In the equation, $\mathrm{BiO}_3^{-}+6 \mathrm{H}^{+}+x ~\mathrm{e}^{-} \rightarrow \mathrm{Bi}^{3+}+3 \mathrm{H}_2 \mathrm{O}$ What is the value of $x$ ?

Balancing the half-reaction $\mathrm{BiO}_3^{-} + 6\mathrm{H}^{+} + x~\mathrm{e}^{-} \to \mathrm{Bi}^{3+} + 3\mathrm{H}_2\mathrm{O}$ reveals that $x = 2$ through charge conservation or oxidation state analysis.

The oxidation state of bismuth decreases from $+5$ in $\mathrm{BiO}_3^{-}$ to $+3$ in $\mathrm{Bi}^{3+}$, indicating a gain of 2 electrons.

Charge balancing confirms this result: the reactant side charge $(-1 + 6 - x)$ must equal the product side charge $(+3)$, yielding $5 - x = 3$ and thus $x = 2$.

Final answer: $\boxed{2}$