Question

A cell is prepared by dipping a copper rod in 1 M copper sulphate solution and a zinc rod in 1 M zinc sulphate solution. The standard reduction potentials of copper and zinc are 0.34 V and −0.76 V respectively.

1. What will be the cell reaction?
2. What will be the standard emf of the cell?
3. Which electrode will be positive?
4. How can the cell be represented?

Answer 1

Given: Copper electrode in 1 M copper sulphate (CuSO₄) solution. 

Zinc electrode in 1 M zinc sulphate (ZnSO₄) solution. 

\[\text{E}_{\text{Cu}^{2+}/\text{Cu}}^\circ\] = +0.34 V

\[\text{E}_{\text{Zn}^{2+}/\text{Zn}}^\circ\] = −0.76 V 

a. In this electrochemical cell, zinc undergoes oxidation and changes from solid zinc to aqueous zinc ions with the release of two electrons.

\[\ce{Zn_{(s)} −> Zn^2+_{ (aq)} + 2e−}\] 

At the same time, copper undergoes reduction in which copper ions present in the aqueous solution gain two electrons and get deposited as solid copper.

\[\ce{Cu^{2+}_{ (aq)} + 2e -> Cu_{(s)}}\] 

Therefore, the overall cell reaction can be written as zinc in solid state combining with copper ions in aqueous solution to give zinc ions in aqueous solution and solid copper.

\[\ce{Zn_{(s)} + Cu^{2+}_{ (aq)} -> Zn^{2+}_{ (aq)} + Cu_{(s)}}\] 

b. Standard emf of the cell is:

\[\text{E}^\circ_{\text{cell}} = \text{E}^\circ_{\text{cathode}} - \text{E}^\circ_{\text{anode}}\]

= 0.34 V − (−0.76 V)

= 1.10 V

c. In this electrochemical cell, reduction takes place at the cathode, and the copper electrode acts as the positive electrode. Oxidation occurs at the anode, and the zinc electrode serves as the negative electrode. Therefore, the copper electrode is the positive electrode in the cell.

d. Standard cell notation:

\[\ce{Zn_{(s)}|Zn^{2+}_{ (aq)} (1 M)|| Cu^{2+}_{ (aq)} (1 M)|Cu_{(s)}}\]