The measured e.m.f. at `25^(@)C` for the cell reaction ,
`Zn(S)+Cu^(2+)(1.0 M)to Cu(s) +Zn^(2+) (0.1 M)` is 1.3 volt, Calculate `E^(@)` for the cell reaction.

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

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