Calculate the mass of copper deposited if a current of 0.45A flows through \(CuSO_{4}\) solution for 1hour 15mins. [Cu=64.0, S= 32.0, O =16.0, 1F= 96500C]
- 6.40g
-
0.67g
- 0.64g
- 0.45g
The correct answer is: B
Explanation
\(Cu^{2+} + 2e^{-} \to Cu_{(s)}\)
\( 2 \times 96500C = 193000C deposits 64g of Cu\)
\(M \propto It\)
\(Q = It = 0.45A \times 1hr15mins\)
= \( 0.45 \times 75 \times 60\) [converting the time to seconds]
Quantity of electricity passed = 2025C
\(193000C \to 64g Cu\)
\(1C \to \frac{64}{193000}\)
\(2025C \to \frac{64}{193000} \times 2025\)
\(= 0.6715g \approxeq 0.67g\)