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:
Answer & explanation
Correct answer: option 3
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.