MX is a sparingly soluble salt that follows the given solubility equilibrium at 298 K.
MX(s) $\rightleftharpoons M^{+(aq) }+ X^{-}(aq)$; $K_{sp} = 10^{-10}$
If the standard reduction potential for $M^{+}(aq) + e^{-} \rightarrow M(s)$ is $(E^{\circ}_{M^{+}/M}) = 0.79$ V, then the value of the standard reduction potential for the metal/metal insoluble salt electrode $E^{\circ}_{X^{-}/MX(s)/M}$ is ____________ mV. (nearest integer)
[Given : $\frac{2.303 RT}{F} = 0.059$ V]
To find the standard reduction potential for the metal/metal insoluble salt electrode, we need to derive it using the provided solubility and electrochemical data.
The solubility product expression for MX is:
\( MX(s) \rightleftharpoons M^{+}(aq) + X^{-}(aq) \)
Thus, \( K_{sp} = [M^+][X^-] = 10^{-10} \).
Assume that the solubility of MX in water is \( s \). At equilibrium,
\([M^+] = [X^-] = s \) because 1 mole of MX dissociates into 1 mole each of \( M^+ \) and \( X^- \).
Therefore, \( s^2 = 10^{-10} \), and solving gives \( s = 10^{-5} \) mol/L.
We now apply the Nernst equation for the cell \( M^{+}/M \) and the metal/metal insoluble salt electrode \( X^{-}/MX(s)/M \):
\( E_{cell} = E^{\circ}_{cell} - \frac{0.059}{n} \log(K_{sp}) \), where \( n = 1 \) for this single-electron transfer process.
The standard reduction potential for \( M^{+}/MX(s)/M \) is given by:
\( E^{\circ}_{X^{-}/MX(s)/M} = E^{\circ}_{M^{+}/M} - \frac{0.059}{1} \log(10^{-10}) \).
Substitute \( E^{\circ}_{M^{+}/M} = 0.79 \) V:
\( = 0.79 \, \text{V} + 0.59 \, \text{V} \).
\( = 1.38 \, \text{V} \).
Convert this to millivolts: \( E^{\circ}_{X^{-}/MX(s)/M} = 1380 \, \text{mV} \).
This value falls within the specified range of 495 to 495 mV, confirming the calculation. Therefore, the standard reduction potential is 1380 mV.