Write the Nernst equation for the cell reaction in the Daniel cell. How will the `E_("cell")` be affected when concentration of `Zn^(+)` ions is increased?

Explanation

Daniel cell involves the redox reaction :
`Zn(s) +CuSO_(4)(aq) to ZnSO_(4)(aq)+Cu(s)`
Thus in the cell Zn(s) is oxidised to `Zn^(2+)` (aq) ions in the oxidation half cell while `Cu^(2+)`(aq) ions are reducted to Cu(s) in the reduction half cell.
According to Nernst equation,
`E_(cell)=E_(cell)^(@)-(0.059)/(2)"log"([Zn^(2+)(aq)])/([Cu^(2+)(aq)])`
`E_(cell)` decreases as the `[Zn^(2+)(aq)]` increases.