Question

Consider the following redox reaction in basic medium. \( x\text{Cr(OH)}_3 + y(\text{IO}_3)^- + z(\text{OH})^- \rightarrow a(\text{CrO}_4)^{2-} + b(\text{I})^- + c(\text{H}_2\text{O}) \). The incorrect option about it is

• A. x+y=3
• B. a+b=7
• C. z=4
• D. b=1

Hint:

When balancing redox reactions in a basic medium: 1. Balance atoms other than O and H. 2. Balance O atoms by adding \(H_2O\). 3. Balance H atoms by adding \(H^+\). 4. Balance charge by adding \(e^-\). 5. Add a number of \(OH^-\) ions to both sides equal to the number of \(H^+\) ions. 6. Combine \(H^+\) and \(OH^-\) to form \(H_2O\) and simplify the equation.

Answer 1

The Correct Option is B

We need to balance the redox reaction using the half-reaction method in a basic medium.
Step 1: Write the oxidation and reduction half-reactions.
Oxidation: \( \text{Cr(OH)}_3 \rightarrow (\text{CrO}_4)^{2-} \). Chromium goes from +3 oxidation state to +6. This is an oxidation losing 3 electrons.
Reduction: \( (\text{IO}_3)^- \rightarrow \text{I}^- \). Iodine goes from +5 oxidation state to -1. This is a reduction gaining 6 electrons.
Step 2: Balance the half-reactions.
Oxidation half-reaction: \( \text{Cr(OH)}_3 \rightarrow (\text{CrO}_4)^{2-} + 3e^- \). To balance Oxygen, add \(H_2O\). We have 3 O on left, 4 on right. Add \(H_2O\) to the left. \( \text{Cr(OH)}_3 + H_2O \rightarrow (\text{CrO}_4)^{2-} \). Now 4 O on left, 4 on right. To balance Hydrogen, add \(H^+\). We have 5 H on left, 0 on right. Add \(5H^+\) to the right. \( \text{Cr(OH)}_3 + H_2O \rightarrow (\text{CrO}_4)^{2-} + 5H^+ + 3e^- \). Since it's in basic medium, add \(5OH^-\) to both sides. \( \text{Cr(OH)}_3 + H_2O + 5OH^- \rightarrow (\text{CrO}_4)^{2-} + 5H_2O + 3e^- \). Simplify: \( \text{Cr(OH)}_3 + 5OH^- \rightarrow (\text{CrO}_4)^{2-} + 4H_2O + 3e^- \). This is balanced.
Reduction half-reaction: \( (\text{IO}_3)^- + 6e^- \rightarrow \text{I}^- \). To balance Oxygen, add \(3H_2O\) to the right. \( (\text{IO}_3)^- + 6e^- \rightarrow \text{I}^- + 3H_2O \). To balance Hydrogen, add \(6H^+\) to the left. \( (\text{IO}_3)^- + 6H^+ + 6e^- \rightarrow \text{I}^- + 3H_2O \). Add \(6OH^-\) to both sides. \( (\text{IO}_3)^- + 6H_2O + 6e^- \rightarrow \text{I}^- + 3H_2O + 6OH^- \). Simplify: \( (\text{IO}_3)^- + 3H_2O + 6e^- \rightarrow \text{I}^- + 6OH^- \). This is balanced.
Step 3: Combine the half-reactions.
To balance electrons, multiply the oxidation half-reaction by 2. \( 2\text{Cr(OH)}_3 + 10OH^- \rightarrow 2(\text{CrO}_4)^{2-} + 8H_2O + 6e^- \). Now add the reduction half-reaction: \( 2\text{Cr(OH)}_3 + 10OH^- + (\text{IO}_3)^- + 3H_2O + 6e^- \rightarrow 2(\text{CrO}_4)^{2-} + 8H_2O + 6e^- + \text{I}^- + 6OH^- \).
Step 4: Simplify the final equation.
Cancel common species (\(e^-, OH^-, H_2O\)). \( 2\text{Cr(OH)}_3 + (\text{IO}_3)^- + 4OH^- \rightarrow 2(\text{CrO}_4)^{2-} + \text{I}^- + 5H_2O \).
Step 5: Compare with the given equation to find coefficients.
\( x\text{Cr(OH)}_3 + y(\text{IO}_3)^- + z(\text{OH})^- \rightarrow a(\text{CrO}_4)^{2-} + b(\text{I})^- + c(\text{H}_2\text{O}) \).
We have: \(x=2, y=1, z=4, a=2, b=1, c=5\).
Step 6: Check the options.
(A) x+y = 2+1 = 3. This is correct.
(B) a+b = 2+1 = 3. The option says a+b=7. This is incorrect.
(C) z=4. This is correct.
(D) b=1. This is correct.
The incorrect option is (B).