In acidic solution, the sulfate ion can be used to react with a number of metal ions. One such reaction is SO42 - (aq) Sn2 (aq) →H2SO3 (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: SO42 - (aq) Sn2 (aq) - --→H2SO3 (aq) Sn4 (aq) - --

SO₄²⁻ (aq) + Sn²⁺ (aq) + 4H⁺ → H₂SO₃ (aq) + Sn⁴⁺ (aq) + H₂O

Explanation:

At first calculate the oxidation state of that element which undergoes oxidation as well as reduction.

for SO₄²⁻ the oxidation state of sulphur is + 6 and H₂SO₃ the oxidation state of sulphur is + 4

So balance equation is

(Reduction) SO₄²⁻ + 4H⁺ + 2e⁻ → H₂SO₃ + H₂O ... (1)

(oxidation) Sn²⁺ → Sn⁴⁺ + 2e⁻ ... (2)

Adding equation 1 & 2

we get

SO₄²⁻ (aq) + Sn²⁺ (aq) + 4H⁺ → H₂SO₃ (aq) + Sn⁴⁺ (aq) + H₂O