Question

Balance the following reaction.

$S+O{H}^{-}\to S^{2-}+S_2{O}_3^{2-}$

• A. $6O{H}^{-}+4S\to 2S^{2-}+S_2{O}_3^{2-}+3H_{2}O$
• B. $6O{H}^{-}+5S\to 2S^{2-}+3S_2{O}_3^{2-}+2H_{2}O$
• C. $6O{H}^{-}+S\to 2S^{2-}+2S_2{O}_3^{2-}+H_{2}O$
• D. None of these

Answer 1

Answer: (A)

The unbalanced redox equation is
$S+O{H}^{-}\to S^{2-}+S_2{O}_3^{2-}$
Balance all atoms other than H and O.
$3S+O{H}^{-}\to S^{2-}+S_2{O}_3^{2-}$
The oxidation number of S changes from 0 to -2. The change in the oxidation number of S is 2.
The oxidation number of S changes from 0 to 4. The change in the oxidation number of S is 4.
The increase in the oxidation number is balanced with decrease in the oxidation number by multiplying
$S$ and $S^{2-}$ with 2.
$4S+O{H}^{-}\to 2S^{2-}+S_2{O}_3^{2-}$
O atoms are balanced by adding 2 water molecules on LHS.
$4S+O{H}^{-}+2H_{2}O\to 2S^{2-}+S_2{O}_3^{2-}$
Hydrogen atoms are balanced by adding 5 $H^{+}$ atoms on RHS
$4S+O{H}^{-}+2H_{2}O\to 2S^{2-}+S_2{O}_3^{2-}+5H^{+}$.
Since the reaction occurs in basic medium, 5 hydroxide ions are added on the either side
$4S+O{H}^{-}+2H_{2}O+5O{H}^{-}\to 2S^{2-}+S_2{O}_3^{2-}+5H^{+}+5O{H}^{-}$.
On RHS five protons and five hydroxide ions forms five water molecules out of which two cancels out with two water molecule on RHS.
$6O{H}^{-}+4S\to 2S^{2-}+S_2{O}_3^{2-}+3H_{2}O$
This is the balanced chemical equation.