A copper wire is dipped in silver nitrate solution in beaker A and a silver wire is dipped in a solution of copper sulphate kept in a beaker B . If the standard electrode potential for
`Cu^(2+) + 2e^(-) to Cu` is +0.34 for `Ag^(+) + e^(-) to Ag` is 0.80 V
Predict in which beaker the ions present will get reduced ?

Explanation

The probable reaction in beaker A will be
`Cu+2AgNO_(3)toCu(NO_(3))_(2)+2Ag, i.e., Cu+2Ag^(+)toCu^(2+)+2Ag`
`E_(cell)` for this reaction `=-0.34+0.80=+0.46V` i.e., it is +ve. Hence, `Ag^(+)` ions are reduced to Ag.