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Calculate the emf of the cell. Zn|Zn^(2+)(0.001M)||Ag^(+)(0.1M)|Ag The standard potential of Ag//Ag^(+) half - cell is +0.80 V and Zn//Zn^(2+) is -0.76V. |
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Answer» Solution :Step 1 : Write the half-cell reactions of the anode and the cathode. Then ADD the anode and cathode half reactions to obtain the cell reaction and the value of `E_("cell")^(@)` `{:("Cathode",:,2Ag^(+)+2e^(-),rarr,2Ag,E^(@)=+0.80),("Anode",:,""Zn,rarr,Zn^(2+)+2e^(-),E^(@)=-0.76V),("Cell",:,Zn+"2Ag"^(+),rarr,Zn^(2+)+2Ag,E^(@)=1.56V):}` Step 2. K for the cell reaction `=([Zn^(2+)])/([AG^(+)]^(2))` Substituting the given values in the Nernst equation and solving for `E_("cell")`, we have `E_("cell")=E_("cell")^(@)-(0.0591)/(n)logK` `=1.56-(0.0591)/(2)log""([Zn^(2+)])/([Ag^(+)]^(2))` `=1.56-(0.0591)/(2)log""([10^(-3)])/([10^(-1)]^(2))` `=1.56-0.02955` `=1.58955V` Calculation of Equilibrium constant for the cell reaction The Nernst equation for a cell is `E_("cell")=E_("cell")^(@)-(0.0591)/(n)logK` `"or"logK=(E_("cell")^(@))/(0.0591)` |
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