1.

What will be the electrode potential for the given half cell reaction at pH=5 ? 2H_(2)O to O_(2)+4H^(+)+4e^(-),E^(@)=-1.23V (R=8.314" J "mol^(-1)K^(-1), temp.=298 K, oxygen under std. atm. Pressure of 1 bar.)

Answer»


Solution :`2H_(2)O to O_(2) + 4H^(+)+4E^(-),E^(@)=-1.23V`
`therefore E=+1.23-(0.0591)/(4)log[H^(+)]^(4)`
`=+1.23+(0.0591xxpH)`
`=+1.23+0.0591xx5implies1.23+0.2955`
`+1.5255V`


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