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

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