The measured e.m.f. at `25^(@)C` for the cell reaction,
`Zn(s) +XCu^(2+)._((eq))(1.0M) rarr Cu(s) ._((aq))(0.1M)` is 1.3 volt, calculate `E^(@)` for the cell reaction.

Explanation

Using nernst equation `("at" 298K),E_(cell) = E_(cell)^(@) - (0.0591)/(2) log.([Zn^(2+)(aq)])/([Cu^(2+)(aq)])`
Here, `E_(cell) = 1.3 V, [Cu^(2+)(aq)] = 1.0 M, [Zn^(2+)(aq)] = 0.1M, C_(cell) = ?`
Substituting the values,
`1.3 V = E_(cell)^(@) - (0.0591V)/(2) log.(0.1)/(1.0)`
`1.3 = E_(cell)^(@) - 0.02955 V log 10^(-1)`
`1.3 V = E_(cell)^(@) + 0.02955 V log 10`
`E_(cell)^(@) = 1.3 V - 0.02955 V = 1.27 V`