Problem 163 - A Complete Resource Book in Chemistry for JEE Main

Question

When the equation \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+\mathrm{H}^{+}+\mathrm{I}^{-} \longrightarrow \mathrm{Cr}^{3+}+\mathrm{I}_{2}\) Is balanced, give the correct coefficient for \(\mathrm{I}^{-}\)(aq)

Step-by-Step Solution

Answer

The coefficient for \( \mathrm{I}^- \) is 6.

Step 1: Write Down the Unbalanced Equation

First, identify the reactants and products in the chemical equation: \[ \text{Cr}_2\text{O}_7^{2-} + \text{H}^+ + \text{I}^- \rightarrow \text{Cr}^{3+} + \text{I}_2 \] This is the skeleton of the redox reaction that needs to be balanced.

Step 2: Determine Oxidation States

Determine the oxidation states of each element in the reactants and products. \( \text{Cr} \) in \( \text{Cr}_2\text{O}_7^{2-} \) is +6, and in \( \text{Cr}^{3+} \) it is +3. \( \text{I}^- \) is in a -1 oxidation state, and in \( \text{I}_2 \) it is 0.

Step 3: Write Half-Reactions

Write the oxidation and reduction half-reactions. Oxidation: \( 2\text{I}^- \rightarrow \text{I}_2 + 2e^- \). Reduction: \( \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} \).

Step 4: Balance Electrons in Half-Reactions

Multiply the oxidation half-reaction by 3 and the reduction half-reaction by 1 to balance the electrons. This results in: Oxidation: \( 6\text{I}^- \rightarrow 3\text{I}_2 + 6e^- \). Reduction: remains the same.

Step 5: Combine and Cancel Electrons

Combine the balanced half-reactions, ensuring electrons are cancelled: \( \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{I}^- \rightarrow 2\text{Cr}^{3+} + 3\text{I}_2 + 7\text{H}_2\text{O} \).

Step 6: Identify Coefficient for \( \mathrm{I}^- \)

From the balanced equation, the coefficient of \( \mathrm{I}^- \) is 6. This coefficient ensures mass and charge are balanced in the redox reaction.

Key Concepts

Oxidation States

Oxidation states are numbers assigned to elements in a compound to help determine the distribution of electrons among them. Understanding oxidation states is crucial for identifying which elements are oxidized and which are reduced in a redox reaction.
For the equation \[\text{Cr}_2\text{O}_7^{2-} + \text{H}^+ + \text{I}^- \rightarrow \text{Cr}^{3+} + \text{I}_2\]we assess the oxidation states:

• Chromium in \(\text{Cr}_2\text{O}_7^{2-}\) has an oxidation state of +6.
• In \(\text{Cr}^{3+}\), it changes to +3, indicating reduction.
• Iodine in \(\text{I}^-\) starts with an oxidation state of -1 and moves to 0 in \(\text{I}_2\), which shows oxidation.

Identifying these changes helps clarify which elements give or receive electrons, a key step in solving redox reactions.

Half-Reactions

Half-reactions break down the overall redox reaction into simpler parts, focusing separately on oxidation and reduction. This division is essential for visual clarity and aids in the balancing process. Let's explore:
The reaction \[\text{Cr}_2\text{O}_7^{2-} + \text{H}^+ + \text{I}^- \rightarrow \text{Cr}^{3+} + \text{I}_2\]can be split:

• Oxidation half-reaction: \(2\text{I}^- \rightarrow \text{I}_2 + 2e^-\). Here, iodine loses electrons, thereby increasing its oxidation state.
• Reduction half-reaction: \(\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}\). In this step, chromium gains electrons, reducing its oxidation state.

These half-reactions are key for understanding how oxidation and reduction occur independently within the overall process.

Electron Balancing

Electron balancing in redox reactions ensures that the number of electrons lost equals the number gained, maintaining charge balance. This is an essential step for achieving a balanced redox equation.
Taking our half-reactions:

• The oxidation half-reaction produces electrons: \(2\text{I}^- \rightarrow \text{I}_2 + 2e^-\).
• The reduction requires electrons: \(\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}\).

To balance, multiply the oxidation half-reaction by 3 so that the electrons equal 6:\[6\text{I}^- \rightarrow 3\text{I}_2 + 6e^-\]Now, when combining the half-reactions, the electrons cancel out, showcasing the balanced flow:\[\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{I}^- \rightarrow 2\text{Cr}^{3+} + 3\text{I}_2 + 7\text{H}_2\text{O}\]This cancellation confirms the accurate balancing of both mass and charge.