Using Standard Reduction Potentials to Calculate an Equilibrium Constant Use the standard reduction potentials in Table 19.1 to calculate the equilibrium constant at 25 °C for the reaction 6 Br^-(aq) + Cr2O7^2-(aq) + 14 H^+(aq) → 3 Br2(aq) + 2 Cr^3+(aq) + 7 H2O(l)

Question

Using Standard Reduction Potentials to Calculate an Equilibrium Constant
Use the standard reduction potentials in Table 19.1 to calculate the equilibrium constant at 25 °C for the reaction
[latex]6 Br^{-}(aq) + Cr_{2}O^{2-}_{7} (aq) + 14 H^{+}(aq) → 3 Br_{2}(aq) + 2 Cr^{3+}(aq) + 7 H_{2}O(l)[/latex]

STRATEGY
Calculate E° for the reaction from standard reduction potentials, as in Worked Example 19.7. Then use the equation log K = nE°/0.0592 V to determine the equilibrium constant.

Accepted Answer

Find the relevant half-reactions in Table 19.1, and write them in the proper direction for the oxidation of [latex]Br^{-}[/latex] and reduction of [latex]Cr_{2}O^{2-}_{7}[/latex]. Before adding the half-reactions to get the overall reaction, multiply the [latex]Br^{-}/Br_{2}[/latex] half-reaction by a factor of 3 so that the electrons will cancel:

[latex]\begin{array}{rc} \begin{matrix} 3 imes [2 Br^{-}(aq) ightarrow Br_{2}(aq) + 2 e^{-}] \ Cr_{2}O^{2-}_{7}(aq) + 14 H^{+}(aq) + 6 e^{-} ightarrow 2 Cr^{3+}(aq) + 7 H_{2}O(l)\end{matrix} & \begin{matrix} E° = -1.09 V \ E° = 1.36 V \ \end{matrix} \ \hline \begin{matrix} 6 Br^{-}(aq)+ Cr_{2}O^{2-}_{7}(aq) + 14 H^{+}(aq) → 3 Br_{2}(aq) + 2 Cr^{3+}(aq) + 7 H_{2}O(l)\ \end{matrix} & \begin{matrix} E° = 0.27 V \ \end{matrix}\end{array}[/latex]

Note that E° for the [latex]Br^{-}/Br_{2}[/latex] oxidation is the negative of the tabulated standard reduction potential (1.09 V), and remember that we don’t multiply this E° value by a factor of 3 because electrical potential is an intensive property. The E° value for the overall reaction is the sum of the E° values for the half-reactions: -1.09 V + 1.36 V = 0.27 V. To calculate the equilibrium constant, use the relation between log K and nE°, with n = 6:

[latex] \log K =\frac{nE°}{0.0592 V}=\frac{(6)(0.27 V)}{0.0592 V}= 27 K= 1×10^{27} at 25°C [/latex]

CHECK
E° is positive, so K should be greater than 1, in agreement with the solution.

Known	Unknown
Standard reduction potentials (E°) for half-reactions	Equilibrium constant (K)

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
	$Ag^{+}(aq) + e^{-} → Ag(s)$	0.8
	$Fe^{3+}(aq) + e^{-} → Fe^{2+}(aq)$	0.77
	$O_{2}(g) + 2 H^{+}(aq) + 2 e^{-} → H_{2}O_{2}(aq)$	0.7
	$I_{2}(s) + 2 e^{-} → 2 I^{-}(aq)$	0.54
	$O_{2}(g) + 2 H_{2}O(l) + 4 e^{-} → 4 OH^{-}(aq)$	0.4
	$Cu^{2+}(aq) + 2 e^{-} → Cu(s)$	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
	$Cd^{2+}(aq) + 2 e^{-} → Cd(s)$	-0.4
	$Fe^{2+}(aq) + 2 e^{-} → Fe(s)$	-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 19.12: Using Standard Reduction Potentials to Calculate an Equilibr...

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