Balance the following redox reactions:
$S^{2 -} (a q) + I_{2} (s) ⟶ {S O}_{4}^{2 -} (a q) + I^{-} (a q)$

S^{2 -} (a q) + {4 I}_{2} (s) + {4 H}_{2} O (l) ⟶ {8 I}^{-} (a q) {+ S O}_{4}^{2 -} (a q) + {8 H}^{+} (a q)

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S^{2 -} (a q) + {6 I}_{2} (s) + {4 H}_{2} O (l) ⟶ {8 I}^{-} (a q) {+ 4 S O}_{4}^{2 -} (a q) + {8 H}^{+} (a q)

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{3 S}^{2 -} (a q) + {4 I}_{2} (s) + {4 H}_{2} O (l) ⟶ {8 I}^{-} (a q) {+ 5 S O}_{4}^{2 -} (a q) + {8 H}^{+} (a q)

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{2 S}^{2 -} (a q) + {4 I}_{2} (s) + H_{2} O (l) ⟶ {8 I}^{-} (a q) {+ S O}_{4}^{2 -} (a q) + {8 H}^{+} (a q)

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Solution

The correct option is A $S^{2 -} (a q) + {4 I}_{2} (s) + {4 H}_{2} O (l) ⟶ {8 I}^{-} (a q) {+ S O}_{4}^{2 -} (a q) + {8 H}^{+} (a q)$
The unbalanced chemical equation is given below:
$S^{2 -} (a q) + I_{2} (s) ⟶ {S O}_{4}^{2 -} (a q) + I^{-} (a q)$
Balance all atoms except H and O,
$S^{2 -} (a q) + 4 I_{2} (s) ⟶ {S O}_{4}^{2 -} (a q) + 8 I^{-} (a q)$
The oxidation number of S changes from $- 2$ to $+ 6$ .
The net change in the oxidation number is $8$ .
The oxidation number of iodine changes from $0$ to $- 1$ .
The net change in the oxidation number per I atom is $1$ .
Total change in the oxidation number for $8$ I atoms is $8$ .
Thus, the increase in the oxidation number is equal to the decrease in the oxidation number.
Balance O atoms by adding $4$ water molecules to LHS,
$S^{2 -} (a q) + {4 I}_{2} (s) + {4 H}_{2} O (l) ⟶ {8 I}^{-} (a q) {+ S O}_{4}^{2 -} (a q)$
Balance H atoms by adding $8$ $H^{+}$ to the RHS,
$S^{2 -} (a q) + {4 I}_{2} (s) + {4 H}_{2} O (l) ⟶ {8 I}^{-} (a q) {+ S O}_{4}^{2 -} (a q) + {8 H}^{+} (a q)$
This is the balanced chemical equation.

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Q. Use the following

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5 C e^{4 +} (a q) + M n^{2 +} (a q) + 4 H_{2} O (l) \to M n O_{4}^{-} (a q) + 8 H^{+} (a q) M n O_{4}^{-} (a q) + 5 e^{-} \to M n^{2 +} (a q) + 4 H_{2} O (l); E^{o} = + 1.51 V C e^{4 +} + e^{-} \to C e^{3 +} (a q); E^{o} = + 1.61 V

Q. Choose correct statement(s) regarding the following reaction.

{C r}_{2} O_{7}^{2 -} (a q) + 3 {S O}_{3}^{2 -} (a q) + 8 H^{+} ⟶ 2 {C r}^{3 +} (a q) + 3 {S O}_{4}^{2 -} (a q) + 4 H_{2} O

Q. Standard electrode potential data are useful for understanding the suitability of an oxidant in a redox titrations. Some half cell reactions and their standard potentials are given below:

M n O_{4}^{-} (a q .) + 8 H^{+} (a q .) + 5 e^{-} \to M n^{2 +} (a q .) + 4 H_{2} O (l); E^{o} = 1.51 V

C r_{2} O_{7}^{2 -} (a q .) + 14 H^{+} (a q .) + 6 e^{-} \to 2 C r^{3 +} (a q .) + 7 H_{2} O (l); E^{o} = 1.38 V

F e^{3 +} (a q .) + e^{-} \to F e^{2 +} (a q .); E^{o} = 0.77 V

C l_{2} (g) + 2 e^{-} \to 2 C l^{-} (a q .); E^{o} = 1.40 V

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Balance the following redox reactions:S2−(aq)+I2(s)⟶SO2−4(aq)+I−(aq)

Balance the following redox reactions:
$S^{2 -} (a q) + I_{2} (s) ⟶ {S O}_{4}^{2 -} (a q) + I^{-} (a q)$