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: C
Explanation
From faraday's law of electrolysis M = I t Z 0.990 = I x (40 x 60) x 3.3 x 10-4 \( \implies \frac{0.990}{2400 \times 3.3 \times 10^{-4}} = 1.25A \\ \text{Thus the correction to be made } = 1.25 - 1.20 = 0.05A \)
