Question

During electrolysis of $CuS{O}_4$, using $Pt$-electrodes, the $pH$ of solution

• A. Increases
• B. Decreases
• C. Remains unchanged
• D. Can’t be predicted

Answer 1

The correct option is B Decreases

In the presence of inert electrode, the cell reaction is :
$\text{Anode}:2H_{2}O\to 4H^{+}+4e^{–}+O_2$

$\text{Cathode}:\left\{C{u}^{2+}\right(aq.)+2e^{–}\to Cu(s\left)\right\}\times 2$

$\text{Net cell reaction:}2C{u}^{2+}(aq.)+2H_{2}O\to 4H^{+}+2Cu\left(s\right)+O_2\uparrow$

$\therefore pH=–\text{log}\left[H^{+}\right]$

Due to increase in $\left[H^{+}\right]$, $pH$ decreases.
Therefore the correct option is B.