1.

Calculate emf of the cell in which the following reaction takes place :- Ni_(s) + 2Ag^(+) (0.002M) to Ni^(+2) (0.160M) + 2Ag_((s)) Given : E_("cell")^(o) = 1.05V, (2.303RT)/(F) = 0.06, log2 = 0.3

Answer»

`1.05 V`
`0.912 V`
`1.19 V`
`2.05 V`

SOLUTION :`Ni_(g)+2Ag^(+) to Ni^(+2) + 2Ag(s)`
`Q = ([NI^(+2)])/([AG^(+)]^(2)) = (0.16)/((0.002)^(2)) = (0.16)/((2XX10^(-3))^(2)) = (0.16)/(4xx10^(-6))`
`Q = (0.16xx10^(6))/(4) = (16xx10^4)/(4) = 4xx10^(4)`
`E_("cell") = E_("cell")^(0)-(0.06)/(n) logQ`
(from nernst EQUATION)
`E_("cell") = 1.05 - (0.06)/(2)log(4xx10^(4))`
`=1.05 - 0.03{log4+log10^(4)}`
`=1.05 - 0.03{0.6+4 log 10}`
`=1.05 - 0.03 {4.6} = 0.912 V`.


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