Question 19.8: Calculating the Voltage of a Galvanic Cell under Standard-St...

Chemistry

Calculating the Voltage of a Galvanic Cell under Standard-State Conditions
What is the voltage of the galvanic cell shown in the illustration? What is the anode half-reaction, and what is the cathode half-reaction? Which direction do the electrons flow?

STRATEGY
In a galvanic cell, there must be a reduction half-reaction and an oxidation half reaction whose values of E° sum to give a positive value of E° for the cell. Any spontaneous redox reaction must have a positive value of E°. Use Table 19.1 to look up the values of E° for half-reactions and then determine which half-reaction is the reduction and which is the oxidation.

TABLE 19.1 Standard Reduction Potentials at 25 °C
	Reduction Half-Reaction	E° (V)
	$F_{2}(g) + 2 e^{-} →2F^{-}(aq)$	2.87
	$H_{2}O_{2}(aq) + 2 H^{+}(aq) + 2 e^{-} → 2H_{2}O(l)$	1.78
	$MnO^{-}_{4}(aq) + 8 H^{+}(aq) + 5 e^{-}→ Mn^{2+}(aq) + 4 H_{2}O(l)$	1.51
	$Cl_{2}(g) + 2 e^{-} → 2Cl^{-}(aq)$	1.36
	$Cr_{2}O^{2-}_{7}(aq) + 14 H^{+}(aq) + 6 e^{-} → 2 Cr^{3+}(aq) + 7 H_{2}O(l)$	1.36
	$O_{2}(g) + 4 H^{+}(aq) + 4 e^{-} → 2 H_{2}O(l)$	1.23
	$Br_{2}(aq) + 2 e^{-} → 2 Br^{-}(aq)$	1.09
	$Ag^{+}(aq) + e^{-} → Ag(s)$	0.80
	$Fe^{3+}(aq) + e^{-} → Fe^{2+}(aq)$	0.77
	$O_{2}(g) + 2 H^{+}(aq) + 2 e^{-} → H_{2}O_{2}(aq)$	0.70
	$I_{2}(s) + 2 e^{-} → 2 I^{-}(aq)$	0.54
	$O_{2}(g) + 2 H_{2}O(l) + 4 e^{-} → 4 OH^{-}(aq)$	0.40
	$Cu^{2+}(aq) + 2 e^{-} → Cu(s)$	0.34
	$Sn^{4+}(aq) + 2 e^{-} → Sn^{2+}(aq)$	0.15
	$2 H^{+}(aq) + 2 e^{-} → H_{2}(g)$	0
	$Pb^{2+}(aq) + 2 e^{-} → Pb(s)$	-0.13
	$Ni^{2+}(aq) + 2 e^{-} → Ni(s)$	-0.26
	$Cd^{2+}(aq) + 2 e^{-} → Cd(s)$	-0.40
	$Fe^{2+}(aq) + 2 e^{-} → Fe(s)$	-0.45
	$Zn^{2+}(aq) + 2 e^{-} → Zn(s)$	-0.76
	$2 H_{2}O(l) + 2 e^{-} → H_{2}(g) + 2 OH^{-}(aq)$	-0.83
	$Al^{3+}(aq) + 3 e^{-} → Al(s)$	-1.66
	$Mg^{2+}(aq) + 2 e^{-} → Mg(s)$	-2.37
	$Na^{+}(aq) + e^{-} → Na(s)$	-2.71
	$Li^{+}(aq) + e^{-} → Li(s)$	-3.04

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Table 19.1 gives the standard reduction potentials for the two half-reactions we must consider.

$Cd^{2+}(aq) + 2 e^{-} → Cd(s)$ E° = -0.40 V
$Ag^{+}(aq) + e^{-} → Ag(s)$ E° = +0.80 V

One reaction must be reversed to an oxidation because any redox reaction must have one reduction half-reaction and one oxidation half-reaction. The value of E° for the galvanic cell must be positive; therefore, the oxidation half-reaction is the oxidation of $Cd(s) to Cd^{2+}(aq).$

\begin{array}{rc} \begin{matrix} \text{Anode  (oxidation):}    Cd(s) → Cd^{2+}(aq) + 2 e^{-} \\ \text{Cathode  (reduction): }   2 × [Ag^{+}(aq) + e^{-} → Ag(s)] \end{matrix} & \begin{matrix} E° = + 0.40 V \\ E° = + 0.80 V\\ \end{matrix} \\ \hline \begin{matrix} \text{Overall  cell  reaction:}    2 Ag^{+}(aq) + Cd(s) → 2 Ag(s) + Cd^{2+}(aq) \\ \end{matrix} & \begin{matrix} E° = + 1.20 V \\ \end{matrix}\end{array}

Note that the sign of E° was changed when the cadmium half-reaction was changed from a reduction (-0.40 V) to an oxidation (+0.40 V). Also, when the reduction half-reaction for silver was multiplied by the 2, the value of E° did not change. The value of E° for the overall cell reaction is the sum of E° values for the oxidation and reduction half-reactions, 0.80 V + 0.40 V = + 1.20 V. Electrons flow from the cadmium half-cell to the silver half-cell where $Ag^{+}(aq)$ is reduced to Ag(s).