Question

Calculate the reduction potential for the following half cell at ${25}^{\circ }C$:
$Ag\left(s\right)|A{g}^{+}\left({10}^{-5}M\right);E_{A{g}^{+}/Ag}^0=0.80V$

• A. $E_{A{g}^{+}/Ag}=1.5V$
• B. $E_{A{g}^{+}/Ag}=0.8V$
• C. $E_{A{g}^{+}/Ag}=0.5V$
• D. $E_{A{g}^{+}/Ag}=0.25V$

Answer 1

The correct option is C $E_{A{g}^{+}/Ag}=0.5V$

$Ag\left(s\right)|A{g}^{+}\left({10}^{-5}M\right);E_{A{g}^{+}/Ag}^0=0.80V$


The half call electrode reaction is

$A{g}^{+}\left(aq\right)+e^{-}\to Ag\left(s\right)$

Nernst equation for the half cell is

$E_{A{g}^{+}/Ag}=E_{A{g}^{+}/Ag}^0-\frac{2.303RT}{nF}log\frac{\left[M\right(s\left)\right]}{\left[M^{n+}\right(aq\left)\right]}$

n is the number of electrons involved in the half cell reaction.
$\therefore$
n = 1

$E_{A{g}^{+}/Ag}=0.80-\frac{0.059}{1}log\frac{1}{\left[A{g}^{+}\right]}$

$E_{A{g}^{+}/Ag}=0.80-\frac{0.059}{1}log\left(1/{10}^{-5}\right)$

$E_{A{g}^{+}/Ag}=0.80+\frac{0.059}{1}log\left({10}^{-5}\right)$

$E_{A{g}^{+}/Ag}=0.80-5\times 0.059=0.50V$