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Q.

Time required for a current of 10 A to deposit 0.635 g of Cu from CuSO4 solution is about (Given 1F=96500Cmol−1,M(Cu)=63.5gmol−1

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a

220 s

b

193 s

c

249 s

d

181 s

answer is B.

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Detailed Solution

Since m=It M/Fve, we get

t=mFIveM=(0.635g)96500Cmol−110A263.5gmol−1=193s

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