Question

Calculate equilibrium constant for the reaction,
$Ni\left(s\right)+2A{g}^{+}\left(aq\right)\to N{i}^{2+}\left(aq\right)+2Ag\left(s\right)$ at ${25}^{\circ }C$
$E_{Ni}^{\circ }=-0.25V$ and $E_{Ag}^{\circ }=0.799V$.

Answer 1

$N{i}^{+2}+2e^{-}\to Ni$

$E_{Ni}^o=-0.25V$

$A{g}^{+}+e^{-}\to Ag$

$E_{Ni}^o=0.799$

$Ni\to N{i}^{+2}+2e^{-}$ $0.25V$

$2A{g}^{+}+2e^{-}\to 2Ag$ $0.799\times 2V$

$Ni+2A{g}^{+}\to N{i}^{+2}+2Ag$ $1.598+0.250=1.848v$

$E_{cell}^o=\frac{0.059}{n}10gk$

$1.848=\frac{0.059}{2}10gk$

$\frac{1.848\times 2}{0.059}=10gk$

$10gk=62.644$

$k={10}^{62.644}=4.405\times {10}^{62}$