When `E_(Ag^+//Ag)^(o)=0.8` volt and `E_(Zn^(2+)//Zn)^(o)=-0.76` volt, which of the following is correctA. `Ag^(+)` can be reduced by `H_(2)`B. Ag can oxidise `H_(2)` into `H^(+)`C. `Zn^(2+)` can be reduced by `H_(2)`D. Ag and reduce `Zn^(2+)` ion

When `E_(Ag^+//Ag)^(o)=0.8` volt and `E_(Zn^(2+)//Zn)^(o)=-0.76` volt,

1.	When `E_(Ag^+//Ag)^(o)=0.8` volt and `E_(Zn^(2+)//Zn)^(o)=-0.76` volt, which of the following is correctA. `Ag^(+)` can be reduced by `H_(2)`B. Ag can oxidise `H_(2)` into `H^(+)`C. `Zn^(2+)` can be reduced by `H_(2)`D. Ag and reduce `Zn^(2+)` ion
Answer» Correct Answer - A Because `H_(2)` has greater reduction potential so it reduced at `Ag^(+)`.

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When `E_(Ag^+//Ag)^(o)=0.8` volt and `E_(Zn^(2+)//Zn)^(o)=-0.76` volt, which of the following is correctA. `Ag^(+)` can be reduced by `H_(2)`B. Ag can oxidise `H_(2)` into `H^(+)`C. `Zn^(2+)` can be reduced by `H_(2)`D. Ag and reduce `Zn^(2+)` ion

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