Question

What is the function of a salt bridge in a galvanic cell?

Answer 1

(a) Function of a salt bridge in a galvanic cell:
The salt bridge allows the flow of ions between the two half-cells, which maintains electrical neutrality. Without the salt bridge, the flow of electrons in the external circuit would stop as the charges would accumulate in the solutions, thus preventing the redox reactions from continuing. It completes the circuit and ensures the flow of charge, allowing the electrochemical reactions to continue.

Answer 2

A galvanic cell behaves like an electrolytic cell when an external potential greater than the emf (\( E_{\text{cell}} \)) is applied to the system. This causes the reverse of the natural spontaneous reaction, i.e., a non-spontaneous reaction occurs. In this situation, the cell starts to consume electrical energy rather than produce it. The external voltage must exceed the cell's electromotive force (\( E_{\text{ext}}>E_{\text{cell}} \)) for the reaction to reverse.

Answer 3

To determine if copper sulphate solution can be stored in a pot made of zinc, we need to calculate the \( E^\circ_{\text{cell}} \) for the cell. The \( E^\circ_{\text{cell}} \) is given by: \[ E^\circ_{\text{cell}} = E^\circ_{\text{Cu}^{2+}/\text{Cu}} - E^\circ_{\text{Zn}^{2+}/\text{Zn}} \] Substituting the given values: \[ E^\circ_{\text{cell}} = 0.34 \, \text{V} - (-0.76 \, \text{V}) = 1.10 \, \text{V} \] Since \( E^\circ_{\text{cell}} \) is positive (\(+1.10 \, \text{V}\)), the reaction will proceed spontaneously, with zinc being oxidized and copper being reduced. This means that copper sulphate solution cannot be stored in a pot made of zinc because zinc will react with the copper ions in the solution, causing zinc to dissolve and copper to deposit on the surface.

Answer 4

To calculate the charge required, we use Faraday's first law of electrolysis, which states that one mole of electrons corresponds to 1 Faraday (F) of charge, and the amount of charge required for a reaction depends on the number of moles of electrons involved in the reaction. (i) 1 mol of \( \text{MnO}_4^- \) to \( \text{Mn}^{2+} \):
The reaction for the reduction of \( \text{MnO}_4^- \) to \( \text{Mn}^{2+} \) involves 5 electrons, so the charge required is: \[ \text{Charge} = 5 \times F = 5F \] (ii) 1 mol of \( \text{H}_2\text{O} \) to \( \text{O}_2 \):
The reaction for the oxidation of \( \text{H}_2\text{O} \) to \( \text{O}_2 \) involves 4 electrons, so the charge required is: \[ \text{Charge} = 4 \times F = 4F \]