The reduction potential of the two half cell reactions (occuring in an electrochemical cell) are `PbSO_(4)+ 2e^(-)rarrPb+SO_(4)^(2-) (E^(@)=-0.31V)` `Ag^(+)(aq)+e^(-)rarrAg(s)(E^(@)=+0.80V)` The fessible reaction will beA. `Pb^(2+)+2Agrarr2Ag^(+)+Pb`B. `Pb^(2+)+H_(2)rarr2H^(+)+Pb`C. `2H^(+)+2Agrarr2Ag^(+)+H_(2)`D. `2Ag^(+)+PbrarrPb^(2+)+2Ag`

The reduction potential of the two half cell reactions (occuring in an

1.	The reduction potential of the two half cell reactions (occuring in an electrochemical cell) are `PbSO_(4)+ 2e^(-)rarrPb+SO_(4)^(2-) (E^(@)=-0.31V)` `Ag^(+)(aq)+e^(-)rarrAg(s)(E^(@)=+0.80V)` The fessible reaction will beA. `Pb^(2+)+2Agrarr2Ag^(+)+Pb`B. `Pb^(2+)+H_(2)rarr2H^(+)+Pb`C. `2H^(+)+2Agrarr2Ag^(+)+H_(2)`D. `2Ag^(+)+PbrarrPb^(2+)+2Ag`
Answer» Correct Answer - D For the cell reaction to be spontaneous `E_("cell")` should be positive. Therefore, two half cell reaction are `Ag^(+)+e^(-)rarrAg (E^(@)=+0.80V)` `PbrarrPb^(2+)+2e^(-)(E^(@)=+0.13V)` and net reaction is `2Ag^(+)+Pbrarr2Ag+Pb^(2+)`

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