We can do this by reversing the second half-reaction and then adding it to the first half-reaction.
Reversing the second half-reaction, we get:
$$\mathrm{M} \longrightarrow \mathrm{M}^{2+} + 2 \mathrm{e}^{-} \quad \mathscr{E}^{\circ}=0.50 \mathrm{~V}$$
Now, we
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