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The electrode pptenticals for Cu^(2+)...

The electrode pptenticals for
` Cu^(2+) (aq) +e^(-) rarr Cu^+ (aq) `
and ` Cu^+ (aq) + e^- rarr Cu (s)`
are `+ 0.15 V` and ` +0. 50 V` repectively. The value of `E_(cu^(2+)//Cu)^@` will be.

A

`0.500" V "`

B

`0.325" V "`

C

`0.650" V "`

D

`0.150" V "`

Text Solution

Verified by Experts

The correct Answer is:
B
(b) `Cu^(2+)(aq)+e^(-) to Cu^(+)(aq),E_(1)^(@)=0.15" V "`
`Cu^(+)(aq)+e^(-) to Cu(s),E_(2)^(@)=0.50" V "`
`Cu_((aq))^(2+)+2e^(-) to Cu(s)`
`DeltaG^(@)=DeltaG_(1)^(@)+DeltaG_(2)^(@)`
`-nFE^(@)=-n_(1)FE_(1)^(@)-n_(2)FE_(2)^(@)`
or `E^(@)=(n_(1)E_(1)^(@)+n_(2)E_(2)^(@))/(n)=(1xx0.15+1xx(0.50))/(2)`
`=(0.15+0.50)/(2)=(0.65)/(2)`
`=0.325" V "`
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