Acidic solution in acidic solution, the iodate ion can be used to react with a number of metal ions. one such reaction is io3−(aq)+sn2+(aq)→i−(aq)+sn4+(aq) since this reaction takes place in acidic solution, h2o(l) and h+(aq) will be involved in the reaction. places for these species are indicated by the blanks in the following restatement of the equation: io3−(aq)+sn2+(aq)+ −−−→i−(aq)+sn4+(aq)+ −−−

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.

i think   the answer is c hope i'm right