Question

The equilibrium constant of the reaction:
$Cu ( s )+2 Ag ^{+}( aq ) \rightarrow Cu ^{2+}( aq )+2 Ag ( s )$
$E^{\circ} =0.46\, V$ at $298\, K$ is:

• A. $2.0 \times 10^{10} $
• B. $4.0 \times 10^{10} $
• C. $4.0 \times 10^{15} $
• D. $2.4 \times 10^{10} $

Answer 1

Answer: (C)

$Cu(s)+2 Ag^{2+}(aq) \rightarrow Cu^{2+}(aq)+2 Ag(s) $
$E^{0}=0.46\, V$ at $298 \,K$
$R T \text{In} K=n F E^{0}$
In $K=\frac{n F E^{0}}{R T}$
In $K=\frac{2 \times 0.46}{0.0591}$
$K=4 \times 10^{15}$