Answer:
[tex]\large \text{IO$_{3}$^{-}$(aq) + 3Sn$^{2+}$(aq) + 6H$^{+}$(aq) $\longrightarrow \,$ I$^{-}$(aq) + 3Sn$^{4+}$(aq) + 3H$_{2}$O(l)}[/tex]
Explanation:
IO₃⁻ + Sn²⁺ ⟶ I⁻ + Sn⁴⁺
1: Separate the equation into two half-reactions.
IO₃⁻ ⟶ I⁻
Sn²⁺ ⟶ Sn⁴⁺
2: Balance all atoms other than H and O.
Done
3: Balance O.
IO₃⁻ ⟶ I⁻ + 3H₂O
Sn²⁺ ⟶ Sn⁴⁺
4: Balance H
IO₃⁻ + 6H ⟶ I⁻ + 3H₂O
Sn²⁺ ⟶ Sn⁴⁺
5: Balance charge.
IO₃⁻ + 6H⁺ + 6e⁻ ⟶ I⁻ + 3H₂O
Sn²⁺ ⟶ Sn⁴⁺ + 2e⁻
6: Equalize electrons transferred.
1 × [IO₃⁻ + 6H⁺ + 6e⁻ ⟶ I⁻ + 3H₂O]
3 × [Sn²⁺ ⟶ Sn⁴⁺ + 2e⁻]
7: Add the two half-reactions.
1 × [IO₃⁻ + 6H⁺ + 6e⁻ ⟶ I⁻ + 3H₂O]
3 × [Sn²⁺ ⟶ Sn⁴⁺ + 2e⁻]
IO₃⁻ + 3Sn²⁺ + 6H⁺ ⟶ I⁻ + 3Sn⁴⁺ + 3H₂O
8: Check mass balance.
On the left: 1 I, 3 O, 3 Sn, 6 H
On the right: 1 I, 3 O, 3 Sn, 6 H
Step 9: Check charge balance.
On the left: 1- + 12+ = 11+
On the right: 1- + 12+ = 11+
The equation is balanced.
[tex]\text{The balanced equation is }\\\large \textbf{IO$_{3}$^{-}$(aq) + 3Sn$^{2+}$(aq) + 6H$^{+}$(aq) $\longrightarrow \,$ I$^{-}$(aq) + 3Sn$^{4+}$(aq) + 3H$_{2}$O(l)}[/tex]
