Question

Find the solubility product of a saturated solution of Ag₂CrO₄ in water at 298 K if the emf of the cell Ag | Ag⁺ (satd. Ag₂CrO₄ soln.)||Ag⁺(0.1M)|Ag is 0.164 V at 298 K.

Answer 1

The cell is Ag|Ag+||Ag+|Ag, so when the concentrations are unity the potential of cathode and anode will be equal and opposite, so E0cell = 0

By Nernst equation

Ecell = E0cell - RTnFln([Ag+]in satd. Ag2CrO4solution[Ag+])

We know that Ecell = 0.164 V, T = 298K, n= 1, [Ag+] (that in the denominator) = 0.1 mol and E0cell = 0

So 0.164 = - 0.059 log 10(10([Ag+]in satd. Ag2CrO4solution))

Gives me [Ag+]in satd. Ag2CrO4 solution = 1.66 * 10-4

Ag2CrO4(aq) <---> 2 Ag+ + CrO42- (*)

So , solubility product is = [Ag+]2[ CrO42-]

from (*), [Ag+]= 1.66 * 10-4, [CrO42-] = 8.3 *10-5

so solubility product = 2.287 * 10-12

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