Question

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

$E_{red}vs.log\left[C{u}^{2+}\right]$ graph is of the type as shown in figure where $OA=0.34V$, then 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. $0.34+0.0591V$
• C. $0.34V$
• D. none of these

Answer 1

Answer: (A)

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

According to Nernst equation:
$E=E^o-\frac{0.059}{n}logQ$
$E_{Cu/C{u}^{2+}}=E_{Cu/C{u}^{2+}}^o-\frac{0.059}{2}log\left[C{u}^{2+}\right]$
If $log\left[C{u}^{2+}\right]=0$, i.e., $\left[C{u}^{2+}\right]=1$,
then $E_{Cu/C{u}^{2+}}=E_{Cu/C{u}^{2+}}^o$
Also,
$E_{Cu/C{u}^{2+}}^o=-E_{C{u}^{2+}/Cu}^o=-0.34$
So, for oxidation cell, emf is
$E_{Cu/C{u}^{2+}}=-0.34-\frac{0.059}{2}log0.1$
$=-0.34+\frac{0.059}{2}$ (as n is 2 here)