Question

Write balanced redox reactions of the following:

4 moles of ${Sn}^{2+}\left(aq\right)$ reduces ${IO}_4^{-}$ to $I^{-}\left(aq\right)$.

• A. ${4Sn}^{2+}+{IO}_4^{-}+8H^{+}⟶{6Sn}^{4+}+I^{-}+4H_{2}O$
• B. ${4Sn}^{2+}+{IO}_8^{-}+8H^{+}⟶{4Sn}^{4+}+I^{-}+4H_{2}O$
• C. ${4Sn}^{2+}+{IO}_4^{-}+8H^{+}⟶{4Sn}^{4+}+I^{-}+4H_{2}O$
• D. none of these

Answer 1

The correct option is C ${4Sn}^{2+}+{IO}_4^{-}+8H^{+}⟶{4Sn}^{4+}+I^{-}+4H_{2}O$

$4$ moles of ${Sn}^{2+}\left(aq\right)$ reduces ${IO}_4^{-}$ to $I^{-}\left(aq\right)$.
The unbalanced redox equation is as follows:
${Sn}^{2+}+{IO}_4^{-}⟶{Sn}^{4+}+I^{-}$
All atoms other than H and O are balanced..
The oxidation number of Sn changes from 2 to 4. The change in the oxidation number of Sn is 2.
The oxidation number of I changes from 7 to -1. The change in the oxidation number of I is 8.
The increase in the oxidation number is balanced with decrease in the oxidation number bu multiplying $S{n}^{2+}$ and $S{n}^{4+}$ with 4.
$4{Sn}^{2+}+{IO}_4^{-}⟶4{Sn}^{4+}+I^{-}$
To balance O atoms add 4 water molecules on RHS.
$4{Sn}^{2+}+{IO}_4^{-}⟶4{Sn}^{4+}+I^{-}+4H_{2}O$
To balance H atoms, add 8 $H^{+}$ on LHS.
$4{Sn}^{2+}+{IO}_4^{-}+8H^{+}⟶4{Sn}^{4+}+I^{-}+4H_{2}O$
This is the balanced chemical equation.