Question

During the electrolysis of 0.1 M $CuS{O}_4$ solution using copper electrodes, a depletion of [$C{u}^{2+}$] occurs near the cathode with a corresponding excess near the anode, owing to inefficient stirring of the solution. If the concentrations of [$C{u}^{2+}$] near the anode & cathode are respectively 0.12 M & 0.08 M, calculate the back e.m.f. developed. Temperature = 298 K.

• A. 22 mV
• B. 5.2 mV
• C. 29 mV
• D. 59 mV

Answer 1

Answer: (B)

5.2 mV

Back emf = emf due to different concentrations of ions near cathode and anode.

using the relation

$E_{electrode}=E^0-\frac{0.059}{n}log\left(\frac{\left[product\right]}{\left[reactant\right]}\right)$

Reaction at cathode:

$C{u}^{2+}\left(\right(aq)+2e^{-}\to Cu(s)$

$E_{cathode}=E^0-\frac{0.059}{2}log\left(\frac{\left[Cu\right]}{\left[C{u}^{2+}\right]}\right)$

$E_{cathode}=-\frac{0.059}{2}log\left(\frac{1}{0.08}\right)$

$E_{cathode}=-0.032V$

Anode reaction:

$Cu(s)\rightarrow Cu^{ 2+ }(aq)+2e^-$

$E_{anode}=E^0-\frac{0.059}{2}log\left(\frac{\left[C{u}^{2+}\right]}{\left[Cu\right]}\right)$

$E_{anode}=-\frac{0.059}{2}log\left(\frac{0.12}{1}\right)$

$E_{anode}=0.027V$

Since potential at cathode is not equal to that at anode the residual potential is the Back emf i.e.:

$\therefore E_b=|E_{cathode}|-|E_{anode}|$

$E_b=0.032-0.027$

$E_b=5\times {10}^{-3}V$

option B is correct