Question

Consider the reaction Zn(s)+2Ag^ + (aq) —-> —-<Zn^ 2+ (aq)+2Ag(s) . If the E ^ 0 = 1.562V and the temperature is 310 K, what is the value of E if the [Ag^ + ] is 0.55 M and the [Zn^ 2+ ] is 1.05 M ?

Answer 1

$E=E_o-\frac{0.06}{n}\frac{[Zn^{2+}]}{[Ag^+]^2}$

$E=1.562-0.03\times\frac{1.05}{(0.55)^2}=1.457 V$