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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"] |
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Answer» Solution :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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