Choose correct statement(s) regarding the following reaction.
${C r}_{2} O_{7}^{2 -} (a q) + 3 {S O}_{3}^{2 -} (a q) + 8 H^{+} ⟶ 2 {C r}^{3 +} (a q) + 3 {S O}_{4}^{2 -} (a q) + 4 H_{2} O$

{C r}_{2} O_{7}^{2 -}

is oxidising agent.

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{S O}_{3}^{2 -}

is reducing agent.

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The oxidisation number of per 'S' atom in

3 {S O}_{3}^{2 -}

is increase by two.

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The oxidation number of per 'Cr' atom in

{C r}_{2} O_{7}^{2 -}

is decreased by three.

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Solution

The correct options are
A ${C r}_{2} O_{7}^{2 -}$ is oxidising agent.
B ${S O}_{3}^{2 -}$ is reducing agent.
C The oxidation number of per 'Cr' atom in ${C r}_{2} O_{7}^{2 -}$ is decreased by three.
D The oxidisation number of per 'S' atom in $3 {S O}_{3}^{2 -}$ is increase by two.
(A) ${C r}_{2} O_{7}^{2 -}$ oxidizes ${S O}_{3}^{2 -}$ to ${S O}_{4}^{2 -}$ . Hence, it acts as oxidizing agent. Thus option A is correct.
(B) ${S O}_{3}^{2 -}$ reduces $C r_{2} O_{7}^{2 -}$ to $C r^{3 +}$ . hence, it acts as reducing agent. Thus option B is correct.
(C) The oxidation number of each S atom increases from +4 to +6. Thus the oxidation number is increased by 2. Thus option C is correct.
(D) The oxidation number of each chromium atom decreases from +6 to +3. Thus it decreases by 3. Thus option D is correct.

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Similar questions

C r_{2} O_{7}^{2 -} (a q) + 3 S O_{3}^{2 -} (a q .) + 8 H^{+} \to 2 C r^{3 +} (a q) + 3 S O_{4}^{2 -} (a q) + 4 H_{2} O

C r_{2} O_{7}^{2 -}

S O_{3}^{2 -}

S O_{3}^{2 -}

C r_{2} O_{7}^{2 -}

{C r}_{2} O_{7}^{2 -} H^{+} - - \to C r

{C r}_{2} O_{7}^{2 -}

{C r}_{2} O_{7}^{2 -} (a q) + H {N O}_{2} (a q) ⟶ {C r}^{3 +} (a q) + {N O}_{3}^{-} (a q)

{C r}_{2} O_{7}^{2 +} (a q) + 14 H^{+} (a q) + 6 {F e}^{2 +} (a q) \to 6 {F e}^{3 +} (a q) + 2 {C r}^{3 +} (a q) + 7 H_{2} O (l)

{F e}^{2 +} (a q) + {C r}_{2} O_{7}^{2 -} (a q) (acid medium) - ------- \to {F e}^{3 +} (a q) + {C r}^{3 +} (a q)

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