Question

For the reaction, $C{u}^{2+}+2e^{-}\to Cu$; log$\left[C{u}^{2+}\right]$ vs $E$ graph is of type as shown in figure where $OA=0.34$V, the electrode potential of the half-cell of $Cu|C{u}^{2+}\left(0.1M\right)$ will be:

• A. $-0.34+\frac{0.0591}{2}V$
• B. $0.34+0.0591$V
• C. $0.34$V
• D. none of these

Answer 1

Answer: (A)

$-0.34+\frac{0.0591}{2}V$

Given,

$E_{cell}=0.34V$

The reaction, $C{u}^{2+}+2e^{-}\to Cu$

$n=2$

We know that,

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

$⟹$ $0.34=E_{cell}^0-\frac{0.0591}{2}log\frac{1}{\left[C{u}^{2+}\right]}$

$⟹$ $E_{cell}^0=-0.34+\frac{0.0591}{2}V$

Hence, the correct answer is option $\text{A}$.