Consider a concentration cell that has both electrodes made of some metal M. Solution A in one compartment of the cell contains 1.0\(M \mathrm{M}^{2+} .\) Solution \(\mathrm{B}\) in the other cell compartment has a volume of 1.00 L. At the beginning of the experiment 0.0100 mole of \(\mathrm{M}\left(\mathrm{NO}_{3}\right)_{2}\) and 0.0100 mole of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) are dissolved in solution \(\mathrm{B}\) (ignore volume changes), where the reaction $$\mathrm{M}^{2+}(a q)+\mathrm{SO}_{4}^{2-}(a q) \rightleftharpoons \mathrm{MSO}_{4}(s)$$ occurs. For this reaction equilibrium is rapidly established, whereupon the cell potential is found to be 0.44 \(\mathrm{V}\) at \(25^{\circ} \mathrm{C} .\) Assume that the process $$\mathrm{M}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{M}$$ has a standard reduction potential of \(-0.31 \mathrm{V}\) and that no other redox process occurs in the cell. Calculate the value of \(K_{\mathrm{sp}}\) for \(\mathrm{MSO}_{4}(s)\) at \(25^{\circ} \mathrm{C} .\)

Step by step solution

01

Determine the reduction potential of M2+/M for both compartments

Using the standard electrode potential, we can find the electrode potential for M2+ compartment and B compartment. For solution A, the standard reduction potential of M2+ is given as -0.31 V. The concentration of M2+ in solution A is 1.0 M. Therefore, using the Nernst equation, we can find the electrode potential of compartment A: \[E_A = E^0_A - \frac{0.0592}{n} \log{\frac{0.010}{1}}\] where \(E^0_A=-0.31 \mathrm{V}\), n=2, and the concentration of M2+ in solution A is 1M. For solution B, we are given the concentration of M(NO3)2 and Na2SO4 which are both 0.0100 moles in 1.0 L of water. At equilibrium, the reaction: \[M^{2+}(aq)+SO_4^{2-}(aq) \rightleftharpoons MSO_4(s)\] occurs. Let the concentration of M2+ remaining at equilibrium be x. Using an ICE table, we get: \[M^2+: 0.010 - x\] \[SO_4^{2-}: 0.010 - x\] Using the Nernst equation again, we can find the electrode potential of compartment B: \[E_B = E^0_B - \frac{0.0592}{n} \log{\frac{x^2}{(0.010 - x)^2}}\] where \(E^0_B=-0.31 \mathrm{V}\), n=2.

02

Calculate the cell potential and find the concentration of M2+ in solution B

As given, the cell potential is 0.44 V. The cell potential can be calculated by the difference in electrode potentials in compartments A and B: \[E_{cell} = E_B - E_A\] \[0.44 = -0.31 - \frac{0.0592}{2} \log{\frac{x^2}{(0.010 - x)^2}} + 0.31 - \frac{0.0592}{2} \log{\frac{0.010}{1}}\] Solve for x, which represents the concentration of M2+ in solution B at equilibrium.

03

Calculate the solubility product constant, Ksp

Now that we have determined the concentration of M2+ in solution B at equilibrium, we can calculate the solubility product constant, Ksp, for the reaction: \[M^{2+}(aq)+SO_4^{2-}(aq) \rightleftharpoons MSO_4(s)\] At equilibrium, the concentrations of M2+ and SO4^2- are both x (from Step 2). The Ksp expression for this reaction is: \[K_{sp} = [M^{2+}][SO_4^{2-}]\] Substitute the concentrations obtained in Step 2, and calculate Ksp for MSO4 at 25°C.

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