The E^0cell of Pb/Pb^2+ is 0.13V and E^0cell of Cu/Cu^2+ is 0.34V .
Given that ,
There are two voltaic cells , each with a standard hydrogen electrode in one compartment .
One cell contains a Pb/Pb^2+ half-cell and the other contains a Cu/Cu^2+ half-cell .
We know,
The electrode potential of Pb/Pb^2+ is given by,
Pb^2+(aq) +2e^- ==> Pb (s) , E^0=-0.13 V
And the electrode potential of Cu/Cu^2+ is given by ,
Cu^2+ (aq) +2e^- ==>Cu(s) , E^0 =0.34 V
The electrode potential of standard hydrogen electrode is given by,
2H^+(aq) +2e^- ==> H2(g) , E^0= 0.0 V
The E^0cell of Pb/Pb^2+ when it is made with standard hydrogen electrode is given by ,
E^0cell = 0.0 V -(-0.13 V) = 0.13V
Hence the E^0cell of Pb/Pb^2+ is 0.13V .
The E^0cell of Cu/Cu^2+ when it is made with standard hydrogen electrode is given by ,
E^0cell =0.34V - 0.0V = 0.34 V
Hence the E^0cell of Cu/Cu^2+ is 0.34V.
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