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When E(Ag)^(@)+(//Ag)=0.8V and EZn^2+(...

When `E_(Ag)^(@)+_(//Ag)=0.8V` and ` E_Zn^2+_(//Zn)`
`=-0.76V`. Which of the following is correct?

A

`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

Text Solution

Verified by Experts

The correct Answer is:
A
Because `H_(2)` has greater reduction potential so it reduced at `Ag^(+)`.
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