Question

Given the half - cell reactions,

$Cu^{+}(aq) + e^{-} \rightarrow Cu(s),E_{1}^{0} = + 0.52V$; $Cu^{2 +}(aq) + e^{-} \rightarrow \overset{+}{Cu}(aq),E_{2}^{0} = + 0.16V$

the equilibrium constant for the disproportionation reaction

$2Cu^{+}(aq) \rightarrow Cu(s) + Cu^{2 +}(aq)\text{at}298K$is

• A. $6 \times 10^{4}$
• B. $6 \times 10^{6}$
• C. $1.2 \times 10^{6}$
• D. $1.2 \times 10^{- 6}$

Answer 1

Answer: (C)

$1.2 \times 10^{6}$

Answer 2

$E^{0} = E_{1}^{0} - E_{2}^{0} = 0.52 - 0.16 = 0.36V$

$\log K_{eq} = \frac{nE^{0}}{0.0592} = \frac{1 \times 0.36}{0.0592} = 6.081$, $\therefore K_{eq} = 1.20 \times 10^{6}$