Question

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

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

Answer 1

Answer: (A)

$Ag^{+}+e^{-} \to Ag$
$E_{cell}=E^{\circ}_{cell}-\frac{0.0591}{1} log \frac{1}{\left[Ag^{+}\right]}$
$E_{cell}=0.80-\frac{0.0591}{1} log \frac{1}{0.1}$
$=0.80-0.0591$
$=0.741\,V$