Chlorine undergoes disproportionation in alkaline medium as shown below: The values of a, b, c and d in a balanced…

Q: Chlorine undergoes disproportionation in alkaline medium as shown below: The values of a, b, c and d in a balanced redox reaction are respectively :

The disproportionation of Cl2 in cold, dilute alkaline medium yields chloride (Cl^-) and hypochlorite (ClO^-) ions. Oxidation half-reaction: Cl2 arrow 2ClO^- + 2e^- Balancing O and H in basic medium: Cl2 + 4OH^- arrow 2ClO^- + 2H2O + 2e^- Reduction half-reaction: Cl2 + 2e^- arrow 2Cl^- Adding both half-reactions (electrons are already equal): 2Cl2 + 4OH^- arrow 2ClO^- + 2Cl^- + 2H2O Dividing the entire equation by 2 to get the simplest integer coefficients: 1Cl2 + 2OH^- arrow 1ClO^- + 1Cl^- + 1H

Year	2026	2025	2024	Total
Questions	4	4	6	14

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Core Logic

The disproportionation of $Cl₂$ $Cl_2$ in cold, dilute alkaline medium yields chloride ( $Cl^-$ $Cl^-$) and hypochlorite ( $ClO^-$ $ClO^-$) ions.

Oxidation half-reaction: $Cl₂ arrow 2ClO^- + 2e^-$ $Cl_2 \rightarrow 2ClO^- + 2e^-$ Balancing O and H in basic medium: $Cl₂ + 4OH^- arrow 2ClO^- + 2H₂O + 2e^-$

Reduction half-reaction: $Cl₂ + 2e^- arrow 2Cl^-$ $Cl_2 + 2e^- \rightarrow 2Cl^-$

Adding both half-reactions (electrons are already equal): $2Cl₂ + 4OH^- arrow 2ClO^- + 2Cl^- + 2H₂O$

Dividing the entire equation by 2 to get the simplest integer coefficients: $1Cl₂ + 2OH^- arrow 1ClO^- + 1Cl^- + 1H₂O$

Step 1: Coefficient Matching

Balancing Redox Reactions diagram for Q79 - JEE Main 2024 Morning

By comparing with the given equation: $a Cl₂ + b OH^- arrow c ClO^- + d Cl^- + e H₂O$ We get: $a = 1$ $a = 1$ $b = 2$ $b = 2$ $c = 1$ $c = 1$ $d = 1$ $d = 1$

Thus, the values are 1, 2, 1, and 1.

Chapter Mix

Class 11 Chemistry: Redox Reactions Class 12 Chemistry: The p Block Elements

More Redox Reactions Previous-Year Questions — Page 3

Q87 jee_main_2024_30_january_evening Types of Redox Reactions

Total number of species from the following which can undergo disproportionation reaction

H₂O₂, ClO₃^-, P₄, Cl₂, Ag, Cu⁺¹, F₂, NO₂, K^+

Numerical Answer. Answer: 6 to 6

Solution

Core Logic

For a species to undergo disproportionation, it must contain an element that is present in an intermediate oxidation state. This allows it to act both as an oxidizing agent (by being reduced to a lower state) and a reducing agent (by being oxidized to a higher state).

Let's evaluate each species:

$H₂O₂$ $H_2O_2$: Oxygen is in $-1$ $-1$. Can go to $0$ $0$ ( $O₂$ $O_2$) and $-2$ $-2$ ( $H₂O$ $H_2O$). (Yes)
$ClO₃^-$ $ClO_3^-$: Chlorine is in $+5$ $+5$. Can go to $+7$ $+7$ ( $ClO₄^-$ $ClO_4^-$) and $-1$ $-1$ ( $Cl^-$ $Cl^-$). (Yes)
$P₄$ $P_4$: Phosphorus is in $0$ $0$. Can go to $-3$ $-3$ ( $PH₃$ $PH_3$) and $+1/+3/+5$ $+1/+3/+5$. (Yes)
$Cl₂$ $Cl_2$: Chlorine is in $0$ $0$. Can go to $-1$ $-1$ ( $Cl^-$ $Cl^-$) and $+1$ $+1$ ( $ClO^-$ $ClO^-$). (Yes)
$Ag$ $Ag$: Metal in $0$ $0$ state. Cannot show negative oxidation state. (No)
$Cu⁺¹$ $Cu^{+1}$: Copper is in $+1$ $+1$. Can go to $0$ $0$ ( $Cu$ $Cu$) and $+2$ $+2$ ( $Cu²⁺$ $Cu^{2+}$). (Yes)
$F₂$ $F_2$: Fluorine is the most electronegative, only shows $0$ $0$ and $-1$ $-1$. Cannot be oxidized to a positive state. (No)
$NO₂$ $NO_2$: Nitrogen is in $+4$ $+4$. Can go to $+5$ $+5$ ( $HNO₃$ $HNO_3$) and $+3$ $+3$ ( $HNO₂$ $HNO_2$). (Yes)
$K^+$ $K^+$: Potassium is in its highest oxidation state $+1$ $+1$. Cannot be oxidized further. (No)

Step 1: Final Count

The species that can undergo disproportionation are: $H₂O₂$ $H_2O_2$, $ClO₃^-$ $ClO_3^-$, $P₄$ $P_4$, $Cl₂$ $Cl_2$, $Cu⁺¹$ $Cu^{+1}$, and $NO₂$ $NO_2$. Total count is 6.

Pattern Recognition

Rule out maximum oxidation states ( $K^+$ $K^+$), minimum oxidation states, and elements like $F₂$ $F_2$ which never show positive oxidation states.

Chapter Mix

Class 11 Chemistry: Redox Reactions

Q89 jee_main_2024_30_jan_morning Balancing Redox Reactions

2MnO₄^- + bI^- + cH₂O arrow xI₂ + yMnO₂ + zOH^-

If the above equation is balanced with integer coefficients, the value of z is

Numerical Answer. Answer: 8 to 8

Solution

Core Logic

This is a redox reaction in a slightly basic/neutral medium (as indicated by $MnO₂$ $MnO_2$ product). Separate the reaction into two half-cells and balance using the ion-electron method.

Step 1: Reduction Half-reaction

MnO₄^- arrow MnO₂

Balance O by adding $H₂O$ $H_2O$ and H by adding $OH^-$ $OH^-$ (or balance with $H^+$ $H^+$ then convert):

MnO₄^- + 2H₂O arrow MnO₂ + 4OH^-

Balance charge by adding $e^-$ $e^-$:

MnO₄^- + 2H₂O + 3e^- arrow MnO₂ + 4OH^- (1)

Step 2: Oxidation Half-reaction

I^- arrow I₂

Balance I:

2I^- arrow I₂

Balance charge by adding $e^-$ $e^-$:

2I^- arrow I₂ + 2e^- (2)

Step 3: Combining halves

To cancel electrons, multiply equation (1) by 2 and equation (2) by 3:

2[MnO₄^- + 2H₂O + 3e^- arrow MnO₂ + 4OH^-]

3[2I^- arrow I₂ + 2e^-]

Add them together:

2MnO₄^- + 6I^- + 4H₂O arrow 2MnO₂ + 3I₂ + 8OH^-

Step 4: Conclusion

Comparing this with the given equation:

2MnO₄^- + bI^- + cH₂O arrow xI₂ + yMnO₂ + zOH^-

We see $b = 6, c = 4, x = 3, y = 2, z = 8$ $b = 6, c = 4, x = 3, y = 2, z = 8$. Therefore, the value of z is 8.

Chapter Mix

Class 11 Chemistry: Redox Reactions

Q85 jee_main_2024_31_jan_evening Balancing Redox Reactions

Number of moles of

H^+

ions required by 1 mole of

MnO₄^-

to oxidise oxalate ion to

CO₂

is ________

Numerical Answer. Answer: 8 to 8

Solution

Related Formula

2MnO₄^- + 5C₂O₄²⁻ + 16H^+ arrow 2Mn²⁺ + 10CO₂ + 8H₂O

Core Logic

From the balanced redox equation in an acidic medium, we can see the exact stoichiometry between permanganate, oxalate, and hydrogen ions. For $2$ $2$ moles of $MnO₄^-$ $MnO_4^-$, $16$ $16$ moles of $H^+$ $H^+$ are required.

Step 1: Calculating for 1 mole

For $1$ $1$ mole of $MnO₄^-$ $MnO_4^-$, the number of moles of $H^+$ $H^+$ ions required is:

(16)/(2) = 8

Note: The official NTA answer was given as 4 initially, but our experts confirm the standard balanced stoichiometry requires 8 moles of $H^+$ $H^+$. We are outputting the chemically correct value of 8.

Chapter Mix

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Chlorine undergoes disproportionation in alkaline medium as shown below: $a Cl₂(g) + b OH^-(aq) arrow c ClO^-(aq) + d Cl^-(aq) + e H₂O(l)$ The values of a, b, c and d in a balanced redox reaction are respectively :

Solution & Explanation

Core Logic

Step 1: Coefficient Matching

Chapter Mix

Reference Study Guides

More Redox Reactions Previous-Year Questions — Page 3

Solution

Core Logic

Step 1: Final Count

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Reduction Half-reaction

Step 2: Oxidation Half-reaction

Step 3: Combining halves

Step 4: Conclusion

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculating for 1 mole

Chapter Mix

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Chlorine undergoes disproportionation in alkaline medium as shown below: a Cl₂(g) + b OH^-(aq) arrow c ClO^-(aq) + d Cl^-(aq) + e H₂O(l)$$a Cl_2(g) + b OH^-(aq) \rightarrow c ClO^-(aq) + d Cl^-(aq) + e H_2O(l)$$ The values of a, b, c and d in a balanced redox reaction are respectively :

Solution & Explanation

Core Logic

Step 1: Coefficient Matching

Chapter Mix

Reference Study Guides

More Redox Reactions Previous-Year Questions — Page 3

Solution

Core Logic

Step 1: Final Count

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Reduction Half-reaction

Step 2: Oxidation Half-reaction

Step 3: Combining halves

Step 4: Conclusion

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculating for 1 mole

Chapter Mix

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Chlorine undergoes disproportionation in alkaline medium as shown below: $a Cl₂(g) + b OH^-(aq) arrow c ClO^-(aq) + d Cl^-(aq) + e H₂O(l)$ The values of a, b, c and d in a balanced redox reaction are respectively :