1.

Cu^(2+)+2e^(-) rarr Cu. For this, graph between E_(red) versus ln[Cu^(2+)] is a straight line of intercept 0.34V, then the electrode oxidation potential of the half cell Cu|Cu^(2+)(0.1M) will be

Answer»

`-0.34+(0.0591)/(2)V`
`0.34+0.0591" V"`
`0.34" V"`
NONE of these.

SOLUTION :(a) `E_(Cu//Cu^(2+))=E_(Cu//Cu^(2+))^(@)-(0.0591)/(2)LOG[Cu^(2+)]`
It `log[Cu^(2+)]=0`, then `Cu^(2+)=1`
`E_(Cu//Cu^(2+))=E_(Cu//Cu^(2+))^(@)`
`:. OA=E_(Cu//Cu^(2+))^(@)=-E_(Cu//Cu^(2+))^(@)=-0.34" V"`
Hence `E_(Cu//Cu^(2+))=-0.34-(0.0591)/(2)log 0.1=-0.34+(0.0591)/(2)`


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