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 M 2+ . Solution B in the other cell compartment has a volume of 1.00 L. At the beginning of the experiment 0.0100 mole of M(NO 3 ) 2 and 0.0100 mole of Na 2 SO 4 are dissolved in solution B (ignore volume changes), where the reaction M 2+ ( a q ) + SO 4 2 − ( a q ) ⇌ MSO 4 ( s ) occurs. For this reaction equilibrium is rapidly established, whereupon the cell potential is found to be 0.44 V at 25°C. Assume that the process M 2 + + 2 e − → M has a standard reduction potential of −0.31 V and that no other redox process occurs in the cell. Calculate the value of K sp for MSO 4 ( s ) at 25°C.

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Interpretation Introduction Interpretation: The value of solubility product K SP for MSO 4 ( s ) at 25 ο C is to be calculated. Concept introduction: The solubility product is the mathematical product of a substance’s dissolved ion concentration raised to its power of its stoichiometric coefficients. When sparingly soluble ionic compound releases ions in the solution, it gives relevant solubility product. The solvent is generally water. To determine: The value of solubility product K SP for MSO 4 ( s ) at 25 ο C . Explanation Given The total cell reaction is shown below. The total cell reaction for the given concentration cell is, M + 2 ( a q ) + M ( s ) → M ( s ) + M + 2 ( a q ) The concentration of M +2 in a compartment A is 1.0 M . Now, 0 .0100 mol of M ( NO 3 ) 2 and 0 .0100 mol of Na 2 SO 4 are added to the compartment of B and the same number of moles of solid MSO 4 are obtained. The given cell reaction is M + 2 ( a q ) + M ( s ) → M ( s ) + M + 2 ( a q ) The solubility of M +2 in the compartment B is calculated by the Nernst equation. According to the Nernst equation, E c e l l = E c e l l ο − 0.0592 n log Q For concentration cells, E c e l l ο = 0 Substitute this value in the Nernst equation. E c e l l = − 0.0592 n log [ M + 2 ] B [ M + 2 ] A Where, n is the number of electrons. [ M + 2 ] B is the concentration of M +2 in solution B. [ M + 2 ] A is the concentration of M +2 in solution A. Given E c e l l = 0. 44 V Substitute the required values in the above expression. E c e l l = − 0.0592 n log [ M + 2 ] B [ M + 2 ] A 0

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Chemistry: An Atoms First Approach

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Chemistry: An Atoms First Approach

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Steven S. Zumdahl + 1 other

Publisher: Cengage Learning

ISBN: 9781305079243

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Interpretation Introduction

Interpretation:

The value of solubility product KSP for MSO4(s) at 25 οC is to be calculated.

Concept introduction:

The solubility product is the mathematical product of a substance’s dissolved ion concentration raised to its power of its stoichiometric coefficients. When sparingly soluble ionic compound releases ions in the solution, it gives relevant solubility product. The solvent is generally water.

To determine: The value of solubility product KSP for MSO4(s) at 25 οC .

Explanation of Solution

Given

The total cell reaction is shown below.

The total cell reaction for the given concentration cell is,

M+2(aq)+M(s)→M(s)+M+2(aq)

The concentration of M+2 in a compartment A is 1.0M .

Now, 0.0100 mol of M(NO3)2 and 0.0100 mol of Na2SO4 are added to the compartment of B and the same number of moles of solid MSO4 are obtained.

The given cell reaction is

M+2(aq)+M(s)→M(s)+M+2(aq)

The solubility of M+2 in the compartment B is calculated by the Nernst equation.

According to the Nernst equation,

Ecell=Ecellο−0.0592nlogQ

For concentration cells,

Ecellο=0

Substitute this value in the Nernst equation.

Ecell=−0.0592nlog[M+2]B[M+2]A

Where,

n is the number of electrons.
[M+2]B is the concentration of M+2 in solution B.
[M+2]A is the concentration of M+2 in solution A.

Given

Ecell=0.44 V

Substitute the required values in the above expression.

Ecell=−0.0592nlog[M+2]B[M+2]A0

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