Question

Balance the following reaction.

$I_2+S_2{O}_3^{2-}\to I^{-}+S_4{O}_6^{2-}$

• A. $I_2+2S_2{O}_3^{2-}\to 2I^{-}+S_4{O}_6^{2-}$
• B. $3I_2+S_2{O}_3^{2-}\to 2I^{-}+6S_4{O}_6^{2-}$
• C. $I_2+4S_2{O}_3^{2-}\to 2I^{-}+3S_4{O}_6^{2-}$
• D. None of these

Answer 1

Answer: (A)

$I_2+2S_2{O}_3^{2-}\to 2I^{-}+S_4{O}_6^{2-}$

The unbalanced redox equation is as follows:
$I_2+S_2{O}_3^{2-}\to I^{-}+S_4{O}_6^{2-}$
Balance all atoms other than H and O.
$I_2+2S_2{O}_3^{2-}\to 2I^{-}+S_4{O}_6^{2-}$
The oxidation number of I changes from 0 to -1. The change in the oxidation number of one I atom is 1.
Total change in the oxidation number for 2 I atoms is 2.
The oxidation number of S changes from 2 to 2.5. The change in the oxidation number of one S atom is 0.5.
Total change in the oxidation number for 4 S atoms is 2.
The increase in the oxidation number is balanced with decrease in the oxidation number.
O atoms are balanced. This is the balanced chemical equation.