Question

Value of equilibrium constant for the reaction,
$Cu^{2+}_{\left(aq\right)} +Sn^{2+}_{\left(aq\right)} \to Sn^{4+}_{\left(aq\right)} +Cu_{\left(s\right)} $ is (Given $E^{\circ}_{Cu^{2+}/Cu} = 0.34V, E^{\circ}_{Sn^{4+}/Sn^{2+}} =0.15V$)

• A. $1.64 \times 10^{3}$
• B. $3.82 \times 10^{16}$
• C. $2.69 \times 10^{6}$
• D. $7.23 \times 10^{12}$

Answer 1

Answer: (C)

$E^{\circ}_{\text{cell}} =E^{\circ}_{\text{cathode}} -E^{\circ}_{\text{anode}} = 0.34 -0.15 =0.19 V$
Applying, $E^{\circ}_{\text{cell}} = \frac{0.0591}{n}$ log $K$
log $K_{c} = \frac{0.19 \times2}{0.0591} =6.43$
Taking antilog, $K_{c} = 2.69 \times10^{6}$