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$\mathrm{BrO}_3^{-}$changes into $\mathrm{Br}_2$ in an acidic medium of a unbalanced equation. How many electron should be present on the balanced equation?
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The correct answer is:
10 electron in left
The equation when $\mathrm{BrO}_3^{-}$changes into $\mathrm{Br}_2$ in an acidic medium is as follows
$2 \mathrm{BrO}_3^{-}+12 \mathrm{H}^{+} \longrightarrow \mathrm{Br}_2+6 \mathrm{H}_2 \mathrm{O}$
Add $10 \mathrm{e}^{-}$in LHS to balance the charge on both sides
$2 \mathrm{BrO}_3^{-}+10 \mathrm{e}^{+}+12 \mathrm{H}^{+} \longrightarrow \mathrm{Br}_2+6 \mathrm{H}_2 \mathrm{O} \text {. }$
$2 \mathrm{BrO}_3^{-}+12 \mathrm{H}^{+} \longrightarrow \mathrm{Br}_2+6 \mathrm{H}_2 \mathrm{O}$
Add $10 \mathrm{e}^{-}$in LHS to balance the charge on both sides
$2 \mathrm{BrO}_3^{-}+10 \mathrm{e}^{+}+12 \mathrm{H}^{+} \longrightarrow \mathrm{Br}_2+6 \mathrm{H}_2 \mathrm{O} \text {. }$
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