Question

Solubility product of \( \text{MX}(s) \) is \( 10^{-10} \) and \( E^{\circ}_{\text{M}^{+}/\text{M}} = 0.71 \, \text{V} \). Find out \( E^{\circ}_{\text{M}/\text{MX}^{-}} \).

• A. 0.119 V
• B. -0.119 V
• C. 1.301 V
• D. -1.301 V

Hint:

In electrochemical reactions, the Nernst equation can be used to calculate the potential based on concentrations or solubility. For solubility products, the concentration of ions is key to determining the electrochemical potential.

Answer 1

The Correct Option is B

Step 1: Understand the relationship between solubility product and electrode potential.
The relationship between the solubility product (\( K_{\text{sp}} \)) and the electrode potential can be derived using the Nernst equation. The equation for the half-cell reaction is: \[ E = E^{\circ} - \frac{0.0591}{n} \log Q \] where: - \( E^{\circ} \) is the standard electrode potential, - \( n \) is the number of electrons involved in the reaction, - \( Q \) is the reaction quotient.
Step 2: Calculate the solubility from the solubility product.
For the dissociation of \( \text{MX}(s) \) in water, the solubility product is: \[ K_{\text{sp}} = [\text{M}^{+}][\text{X}^{-}] = s^2 \] where \( s \) is the solubility. Given \( K_{\text{sp}} = 10^{-10} \), we can solve for \( s \): \[ s = \sqrt{K_{\text{sp}}} = \sqrt{10^{-10}} = 10^{-5} \, \text{mol/L} \]
Step 3: Apply the Nernst equation.
For the reaction: \[ \text{M}^{+} + \text{X}^{-} \rightleftharpoons \text{MX}(s) \] we know that the standard electrode potential for the oxidation of \( \text{M}^{+} \) to \( \text{M} \) is \( E^{\circ}_{\text{M}^{+}/\text{M}} = 0.71 \, \text{V} \). The Nernst equation for the reverse reaction (reduction of \( \text{MX}^{-} \)) is: \[ E = E^{\circ} - \frac{0.0591}{1} \log \left( \frac{1}{s^2} \right) \] Substituting \( s = 10^{-5} \): \[ E = 0.71 \, \text{V} - \frac{0.0591}{1} \log \left( \frac{1}{(10^{-5})^2} \right) \] \[ E = 0.71 \, \text{V} - 0.0591 \log(10^{10}) \] \[ E = 0.71 \, \text{V} - 0.0591 \times 10 \] \[ E = 0.71 \, \text{V} - 0.591 \] \[ E = 0.119 \, \text{V} \] Thus, \( E^{\circ}_{\text{M}/\text{MX}^{-}} = -0.119 \, \text{V} \).