Write balanced net ionic equation for the following reactions in acidic solution.
$S_{4} O {_{6}}^{2 -} (a q) + A l (s) \to H_{2} S (a q) + A l^{3 +} (a q)$

S_{4} O {_{6}}^{2 -} (a q) + 6 A l (s) + 20 H^{+} \to 4 H_{2} S (a q) + 6 A l^{3 +} (a q) + 6 H_{2} O

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2 S_{4} O {_{6}}^{2 -} (a q) + 6 A l (s) + 20 H^{+} \to 4 H_{2} S (a q) + 2 A l^{3 +} (a q) + 6 H_{2} O

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S_{4} O {_{6}}^{2 -} (a q) + 5 A l (s) + 20 H^{+} \to 4 H_{2} S (a q) + 3 A l^{3 +} (a q) + 6 H_{2} O

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None of the above

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Solution

The correct option is A $S_{4} O {_{6}}^{2 -} (a q) + 6 A l (s) + 20 H^{+} \to 4 H_{2} S (a q) + 6 A l^{3 +} (a q) + 6 H_{2} O$
The unbalanced chemical equation is as shown.
$S_{4} O {_{6}}^{2 -} (a q) + A l (s) \to H_{2} S (a q) + A l^{3 +} (a q)$
Balance S atoms by adding coefficient 4 to hydrogen sulphide.
$S_{4} O {_{6}}^{2 -} (a q) + A l (s) \to 4 H_{2} S (a q) + A l^{3 +} (a q)$
Al is oxidized and S is reduced.
The oxidation number of Al increases from 0 to +3. Total increase in the oxidation number is 3.
The oxidation number of S decreases from +2.5 to -2. Decrease in the oxidation number of 1 S atom is 4.5. Total decrease in the oxidation number of 4 S atoms is 18.
Balance total increase in the oxidation number by total decrease in the oxidation number by multiplying Al and Al(III) ions with 6.
$S_{4} O {_{6}}^{2 -} (a q) + 6 A l (s) \to 4 H_{2} S (a q) + 6 A l^{3 +} (a q)$
Balance O atoms by adding 6 water molecules to RHS.
$S_{4} O {_{6}}^{2 -} (a q) + 6 A l (s) \to 4 H_{2} S (a q) + 6 A l^{3 +} (a q) + 6 H_{2} O (l)$
Balance H atoms by adding 20 protons on the RHS.
$S_{4} O {_{6}}^{2 -} (a q) + 6 A l (s) + 20 H^{+} (a q) \to 4 H_{2} S (a q) + 6 A l^{3 +} (a q) + 6 H_{2} O (l)$
This is the balanced net ionic equation for the given reaction in acidic solution.

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S_{4} O_{6}^{2 -} (a q) + A l (s) \to H_{2} S (a q) + A l^{3 +} (a q)

S_{2} O {_{6}}^{2 -} (a q) + C r_{2} O {_{7}}^{2 -} (a q) \to S_{4} O {_{6}}^{2 -} (a q) + C r^{3 +} (a q)

H^{+}

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H^{+}

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