Question 17.3: GRADED Consider the voltaic cell in which the reaction is 2A...

GRADED

Consider the voltaic cell in which the reaction is

2Ag^{+}(aq)  +  Cd(s)  →  2Ag(s)  +  Cd^{2+}(aq)

ⓐ Use Table 17.1 to calculate E° for the voltaic cell.

Table 17.1 Standard Potentials in Water Solution at 25°C

Lithium is the strongest reducing agent.               O = strongest oxidizing agent; R = strongest reducing agent.           Fluorine is the strongest oxidizing agent.   Lithium and fluorine are very dangerous materials to work with.	AcidicSolution,[H^{+}] = 1 M
	E°_{red}(V)
	-3.040 -2.936 -2.906 -2.869 -2.714 -2.357 -1.680 -1.182 -0.762 -0.744 -0.409 -0.408 -0.402 -0.356 -0.336 -0.282 -0.236 -0.152 -0.141 -0.127 0.000 0.073 0.144 0.154 0.155 0.161 0.339 0.518 0.534 0.769 0.796 0.799 0.908 0.964 1.001 1.077 1.229 1.229 1.330 1.360 1.458 1.498 1.512 1.687 1.763 1.953 2.889		→Li(s) → K(s) → Ba(s) → Ca(s) →Na(s) → Mg(s) → Al(s) → Mn(s) → Zn(s) → Cr(s) → Fe(s) → Cr^{2+}(aq) → Cd(s) → Pb(s) + SO_{4}^{2-}(aq) → Tl(s) → Co(s) →Ni(s) → Ag(s) + I^{-}(aq) → Sn(s) → Pb(s) →H_{2}(g) → Ag(s) + Br^{-}(aq) →H_{2}S(aq) → Sn^{2+}(aq) → SO_{2}(g)  +  2H_{2}O → Cu^{+}(aq) → Cu(s) → Cu(s) → 2I^{-}(aq) → Fe^{2+}(aq) → 2Hg(l) → Ag(s) →Hg_{2}^{2+}(aq) →NO(g) + 2H_{2}O → Au(s)  +  4Cl^{-}(aq) → 2Br^{-}(aq) → 2H_{2}O → Mn^{2+}(aq)  +  2H_{2}O → 2Cr^{3+}(aq)  +  7H_{2}O → 2Cl^{-}(aq) →\frac{1}{2} Cl_{2}(g)  +  3H_{2}O → Au(s) → Mn^{2+}(aq)  +  4H_{2}O → PbSO_{4}(s)  +  2H_{2}O → 2H_{2}O → Co^{2+}(aq) → 2F^{-}(aq)	Li^{+}(aq)  +  e^{-} K^{+}(aq)  +  e^{-} Ba^{2+}(aq)  +  2e^{-} Ca^{2+}(aq)  +  2e^{-} Na^{+}(aq) +  e^{-} Mg^{2+}(aq)  +  2e^{-} Al^{3+}(aq)  +  3e^{-} Mn^{2+}(aq)  +  2e^{-} Zn^{2+}(aq)  +  2e^{-} Cr^{3+}(aq)  +  3e^{-} Fe^{2+}(aq)  +  2e^{-} Cr^{3+}(aq)  +  e^{-} Cd^{2+}(aq)  +  2e^{-} PbSO_{4}(s)  +  2e^{-} Tl^{+}(aq)  +  e^{-} Co^{2+}(aq)  +  2e^{-} Ni^{2+}(aq)  +  2e^{-} AgI(s)  +  e^{-} Sn^{2+}(aq)  +  2e^{-} Pb^{2+}(aq)  +  2e^{-} 2H^{+}(aq)  +  2e^{-} AgBr(s)  +  e^{-} S(s)  +  2H^{+}(aq)  +  2e^{-} Sn^{4+}(aq)  +  2e^{-} SO_{4}^{2-}(aq)  +  4H^{+}(aq) +  2e^{-} Cu^{2+}(aq)  +  e^{-} Cu^{2+}(aq)  +  2e^{-} Cu^{+}(aq)  +  e^{-} I_{2}(s)  +  2e^{-} Fe^{3+}(aq)  +  e^{-} Hg_{2}^{2+}(aq)  +  2e^{-} Ag^{+}(aq)  +  e^{-} 2Hg^{2+}(aq)  +  2e^{-} NO_{3}^{-}(aq)  +  4H^{+}(aq) +  3e^{-} AuCl_{4}^{-}(aq)  +  3e^{-} Br_{2}(l)  +  2e^{-} O_{2}(g)  +  4H^{+}(aq)  +  4e^{-} MnO_{2}(s)  +  4H^{+}(aq)  +  2e^{-} Cr_{2}O_{7}^{2-}(aq)  +  14H^{+}(aq)  +  6e^{-} Cl_{2}(g)  +  2e^{-} ClO_{3}^{-}(aq)  +  6H^{+}(aq)  +  5e^{-} Au^{3+}(aq)  +  3e^{-} MnO_{4}^{-}(aq)  +  8H^{+}(aq) +  5e^{-} PbO_{2}(s)  +  SO_{4}^{2-}(aq)  +  4H^{+}(aq)  +  2e^{-} H_{2}O_{2}(aq)  +  2H^{+}(aq)  +  2e^{-} Co^{3+}(aq)  +  e^{-} F_{2}(g)  +  2e^{-}
	Basic Solution, [OH^{-}] = 1 M
	E°_{red}(V)
	-0.891 -0.828 -0.547 -0.445 -0.140 0.004 0.398 0.401 0.614 0.890	→ Fe(s) + 2 OH^{-}(aq) →H_{2}(g) + 2 OH^{-}(aq) → Fe(OH)_{2}(s) + OH^{-}(aq) → S^{2-}(aq) →NO(g) + 4 OH^{-}(aq) →NO_{2}^{-}(aq) + 2 OH^{-}(aq) → ClO_{3}^{-}(aq) + 2 OH^{-}(aq) → 4 OH^{-}(aq) → Cl^{-}(aq) + 6 OH^{-}(aq) → Cl^{-}(aq) + 2 OH^{-}(aq)		Fe(OH)_{2}(s)  +  2e^{-} 2H_{2}O  +  2e^{-} Fe(OH)_{3}(s)  +  e^{-} S(s)  +  2e^{-} NO_{3}^{-}(aq)  +  2H_{2}O  +  3e^{-} NO_{3}^{-}(aq)  +  H_{2}O +  2e^{-} ClO_{4}^{-}(aq)  +  H_{2}O  +  2e^{-} O_{2}(g)  +  2H_{2}O  +  4e^{-} ClO_{3}^{-}(aq)  +  3H_{2}O  +  6e^{-} ClO^{-}(aq)  +  H_{2}O  +  2e^{-}

ⓑ If the value zero is arbitrarily assigned to the standard voltage for the reduction of Ag^{+} ions to Ag, what is E°_{red} for the reduction of Cd^{2+} ions to Cd?

ⓐ

STRATEGY

1. Assign oxidation numbers to each element so you can decide which element is reduced and which one is oxidized.
2. Write the oxidation and reduction half-reactions together with the corresponding E°_{ox}  and E°_{red}. Recall that E°_{ox} = -(E°_{red}).
3. Add both half-reactions (make sure you cancel electrons) and take the sum of E°_{ox} and E°_{red} to obtain E° for the cell.