Question

When a solution of ${CuSO_4}$ is electrolysed for 15 minutes with a current of 1 amp, then the mass of Cu deposited at the cathode will be

• A. 30 gm
• B. 0.3 gm
• C. 0.03 gm
• D. 3 gm

Answer 1

Answer: (B)

0.3 gm

Answer 2

$I = {1.0\, A}, t = {15\, min} = 15 \times 60 { = 900\, s}$
Quantity of electricity passed = $I \times t$
$= 1.0 \times 900$ coulombs
${= 900 \,C}$
The reaction occurring at the cathode is
${Cu^{2+} + 2e^- -> Cu}$
Thus, ${2F} i.e. 2 \times 96500 \, {C\, deposit\, Cu}$
${ = 1 \, mole = 63.5 \, g}$
and ${900\, C}$ will deposit ${Cu = \frac{63.5 }{2 \times 96500} \times 900}$
$= 0.26 \, {g}$
$\approx 0.3 \, {g}$