Choose the reactions given by:

(A) \(Cr_2O_7^{2-} + 14H^+ + 6Fe^{2+} \longrightarrow 2Cr^{3+} + 6Fe^{3+} + 7H_2O\)

(B) \(Cr_2O_7^{2-} + 14H^+ + 3H_2S \longrightarrow 2Cr^{3+} + SO_4^{2-} + 7H_2O\)

(C) \(Cr_2O_7^{2-} + 14H^+ + 3Sn^{2+} \longrightarrow 2Cr^{3+} + 3Sn^{4+} + 7H_2O\)

(D) \(Cr_2O_7^{2-} + 14H^+  \longrightarrow 3Cr^{3+} + 2H_2O\)

(E) \(Cr_2O_7^{2-} + 2OH^- \longrightarrow 2CrO_4^{2-} + H_2O\)

Choose the correct answers from the options given below:

A, C, E only

The correct answer is option 3. A, C, E only.

Let us dive into each reaction to understand why certain reactions are correct or incorrect.

(A) \( \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{Fe}^{2+} \longrightarrow 2\text{Cr}^{3+} + 6\text{Fe}^{3+} + 7\text{H}_2\text{O} \)

Reaction Type: Redox reaction (dichromate reduction and iron(II) oxidation).

Oxidation: \( \text{Fe}^{2+} \) is oxidized to \( \text{Fe}^{3+} \).

Reduction: \( \text{Cr}_2\text{O}_7^{2-} \) is reduced to \( \text{Cr}^{3+} \).

Balanced: This reaction is correctly balanced. The dichromate ion (\( \text{Cr}_2\text{O}_7^{2-} \)) is reduced by accepting electrons while iron(II) is oxidized by losing electrons. The stoichiometry of the equation shows that for every 6 moles of \( \text{Fe}^{2+} \) oxidized, 2 moles of \( \text{Cr}_2\text{O}_7^{2-} \) are reduced, producing 6 moles of \( \text{Fe}^{3+} \) and 7 moles of \( \text{H}_2\text{O} \).

(B) \( \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 3\text{H}_2\text{S} \longrightarrow 2\text{Cr}^{3+} + \text{SO}_4^{2-} + 7\text{H}_2\text{O} \)

Reaction Type: Incorrect.

The correct reaction will be:

\( \text{Cr}_2\text{O}_7^{2-} + 8\text{H}^+ + 3\text{H}_2\text{S} \longrightarrow 2\text{Cr}^{3+} + \text{S} + 7\text{H}_2\text{O}\)

Issue: The ion produced after the oxidation of \(H_2S\) is incorrectly given in the option.

(C) \( \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 3\text{Sn}^{2+} \longrightarrow 2\text{Cr}^{3+} + 3\text{Sn}^{4+} + 7\text{H}_2\text{O} \)

Reaction Type: Redox reaction (dichromate reduction and tin(II) oxidation).

Oxidation: \( \text{Sn}^{2+} \) is oxidized to \( \text{Sn}^{4+} \).

Reduction: \( \text{Cr}_2\text{O}_7^{2-} \) is reduced to \( \text{Cr}^{3+} \).

Balanced: This reaction is correctly balanced. For every 3 moles of \( \text{Sn}^{2+} \) oxidized to \( \text{Sn}^{4+} \), 1 mole of \( \text{Cr}_2\text{O}_7^{2-} \) is reduced to 2 moles of \( \text{Cr}^{3+} \), with 7 moles of \( \text{H}_2\text{O} \) produced.

(D) \( \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ \longrightarrow 3\text{Cr}^{3+} + 2\text{H}_2\text{O} \)

Reaction Type: Incomplete/Incorrect redox reaction.

Issue: The dichromate ion (\( \text{Cr}_2\text{O}_7^{2-} \)) should reduce to \( \text{Cr}^{3+} \), but this reaction suggests it produces 3 moles of \( \text{Cr}^{3+} \), which is incorrect. The balanced reduction reaction should yield 2 moles of \( \text{Cr}^{3+} \) per mole of \( \text{Cr}_2\text{O}_7^{2-} \). Additionally, it does not account for the required stoichiometry of hydrogen ions and the correct amount of water.

(E) \( \text{Cr}_2\text{O}_7^{2-} + 2\text{OH}^- \longrightarrow 2\text{CrO}_4^{2-} + \text{H}_2\text{O} \)

Reaction Type: Redox reaction in a basic medium.

Oxidation: \( \text{Cr}_2\text{O}_7^{2-} \) is oxidized to \( \text{CrO}_4^{2-} \).

Balanced: This reaction is correctly balanced for a basic medium. The dichromate ion is oxidized to chromate ion (\( \text{CrO}_4^{2-} \)) in the presence of hydroxide ions, with water as a byproduct.

Summary

From the detailed analysis, the correct reactions are:

(A): Valid redox reaction with correct stoichiometry.

(C): Valid redox reaction with correct stoichiometry.

(E): Correct reaction in a basic medium with correct stoichiometry.

The incorrect reactions are:

(B) : Incorrect due to incorrect stoichiometry.

(D): Incorrect due to incorrect stoichiometry.

Thus, the correct answer is 3. A, C, E only.