The Question is from neet-chemistry basic-chemistry-concepts concentration-methods, and it was created by the author A.RAMA NAGESWARARAO

Question

The number of moles of ${K_2}C{r_2}{O_7}$ that will be needed to react with one mole of nitrite ion in acidic solution is

A.RAMA NAGESWARARAO · Accepted Answer

## CORE INFORMATION
✓ **Answer**: B
## CONCEPT FRAMEWORK
**Primary Concept**: Balancing Redox Reactions in Acidic Medium
**Related Topics**: Oxidation Numbers, Stoichiometry, Mole Ratios
**Prerequisites**: Understanding of redox reactions, balancing equations, and mole concepts.
## SOLUTION PATHWAY
**Given Information**:
  - 1 mole of nitrite ion ($NO_2^-$)
  - Reaction occurs in acidic solution
  - Reactant: $K_2Cr_2O_7$ (dichromate ion, $Cr_2O_7^{2-}$)
**Method & Steps**:
  1. **Write the unbalanced half-reactions:**
    - Oxidation: $NO_2^-  ightarrow NO_3^-$
    - Reduction: $Cr_2O_7^{2-} ightarrow Cr^{3+}$
     Reasoning: Identify the species being oxidized and reduced.
  2. **Balance the atoms (except H and O) in each half-reaction:**
    - Oxidation: $NO_2^-  ightarrow NO_3^-$
    - Reduction: $Cr_2O_7^{2-} ightarrow 2Cr^{3+}$
       Working: Balance Cr atoms in the reduction half-reaction.
  3. **Balance oxygen atoms by adding $H_2O$:**
    - Oxidation: $NO_2^- + H_2O ightarrow NO_3^-$
    - Reduction: $Cr_2O_7^{2-} ightarrow 2Cr^{3+} + 7H_2O$
       Working: Add water to the side deficient in oxygen.
  4. **Balance hydrogen atoms by adding $H^+$:**
    - Oxidation: $NO_2^- + H_2O ightarrow NO_3^- + 2H^+$
    - Reduction: $Cr_2O_7^{2-} + 14H^+ ightarrow 2Cr^{3+} + 7H_2O$
       Working: Add H+ to balance hydrogen atoms.
  5. **Balance the charge by adding electrons:**
    - Oxidation: $NO_2^- + H_2O ightarrow NO_3^- + 2H^+ + 2e^-$
    - Reduction: $Cr_2O_7^{2-} + 14H^+ + 6e^- ightarrow 2Cr^{3+} + 7H_2O$
       Working: Add electrons to balance the charge on each side of the half-reaction.
  6. **Multiply the half-reactions by suitable factors to equalize the number of electrons:**
    - Oxidation: $3(NO_2^- + H_2O ightarrow NO_3^- + 2H^+ + 2e^-)$  -> $3NO_2^- + 3H_2O ightarrow 3NO_3^- + 6H^+ + 6e^-$
    - Reduction: $Cr_2O_7^{2-} + 14H^+ + 6e^- ightarrow 2Cr^{3+} + 7H_2O$
       Working: Multiply the oxidation half-reaction by 3 to make the electrons equal.
  7. **Add the half-reactions and cancel out common terms:**
    - $3NO_2^- + 3H_2O + Cr_2O_7^{2-} + 14H^+ + 6e^- ightarrow 3NO_3^- + 6H^+ + 6e^- + 2Cr^{3+} + 7H_2O$
    - Simplified: $3NO_2^- + Cr_2O_7^{2-} + 8H^+ ightarrow 3NO_3^- + 2Cr^{3+} + 4H_2O$
       Working: Combine and cancel electrons and water and H+ where applicable.
  8. **Determine the mole ratio of $K_2Cr_2O_7$ to $NO_2^-$:**
    - From the balanced equation, 1 mole of $Cr_2O_7^{2-}$ reacts with 3 moles of $NO_2^-$.
    - Therefore, 1 mole of $NO_2^-$ requires $\frac{1}{3}$ mole of $K_2Cr_2O_7$ (since $K_2Cr_2O_7$ provides $Cr_2O_7^{2-}$).
      Result: The required mole ratio is 1:3, so 1/3 mole of dichromate is required for 1 mole of nitrite.
## OPTION ANALYSIS
[A] :x: Incorrect mole ratio.
[B] ✓: Correct mole ratio calculated from the balanced equation.
[C] :x: Incorrect mole ratio.
[D] :x: Incorrect mole ratio.
## LEARNING AIDS
**Common Mistakes**:
  1. **Incorrect balancing of half-reactions**: Ensure atoms and charges are balanced correctly.
  2. **Incorrectly determining mole ratios**: Carefully read the balanced equation to extract mole ratios.
**Quick Tips**:
  1. **Use Oxidation Numbers**: Determine changes in oxidation numbers to confirm which species is reduced and oxidized.
 2. **Systematic Balancing**: Follow a logical step-by-step process for balancing redox equations.

A.RAMA NAGESWARARAO · Answer

$\frac{2}{3}$

A.RAMA NAGESWARARAO · Answer

The correct Option is "$\frac{1}{3}$"

A.RAMA NAGESWARARAO · Answer

$\frac{1}{6}$

A.RAMA NAGESWARARAO · Answer

$\frac{1}{2}$

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