Question

The above fuel cell is used completely as an electrolytic cell with Cu voltameter of resistance 26.94 $\Omega$ using Pt electrodes. Initially Cu voltameter contains 1 litre solution of 0.5 M CuSO${}_4$ [H${}^{+}$] in solution after electrolysis (Assuming no change in volume of solution).

• A. 0.015 M
• B. 0.03 M
• C. 0.025 M
• D. 0.01 M

Answer 1

The correct option is A 0.015 M

$i=\frac{V}{R}=\frac{1.3}{26.94}=0.04825A$

$\text{time of electrolysis}=\frac{5}{10\times {10}^{-3}}=\text{500 minutes}.$

$C{u}^{2+}\left(aq\right)+H_{2}O\to C{u}^{+}\left(aq\right)+\frac{1}{2}{O}_2+2H^{+}\left(aq\right)$

$\therefore \text{Moles of}{H}^{+}formed=\frac{i\times t}{F}=\frac{0.04825\times 500\times 60}{96500}=0.015moles$