Using the Nernst Equation to Calculate the Cell Potential under Nonstandard-State Conditions Consider a galvanic cell that uses the reaction Zn(s) + 2 H^+(aq) → Zn^2+(aq) + H2(g) Calculate the cell potential at 25 °C when [H^+] = 1.0 M, [Zn^2+] = 0.0010 M, and PH2 = 0.10 atm.

Question

Using the Nernst Equation to Calculate the Cell Potential under Nonstandard-State Conditions
Consider a galvanic cell that uses the reaction

[latex]Zn(s) + 2 H^{+}(aq) → Zn^{2+}(aq) + H_{2}(g)[/latex]

Calculate the cell potential at 25 °C when [latex][H^{+}] = 1.0 M, [Zn^{2+}] = 0.0010 M,[/latex] and [latex]P_{H_{2}} = 0.10 atm.[/latex]

STRATEGY
We can calculate the standard cell potential E° from the standard reduction potentials in Table 19.1 and then use the Nernst equation to find the cell potential E under the cited conditions.

Accepted Answer

The standard cell potential is

[latex]E° = E°_{Zn→Zn^{2+}} + E°_{H^{+}→ H_{2}} = (0.76 V) + 0 V = 0.76 V[/latex]

The cell potential at 25 °C under nonstandard-state conditions is given by the Nernst equation:

[latex]E = E° -\frac{0.0592 V}{n}\log Q = E° -(\frac{0.0592 V}{n})(\log \frac{[Zn^{2+}](P_{H_{2}}) }{[H^{+}]^{2}})[/latex]

where the reaction quotient contains both molar concentrations of solutes and the partial pressure of a gas (in atm). As usual, zinc has been omitted from the reaction quotient because it is a pure solid. For this reaction, 2 mol of electrons are transferred, so n = 2. Substituting into the Nernst equation the appropriate values of E°, n, [latex][H^{+}] , [Zn^{2+}], and P_{H_{2}}[/latex] gives

[latex]E = (0.76 V) -(\frac{0.0592 V}{2})(\log\frac{(0.0010)(0.10)}{(1.0)^{2}})[/latex]

[latex]\= (0.76 V) -(\frac{0.0592 V}{2})(-4.0)[/latex]

[latex]\= 0.76 V + 0.12 V = 0.88 V at 25 °C[/latex]

CHECK
We expect that the reaction will have a greater tendency to occur under the cited conditions than under standard-state conditions because the product concentrations are lower than standard-state values. We therefore predict that the cell potential E will be greater than the standard cell potential E°, in agreement with the solution.

Known	Unknown
Concentrations $([H^{+}] = 1.0 M, [Zn^{2+}] = 0.0010 M, P_{H_{2}} = 0.10 atm)$	E
Standard reduction potentials (Table 19.1)

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

Question 19.9: Using the Nernst Equation to Calculate the Cell Potential un...

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