_{2}(g) → 2H^+ + 2e^- Reduction: Sn^{4+} + 2e^- → Sn^{2+} 4. $2Fc^{3+}+Sn^{2+}\rightarrow 2Fc^{2+}+Sn^{4+}$ Oxidation: Sn^{2+} → Sn^{4+} + 2e^- Reduction: 2Fc^{3+} + 2e^- → 2Fc^{2+} 5. $Sn^{2+}+Co(s)\rightarrow Sn(s)+Co^{2+}$ Oxidation: Sn^{2+} → Sn(s) + 2e^- Reduction: Co(s) → Co^{2+} + 2e^- (c) To arrange the half-reactions in order of decreasing effectiveness as electron acceptors, we need to consider the standard reduction potentials (E°) of the half-reactions. The higher the standard reduction potential, the more effective the species is at accepting electrons. Here are the standard reduction potentials for the half-reactions: 1. Ag^+ + e^- → Ag(s) E° = +0.80 V 2. Fc^{3+} + e^- → Fc^{2+} E° = +0.77 V 3. Sn^{4+} + 2e^- → Sn^{2+} E° = +0.15 V 4. Co^{2+} + 2e^- → Co(s) E° = -0.28 V 5. H^+ + e^- → H_{2}(g) E° = 0.00 V Based on the standard reduction potentials, the order of decreasing effectiveness as electron acceptors is: 1. Ag^+ + e^- → Ag(s) 2. Fc^{3+} + e^- → Fc^{2+} 3. Sn^{4+} + 2e^- → Sn^{2+} 4. Co^{2+} + 2e^- → Co(s) 5. H^+ + e^- → H_{2}(g)

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