What will happen to the value of emf of the cell,                    Zn|Zn²⁺(0.1M)||Cu²⁺(0.1M)|Cu if the concentration of the electrolyte in the anode compartment is increased?

Dear Student,

Please find below the solution to the asked query:

Chemical Reaction:
At anode (oxidation): Zn ----> Zn²⁺ + 2e^-

 At Cathode (reduction):  Cu²⁺ + 2e^- ----> Cu

Zn + Cu²⁺ ----->  Zn²⁺ + Cu

$E=E^o-\frac{0.0591}{n}\log \frac{\left[Z{n}^{2+}\right]}{\left[C{u}^{2+}\right]}$

Hence, increasing the concentration at anode compartment, will increase the concentration of Zn²⁺ ions, and therefore, from above equation we can see that the negative value on the right hand side of the equation increases, and the emf (E) of a cell decreases.

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