MCQ
A current of $0.25\,A$ is passed through $CuS{O_4}$ solution placed in voltameter for $45 $ minutes. The amount of $Cu$ deposited on cathode is ............. $\mathrm{g}$ (At weight of $Cu = 63.6$)
- A$0.22$
- ✓$0.20$
- C$0.25$
- D$0.30$
✓
Answer
Correct option: B.
$0.20$
b
Given :-
Given :-
Time $=45 min =45 \times 60 sec =2700\,sec$
Current$=0.25\,A$
$C u^{2+}+2 e^{-} \rightarrow C u$
At.weight of $Cu =63.6$ Electricity passed $=45 \times 60 \times 0.25=675\, C$ electricity deposit copper $=x \times 2 \times 96500\, C$ of electricity will $M=Z \times I \times T$
$x=0.20 \,g$
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