Question

Which of the following statement is correct about the balanced equation given below? \[ Cr_2O_7^{2-}(aq)+14H^+(aq)+3S^{2-}(aq) \rightarrow 2Cr^{3+}(aq)+3S(s)+7H_2O(l) \]

• A. \(Cr_2O_7^{2-}\) reduces the \(S^{2-}\)
• B. Oxidation number of Cr changes from +7 to +3
• C. Oxidation number of S remains -2
• D. \(Cr_2O_7^{2-}\) oxidises the \(S^{2-}\)

Hint:

Dichromate ion \((Cr_2O_7^{2-})\) is a powerful oxidizing agent in acidic medium.

Answer 1

The Correct Option is D

Concept: Oxidation involves increase in oxidation number, while reduction involves decrease in oxidation number.

Step 1: Find oxidation state of chromium.
In \[ Cr_2O_7^{2-} \] let oxidation number of Cr be \(x\). \[ 2x+7(-2)=-2 \] \[ 2x=12 \] \[ x=+6 \] After reaction: \[ Cr^{3+} \] Oxidation number becomes \[ +3. \] Thus chromium is reduced.

Step 2: Find oxidation state of sulphur.
Initially: \[ S^{2-} \] Oxidation number \[ =-2 \] Finally: \[ S(s) \] Oxidation number \[ =0 \] Thus sulphur is oxidized.

Step 3: Identify oxidizing agent.
The species causing oxidation is called oxidizing agent. Hence, \[ Cr_2O_7^{2-} \] oxidizes \[ S^{2-}. \] Therefore, \[ \boxed{Cr_2O_7^{2-}\text{ oxidises }S^{2-}} \]