Question

A current of strength $2.5$ amp was passed through $CuSO _{4}$ solution for 6 minutes $26$ seconds. The amount of copper deposited is : (At. wt. of $Cu =63.5,1\, F =96500$ coulomb $)$

• A. 0.3175 g
• B. 3.175 g
• C. 0.635 g
• D. 6.35 g

Answer 1

Answer: (A)

0.3175 g

Answer 2

Number of coulombs passed $=i \times t$ $t=6 \times 60+26=360+26$ $=386$ seconds $=2.5 \times 386=965$ $2 \times 96500 C$ charge deposits $=63.5 g$ $1\, C$ charge deposits $=\frac{63.5}{2 \times 96500}$ $965\, C$ charge deposits $=\frac{63.5 \times 965}{2 \times 96500}$ $=0.3175\, g$