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Calculate the E.M.F. of the zinc - silve...

Calculate the E.M.F. of the zinc - silver cell at `25^(@)C` when `[Zn^(2+)]=0.10M and [Ag^(+)]=10M.[E^(@) " cell at "25^(@)C="1.56 volt"]`

Text Solution

Verified by Experts

The cell reaction in the zinc - silver cell would be
`2Ag^(+)+Zn hArr 2Ag+Zn^(2+)`
The Nernst equation for the above all reaction may be written as :
`E_("cell")=E_("cell")^(@)-(RT)/(nF)ln([Ag]^(2)[Zn^(2+)])/([Ag^(+)]^(2)[Zn])`
(since concentrations of solids are taken as unity)
`=E_("cell")^(@)-ln.([Zn^(2+)])/([Ag^(+)]^(2))`
Substituting the various values in Nernst equation, we have
`E_("cell")=1.56-(2.303xx8.314xx298)/(2xx96495)log""(0.1)/((10)^(2))`
`="1.648 volts."`
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