In the calibration of an ammeter using faraday's laws of electrolysis, the ammeter reading is kept constant at 1.2A. If 0.990g of copper is deposited in 40 minutes , the correction to be applied to the ammeter is

A.

0.03A

B.

0.04A

C.

0.05A

D.

0.06A

Correct answer is C

From faraday's law of electrolysis

M = I t Z
0.990 = I x (40 x 60) x 3.3 x 10-4

\( \implies \frac{0.990}{2400 \times 3.3 \times 10^{-4}} = 1.25A \\
\text{Thus the correction to be made } = 1.25 - 1.20 = 0.05A \)