Solubility product of MX(s) is ${10}^{-10}=E_{M^{+}/M}^{°}=0.71V,\text{Find out }{E}_{M/{\mathrm{MX}}^{-}}^{°}$

The correct option is - 0.119 V.

To find the standard reduction/oxidation potential for the half-cell involving the sparingly soluble salt $\mathrm{MX}$, let us first write down the half-reaction and relate it to the solubility product and the standard electrode potential of the metal half-cell.

Step 1: Write down the half-cell reactions and potentials

The standard reduction potential for the metal electrode is given as:
$E_{M^{+}/M}^{°}=0.71\text{ V}$

The half-reaction for the electrode $M/{\mathrm{MX}}^{-}$ (or oxidation half-reaction involving the sparingly soluble salt) can be represented as:
$M\left(s\right)+X^{-}\left(\mathrm{aq}\right)\to \mathrm{MX}\left(s\right)+e^{-}$

Alternatively, looking at the reduction half-reaction:
$\mathrm{MX}\left(s\right)+e^{-}\to M\left(s\right)+X^{-}\left(\mathrm{aq}\right)$

Step 2: Relate the potentials using the solubility product

The relationship between the standard reduction potential of the salt electrode $E_{X/\mathrm{MX}/M}^{°}$ and the metal-ion electrode $E_{M^{+}/M}^{°}$ is given by Nernst equation at equilibrium:
$E_{{\mathrm{MX/M,\; X}}^{-}}^{°}=E_{M^{+}/M}^{°}+\frac{0.0591}{n}\log \left(K_{\mathrm{sp}}\right)$

Given data:
Solubility product, $K_{\mathrm{sp}}={10}^{-10}$
Standard reduction potential, $E_{M^{+}/M}^{°}=0.71\text{ V}$
Number of electrons transferred, $n=1$

Step 3: Calculate the standard reduction potential

Substituting the values into the equation:
$E_{{\mathrm{MX/M,\; X}}^{-}}^{°}=0.71+0.0591\times \log ({10}^{-10})$

Since $\log ({10}^{-10})=-10$:
$E_{{\mathrm{MX/M,\; X}}^{-}}^{°}=0.71+0.0591\times (-10)$

$E_{{\mathrm{MX/M,\; X}}^{-}}^{°}=0.71-0.591=0.119\text{ V}$

Step 4: Determine the oxidation potential

The standard oxidation potential is the negative of the standard reduction potential:
$E_{M/{\mathrm{MX}}^{-}}^{°}=-E_{{\mathrm{MX/M,\; X}}^{-}}^{°}=-0.119\text{ V}$

Therefore, the value of $E_{M/{\mathrm{MX}}^{-}}^{°}$ is - 0.119 V.