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Calculate the e.m.f. of the cell in whic...

Calculate the e.m.f. of the cell in which the following reaction takes place :
`Ni(s) +2Ag^(+)(0.002 M)to Ni^(2+)(0.160 M)+2Ag(s)`
Given `E_(cell)^(@)`=1.05 v

Text Solution

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At an: `Ni(s) to Ni^(2+) + 2e^(-)`
At cathode: `2Ag^(+) + 2e^(-) to 2Ag`
The net reactions:
`Ni(s) + 2Ag^(+) to Ni^(3+) + 2Ag(s)`
`E_("cell") = E_("cell")^(@) -(0.0591)/n log (["Anode"])/(["Cathode"])`
`E_("cell") =1.05 -(0.0591)/n log [Ni^(2+)]/[Ag^(+)]^(2)`
`E_("cell") = 1.05 -(0.0591)/2 log (0.16)/(0.002)^(2) = 0.9142` V
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