Question

Calculate $E^{\circ }$ of the following half-cell reaction at $298K$:
$Ag(N{H}_3{)}_2^{+}+e^{-}\to Ag+2N{H}_3$
$A{g}^{+}+e^{-}\to Ag;E_{A{g}^{+}/Ag}^{\circ }=0.80V$
$Ag(N{H}_3{)}_2^{+}\rightleftharpoons A{g}^{+}+2N{H}_3;K=6\times {10}^{-8}$.

Answer 1

$AnodeAg\left(s\right)\to A{g}^{+}+e^{-};E^{\circ }=0.80volt$
$\underset{–}{Cathode\left[Ag\right(N{H}_3{)}_2{]}^{+}+e^{-}\to Ag\left(s\right)+2N{H}_3;E^{\circ }=Vvolt}$
$\underset{–}{Ag\left(s\right)+\left[Ag\right(N{H}_3)2{]}^{+}\rightleftharpoons Ag(s)+A{g}^{+}+2N{H}_3}$
$Q=\frac{\left[A{g}^{+}\right][N{H}_3{]}^2}{\left[Ag\right(N{H}_3{)}_2{]}^{+}}=6\times {10}^{-8(n=1)}$
$E_{cell}^{\circ }=(V-0.80)$
At equilibrium, $E=0$
$\therefore E=E^{\circ }-\frac{0.0591}{n}\log Q$
$0=(V-0.80)-\frac{0.0591}{1}\log (6\times {10}^{-8})$
$V=0.373volt$.