At temperature of 298 K, the emf of the following electrochemical cell
`Z n _((s))`
`| Zn ^(2+) ( 0.1 M ) || Cu ^(2+) (0.01M)| Cu _((s))`
be _________V.
(Given `E _("cell") ^(@) = 1.10V`)

Answer

`1.07`

Explanation

For the given electrochemical cell, the cell reaction will be as follows:
`Zn to Zn ^(2+) + 2 e ^(-)` (Oxidation at anode)
`Cu ^(2**) + 2 e ^(-) to Cu Cu ^(2**) + 2 e ^(-) to Cu`
(Reduction at cathode)
`Cu ^(2+) + Zn to Cu + Zn ^(2+) `( Overall cel reaction)
The e.m.f. of the cell is given by
`E _("cell") = E _("cell") ^(**) - ( 0.0591)/(n)log _(10) ""([Zn ^(2+)])/([Cu ^(2+)])`
Substituting the values in above equation,
`E _("cell") = 1. 10 - ( 0. 0 591)/(2)`
`log _(10) "" ((0.1))/( ( 0.01))`
`= 1. 10 - ( 0. 0 591)/(2)`
`log _(2) 10`
`= 1. 10 - ( 0. 0 591)/(2)`
`= 1, 10-0.0296`
`= 1. 07 V`