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The solution of CuSO(4) in which Cu cord...

The solution of `CuSO_(4)` in which `Cu` cord is dipped is diluted to 10 times, the reduction electrode potential will `:`
`a.` Decrease by `0.03V" "b.` Decrease by `0.059V`
`c.` Increase by `0.03 V" "d.` Increase by `-.059V`

A

increases by `0.030` volt

B

decreases by `0.030` volt

C

increases by `0.059` volt

D

decreases by `0.0059` volt

Text Solution

Verified by Experts

The correct Answer is:
B
`Cu^(2+)+2e^(-)rarr Cu` `{:(V_(1)=1, C_(1)=1M),(V_(2)=10,C_(2)=0.1 M):}`
`E_("cell")=E^(@)-(0.591)/(2)"log"(1)/(1)` `E_("cell")=E^(@)`
`E_("cell")=E^(@)-(0.0591)/(2)"log"[(1)/(0.1)]`
`= E^(@)-0.059//2 = E^(@)-0.030`
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