For the metal/metal insoluble salt electrode, the reduction half reaction is:
\[ MX(s) + e^- \rightarrow M(s) + X^-(aq) \]
This reaction can be obtained by combining:
\[ M^+ + e^- \rightarrow M(s) \quad E^\Theta = 0.79 \, V \]
and the solubility equilibrium:
\[ MX(s) \rightleftharpoons M^+ + X^- \]
From the equilibrium expression:
\[ K_{sp} = [M^+][X^-] \]
For standard electrode potential of the insoluble salt electrode:
\[ E^\Theta = E^\Theta_{M^+/M} + \frac{0.059}{1} \log K_{sp} \]
Substituting the given values:
\[ E^\Theta = 0.79 + 0.059 \log (10^{-10}) \]
\[ \log (10^{-10}) = -10 \]
\[ E^\Theta = 0.79 + 0.059(-10) \]
\[ E^\Theta = 0.79 - 0.59 \]
\[ E^\Theta = 0.20 \, V \]
Convert into mV:
\[ 0.20 \, V = 200 \, mV \]
Final Answer = 200