The following cell was constructed to find the difference in Ksp between two nat

Question

The following cell was constructed to find the difference in Ksp between two naturally occurring forms of CaCO3(s), called calcite and aragonite. Pb(s) | PbCO3(s) | CaCO3 (s, calcite) | buffer (pH 7.00) || buffer (pH 7.00) | CaCO3 (s, aragonite) | PbCO3(s) | Pb(s) Each compartment of the cell contains a mixture of solid PbCO3 (Ksp = 7.4 x 10-14) and either calcite or aragonite, both of which have Ksp ~ 5 x 10-9. Each solution was buffered to pH 7.00 with an inert buffer, and the cell was completely isolated from atmospheric CO2. The measured cell voltage was -1.8 mV. Find the ratio of solubility products, Ksp (for calcite) / Ksp (for aragonite).

Please let me know the solution detailed. I know the answer is 0.76 but don't know the reaseon. What reaction happens in each electrode?

Explanation / Answer

At anode oxidation takes place, so the reaction will be Pb ---------> Pb+2 + 2e-
at the cathode reduction takes place, so the reaction will be reversed.
E = Eo - 0.059/n log Q
E = -1.8mV = -1.8*10-3 V ; Eo = 0 V
[Pb+2] 2 = Ksp ; [Pb+2] = Ksp
n = 2
-1.8*10-3 = 0 - 0.059 /2 log [Ksp]cal / [Ksp]arg
1.8*10-3 = 0.0295 log [Ksp]1/2cal / [Ksp]1/2arg
0.061016 =  log [Ksp]1/2cal / [Ksp]1/2arg
100.061016 = [ Ksp]1/2cal / [Ksp]1/2arg
Ksp (for calcite) / Ksp (for aragonite) = 1.15092 = 1.3246 (this should be the answer)

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