Question

The standard reduction potential of $E_{B{i}^{3+}/Bi}^o$ and $E_{C{u}^{2+}/Cu}^o$ are 0.226 V and 0.344 V respectively. A mixture of salts of Bi and Cu at unit concentration each is electrolysed at 25${}^o$C. To what value can $\left[C{u}^{2+}\right]$ be brought down before bismuth starts to deposit during electrolysis?

• A. ${10}^{-4}M$
• B. ${10}^{-3}M$
• C. ${10}^{-5}M$
• D. ${10}^{-6}M$

Answer 1

The correct option is A ${10}^{-4}M$

We have,
Since, initially
$E_{C{u}^{2+}/Cu}=E_{C{u}^{2+}/Cu}^o+\frac{0.059}{2}log\left[C{u}^{2+}\right]$
$=0.344+\frac{0.059}{2}log1=0.344V$
$E_{B{i}^{3+}/Bi}=0.226+\frac{0.059}{3}log\left[B{i}^{3+}\right]$
$=0.226+\frac{0.059}{3}log1=0.226V$
Thus, passage of current would initially deposit $C{u}^{2+}$ till $E_{C{u}^{2+}/Cu}$
becomes 0.226 V because then only $B{i}^{3+}$ will be deposited.
Thus, $E_{C{u}^{2+}/Cu}=E_{C{u}^{2+}/Cu}^o+\frac{0.059}{2}log\left[C{u}^{2+}\right]$
$0.226=0.344+\frac{0.059}{2}log\left[C{u}^{2+}\right]$
$\therefore \left[C{u}^{2+}\right]={10}^{-4}M$