Question

Zn+ag2o+h2o gives zn2+ +2ag+2oh- find e0 of reaction

Answer 1

given redox reaction,

$Zn+Ag_2o\rightleftharpoons Zn^{2+}+2Ag+2OH^-$

here, half oxidation reaction : $Zn\rightleftharpoons Zn^{2+}+2e^-$

half reduction reaction : $Ag^{1+}\rightleftharpoons Ag-e^-$

we know, $\bf{E_{cell}=E_{reduction}-E_{oxidation}}$

from Nernst's equation,

$E_{Ag^+/Ag}=E^0_{Ag^+/Ag}-\frac{RT}{2F}ln\frac{1}{[Ag^+]}$

similarly, $E_{Zn/Zn^{2+}}=E^0_{Zn/Zn^{2+}}-\frac{RT}{2F}ln\frac{1}{[Zn^{2+}]}$

so, $E_{cell}=E_{Ag+/Ag}-E_{Zn/Zn^{2+}}$

= $E^0_{Ag^+/Ag}-\frac{RT}{2F}ln\frac{1}{[Ag^+]}-E^0_{Zn/Zn^{2+}}+\frac{RT}{2F}ln\frac{1}{[Zn^{2+}]}$

= $E^0_{cell}-\frac{RT}{2F}ln\frac{[Zn^{2+}]}{[Ag^+]}$