1.

When E_(Ag^+//Ag)^(o)=0.8 volt and E_(Zn^(2+)//Zn)^(o)=-0.76 volt, which of the following is correct

Answer»

`AG^(+)` can be reduced by `H_(2)`
Ag can oxidise `H_(2)` into `H^(+)`
`Zn^(2+)` can be reduced by `H_(2)`
Ag and reduce `Zn^(2+)` ion

Solution :Because `H_(2)` has greater reduction potential so it reduced at `Ag^(+)`.


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