Question

What will be the reduction potential for the following half-cell reaction at $298$K?
(Given: $\left[A{g}^{+}\right]=0.1$M and $E_{cell}^o=+0.80$V)

• A. $0.741$V
• B. $0.80$V
• C. $-0.80$V
• D. $-0.741$V

Answer 1

The correct option is A $0.741$V

A galvanic cell or simple battery is made of two electrodes. Each of the electrodes of a galvanic cell is known as a half cell. In a battery, the two half cells form an oxidizing-reducing couple. When two half cells are connected via an electric conductor and salt bridge, an electrochemical reaction is started.

Nernst equation is,

$E_{cell}=E_{cell}^0-\frac{0.0591}{n}logQ$

Here the reaction is,

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

$n=1$

$E_{cell}^0=0.80V$

Hence,

$E_{cell}=E_{cell}^0-\frac{0.0591}{n}log\frac{1}{\left[A{g}^{+}\right]}$

$E_{cell}=0.80-\frac{0.0591}{1}log\frac{1}{0.1}$

$=0.741V$