For the cell to be spontaneous (oxygen to evolve at anode), $E_{cell}\ge 0$.

Reaction at anode (oxidation): $H_{2}O\to 2H^{+}+\frac{1}{2}{O}_2+2e^{-}$, so $E_{anode}^{oxi}=-E_{O_2/H_{2}O}^{\circ }=-1.23$ V (at standard).

Cathode: $M^{2+}+2e^{-}\to M$, $E_{cathode}=0.994$ V (at 1 M $M^{2+}$).

Using the Nernst correction for the anode at non-standard $[H^{+}]$:

$E_{anode}^{oxi}=-1.23-\frac{0.059}{2}\log \frac{[H^{+}{]}^2{P}_{O_2}^{1/2}}{1}=-1.23+0.059pH$ (since $P_{O_2}=1$).

At the threshold $E_{cell}=E_{cathode}+E_{anode}^{oxi}=0$:

$0.994+(-1.23+0.059pH)=0\Rightarrow 0.059pH=0.236\Rightarrow pH\approx 4$.