Question

Which of the following does not evolve oxygen at anode when electroysis is carried?

Given:
$E_{Cu/C{u}^{2+}}^0=-0.34V$
$E_{H_{2}O/O_2}^0=-1.23V$

• A. $dil.H_{2}S{O}_4$ with Pt electrode
• B. Fused $NaOH$ with $Pt$ electrode
• C. Acidic water with $Pt$ electrode
• D. $dil.H_{2}S{O}_4$ with $Cu$ electrode

Answer 1

The correct option is D $dil.H_{2}S{O}_4$ with $Cu$ electrode

In (a), (b) and (c) $Pt$ electrode is used. It is an inert electrode.
Inert electrode are metal electrode in which, metal doesn’t involve in the redox reaction of the cell.
Eg. $Pt$ and graphite electrodes.
Here, electrode materials can conduct electrons into or out of the cell but cannot take part in the half-reactions.
In such cell reactions, no reactants or products are capable of serving as electrodes.
The species involved in the half reactions cannot act as electrodes.

In option (d), copper electrode is used. It is an active electrode.
Active electrode is a metal electrodes in which metals themselves are components of the half reactions.
Eg. $Cu,Zn,Ag,Hg,etc.,$
In Daniell cell, active electrodes are used. Zinc rod at anodic half cell and copper rod at cathodic half cells are active electrodes.

For electrolysis of $dil.H_{2}S{O}_4$ with copper electrode,
$E_{Cu/C{u}^{2+}}^0=-0.34V$
$E_{H_{2}O/O_2}^0=-1.23V$
Comparing the oxidation potential, $Cu$ has higher tendency to get oxidised compared to $O{H}^{-}\left(aq\right)$

In the electrolysis of dil. $H_{2}S{O}_4$ using $Cu$ electrodes, the reaction at anode occurs is
$Cu\left(s\right)\to C{u}^{2+}\left(aq\right)+2e^{-}$
Here, the copper electrode is not inert and so participates in the reaction.

In all other reaction, inert electrode i.e. $Pt$ is used. Thus, $O{H}^{-}/H_{2}O$ migrate toward the anode and gets oxidised to evolve oxygen.
$2H_{2}O\left(l\right)\rightleftharpoons 4H^{+}\left(aq\right)+O_2\left(g\right)+4e^{-}$
$E_{H_{2}O/O_2}^0=-1.23V$

Thus, option (c) is correct.