Question

In acidic medium, \( K_2Cr_2O_7 \) shows oxidizing action as represented in the half-reaction: \(Cr_2O_7^{2-} + XH^+ + Ye^- \rightarrow 2A + ZH_2O\)
\( X \), \( Y \), \( Z \), and \( A \) are respectively:

• A. 8, 6, 4, and \( Cr_2O_3 \)
• B. 14, 7, 6, and \( Cr^{3+} \)
• C. 8, 4, 6, and \( Cr_2O_3 \)
• D. 14, 6, 7, and \( Cr^{3+} \)

Answer 1

The Correct Option is D

The problem asks to balance the given half-reaction for the reduction of dichromate ion (\( K_2Cr_2O_7 \)) in an acidic medium and identify the unknown coefficients \( X, Y, Z \) and the product species \( A \).

Concept Used:

The balancing of a redox half-reaction in an acidic medium follows a set of sequential rules:

1. Balance all atoms other than oxygen (O) and hydrogen (H).
2. Balance the oxygen atoms by adding water (\( H_2O \)) molecules to the side deficient in oxygen.
3. Balance the hydrogen atoms by adding hydrogen ions (\( H^+ \)) to the side deficient in hydrogen.
4. Balance the net charge on both sides by adding electrons (\( e^- \)) to the side with the greater positive charge.

We also need to know the product of the reduction of dichromate. In an acidic medium, the dichromate ion (\( Cr_2O_7^{2-} \)), where chromium is in the +6 oxidation state, is reduced to the chromium(III) ion (\( Cr^{3+} \)).

Step-by-Step Solution:

Step 1: Identify the species A and write the skeleton equation.

In an acidic medium, the dichromate ion (\( Cr_2O_7^{2-} \)) is a strong oxidizing agent and gets reduced to the chromium(III) ion (\( Cr^{3+} \)). Therefore, the species \( A \) is \( Cr^{3+} \). The skeleton half-reaction is:

\[ Cr_2O_7^{2-} \rightarrow 2Cr^{3+} \]

The chromium atoms are already balanced, with 2 on each side.

Step 2: Balance the oxygen atoms.

There are 7 oxygen atoms on the left-hand side (LHS) and none on the right-hand side (RHS). To balance the oxygen atoms, we add 7 water molecules (\( H_2O \)) to the RHS.

\[ Cr_2O_7^{2-} \rightarrow 2Cr^{3+} + 7H_2O \]

Comparing this with the given format \( \dots \rightarrow 2A + ZH_2O \), we can identify \( Z = 7 \).

Step 3: Balance the hydrogen atoms.

Now, there are \( 7 \times 2 = 14 \) hydrogen atoms on the RHS and none on the LHS. To balance the hydrogen atoms in an acidic medium, we add 14 hydrogen ions (\( H^+ \)) to the LHS.

\[ Cr_2O_7^{2-} + 14H^+ \rightarrow 2Cr^{3+} + 7H_2O \]

Comparing this with the given format \( Cr_2O_7^{2-} + XH^+ + \dots \), we can identify \( X = 14 \).

Step 4: Balance the charge.

Calculate the total charge on both sides of the equation.

• Total charge on LHS = \( (1 \times -2) + (14 \times +1) = -2 + 14 = +12 \)
• Total charge on RHS = \( (2 \times +3) + (7 \times 0) = +6 \)

The charge is not balanced. To balance the charge, we add electrons (\( e^- \)) to the side with the higher (more positive) charge. We need to add \( 12 - 6 = 6 \) electrons to the LHS.

\[ Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O \]

Comparing this with the given format \( Cr_2O_7^{2-} + XH^+ + Ye^- \rightarrow \dots \), we can identify \( Y = 6 \).

Final Result:

The completely balanced half-reaction is:

\[ Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O \]

By comparing this with the given equation \( Cr_2O_7^{2-} + XH^+ + Ye^- \rightarrow 2A + ZH_2O \), we find:

• \( X = 14 \)
• \( Y = 6 \)
• \( A = Cr^{3+} \)
• \( Z = 7 \)

Therefore, \( X \), \( Y \), \( Z \), and \( A \) are respectively: 14, 6, 7, and \( Cr^{3+} \).