The electrochemical cell shown below is a concentration cell.
M|M²⁺ (saturated solut ion of a sparingly so luble salt, MX₂ || M²⁺ (0.001 mol dm^–3)| M.
The emf of the cell depends on the difference in concentration of M²⁺ ions at the two electrodes. The emf of the cell at 298 is 0.059 V.
Q.
The value of ΔG (kJ mol^-1) for the given cell is (take 1 F = 96500 C mol^-1)
a)
–5.7
b)
5.7
c)
11.4
d)
–11.4
Correct answer is option 'D'. Can you explain this answer?

Question

The electrochemical cell shown below is a concentration cell.
M|M²⁺ (saturated solut ion of a sparingly so luble salt, MX₂ || M²⁺ (0.001 mol dm^–3)| M.
The emf of the cell depends on the difference in concentration of M²⁺ ions at the two electrodes. The emf of the cell at 298 is 0.059 V.

Q.

The value of ΔG (kJ mol^-1) for the given cell is (take 1 F = 96500 C mol^-1)

a)
–5.7
b)
5.7
c)
11.4
d)
–11.4

Correct answer is option 'D'. Can you explain this answer?

Accepted Answer

At cathode:M⁺²(aq)+2e⁻→M(s)
At anode:M(s)+2X⁻(aq)→MX₂(aq)+2e⁻
n - factor of the cell reaction is 2.

ΔG=−nFE_cell

= −2×96500×0.059

= −11.4 kJmole⁻¹

Answer

Answer

Based on the information provided, it appears that the electrochemical cell shown is a concentration cell. This is indicated by the presence of different concentrations of the same ion in the two half-cells.

In the anode half-cell, there is a saturated solution of a sparingly soluble salt, MX2, which means that the concentration of the ion M2+ is high. In the cathode half-cell, there is a lower concentration of the ion M2+ (0.001 mol dm^-3).

In a concentration cell, the driving force for the cell reaction is the difference in concentration between the two half-cells. The ion flows from the higher concentration side (anode) to the lower concentration side (cathode) in order to equalize the concentrations.

Therefore, based on the information provided, it can be concluded that the electrochemical cell shown is a concentration cell.

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