The electrochemical cell shown below is a concentration cell.
M|M 2+ (saturated solution of a sparingly soluble salt,MX 2 )|| M 2+ (0.001 mol dm –3 ) |M
The emf of the cell depends on the difference in concetration of M 2+ ions at the two electrodes. The emf of the cell at 298 is 0.059 V.
(i) The solubility product (K sp ; in mol 3 dm –9 ) of MX 2 at 298 K based on the information available in the given concentration cell is : (Take 2.303× R × 298/F = 0.059 V)
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
(i) M|M 2+ (aq) || M 2+ (aq) | M
0.001 M
Anode , : M ⎯→ M 2+ (aq) + 2e –
Cathode M 2+ (aq) + 2e – ⎯→ M
M 2+ (aq) c 
M 2+ (aq) a
E cell = 0 – 
0.059 = 
– 2 = 
10 –2 × 10 –3 = M 2+ (aq) a = solubility = s
K sp = 4s 3 = 4 × (10 –5 ) 3 = 4 × 10 –15
(ii) . Δ G = – nFE cell = – 2 × 96500 × 0.059 × 10 –3 kJ/mole
= – 11.4 kJ/mole.
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