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.

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 \times [2 Br^{-}(aq) \rightarrow Br_{2}(aq) + 2 e^{-}] \\ Cr_{2}O^{2-}_{7}(aq) + 14 H^{+}(aq) + 6 e^{-} \rightarrow 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.

Question 19.12: Using Standard Reduction Potentials to Calculate an Equilibr...

Chemistry

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
$6 Br^{-}(aq) + Cr_{2}O^{2-}_{7} (aq) + 14 H^{+}(aq) → 3 Br_{2}(aq) + 2 Cr^{3+}(aq) + 7 H_{2}O(l)$

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.

IDENTIFY

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

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

\begin{array}{rc} \begin{matrix} 3 \times [2  Br^{-}(aq) \rightarrow Br_{2}(aq) + 2  e^{-}] \\ Cr_{2}O^{2-}_{7}(aq) + 14  H^{+}(aq) + 6  e^{-} \rightarrow 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}

Note that E° for the $Br^{-}/Br_{2}$ 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: