Question

$C{u}^{2+}+2e^{-}\to Cu$. for this graph between $n\left[C{u}^{2+}\right]$ versus $E_{red}$ is a straight line of intercept $-\frac{0.34\times 4}{0.0591}$ then he electrode potential of the half-cell $Cu/C{u}^{2+}\left(0.1M\right)$ will be:

• A. $0.34+\frac{0.0591}{2}$
• B. $-0.34-\frac{0.0591}{2}$
• C. $0.34$
• D. $-0.34+\frac{0.0591}{2}$

Answer 1

Answer: (D)

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

Reactions:
$C{u}^{2+}+2e^{-}\to Cu$
$E_{C{u}^{2+}|Cu}=E_{C{u}^{2+}|Cu}^{\ominus }-\frac{0.059}{2}log\frac{1}{\left[C{u}^{2+}\right]}$
$=E_{C{u}^{2+}|Cu}^{\ominus }-\frac{RT}{2F}ln\left[C{u}^{2+}\right]$
Intercept = 0.34 $\Rightarrow E_{C{u}^{2+}|Cu}^{\ominus }=0.34V$
$\Rightarrow E_{C{u}^{2+}|Cu}=0.34+\frac{0.059}{2}log0.1=0.31V$
$\Rightarrow E_{Cu|C{u}^{2+}}=-E_{C{u}^{2+}|Cu}=-0.34+\frac{0.059}{2}V$