Answer
$6.7\times10^{-6}$
Work Step by Step
The oxidation half-reaction at the anode is:
$Ni(s)\rightarrow Ni^{2+}(aq)+2e^{-}$
The reduction half-reaction at the cathode is:
$Cd^{2+}(aq)+2e^{-}\rightarrow Cd(s)$
$E^{\circ}_{cell}=E^{\circ}_{cathode}-E^{\circ}_{anode}$
$=-0.403\,V-(-0.25\,V)$
(values from table 17.1)
$=-0.153\,V$
Number of electrons transferred $n=2$
$T= (25+273)\,K=298\,K$
$\ln K^{\circ}=\frac{nFE^{\circ}_{cell}}{RT}$
$=\frac{2(96485\,C/mol)\times-0.153\,V}{(8.314\,J/mol\cdot K) \times298\,K}=-11.91667$
$K^{\circ}= e^{-11.91667}=6.7\times10^{-6}$
Because the reaction occurs in aqueous solution,
$K^{\circ}=K_{c}=6.7\times10^{-6}$
