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What will be the EMF of the following electrode concentration cel at 25^(@)C Hg-Zn(C_(1)M)|Zn^(2+)(CM)|Hg-Zn(C_(2)M) If the concentrations of zinc amalgam are 2 g per 100 g of mercury and 1 g per 100 g of mercury in the aniodic and the cathodic compartments respectively. |
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Answer» SOLUTION :The half-cell reaction are At anode: `Zn(C_(1))TOZN^(2+)(C)+2E^(-)` `underset("At anode:"Zn^(2+)(C)+2e^(-)toZn(C_(2)))` net reaction: `Zn(C_(1))toZn(C_(2))` Thus, it is a CONCENTRATION cell for which `E_(cell)=(0.0591)/(2)"log"(C_(1))/(C_(2))=(0.0591)/(2)"log"(2)/(1)=(0.0591)/(2)xx0.3010=8.89xx10^(-3)V` |
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