Redox Reactions and Electrochemistry JEE 2023: H2S Solution

What is the redox reactions and electrochemistry JEE 2023 numerical question?

The answer is 18: 5 mol of hydrogen sulfide produces 8 mol of water and transfers 10 mol of electrons. This redox reactions and electrochemistry JEE 2023 question appeared in JEE Advanced 2023, Paper 2, Chemistry, in numerical-answer format.

A 5 mol sample of hydrogen sulfide is completely consumed by aqueous potassium permanganate in acidic conditions. Find the sum of the moles of water formed and electrons transferred, defined as:

x = n (H_{2} O), y = n (e^{-})

The required numerical entry is: $x + y$

The question bank classifies this as medium difficulty, not an official exam classification. No options were supplied because this is a numerical-answer question.

How do you build the two half-reactions in acidic medium?

Use the official products: permanganate is reduced to manganese(II), while hydrogen sulfide is oxidised to elemental sulfur. Balance oxygen with water, hydrogen with hydrogen ions, and charge with electrons, in that order. The half-reactions keep water production and electron transfer in separate accounts.

For manganese, the oxidation-number change is:

M n : + 7 \to + 2 (gain of 5 electrons)

Start with the reduction skeleton:

M n O_{4}^{-} \to M n^{2 +}

Four oxygen atoms require four water molecules on the right:

M n O_{4}^{-} \to M n^{2 +} + 4 H_{2} O

Balance those eight hydrogen atoms by adding hydrogen ions on the left:

M n O_{4}^{-} + 8 H^{+} \to M n^{2 +} + 4 H_{2} O

Before electrons are added, the charges are:

Q_{l e f t} = - 1 + 8 = + 7, Q_{r i g h t} = + 2

Five electrons on the left bring its charge down to match the right:

M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O

For sulfur, the oxidation-number change is:

S : - 2 \to 0 (loss of 2 electrons)

Start with the oxidation skeleton, then balance hydrogen: $H_{2} S \to S$ $H_{2} S \to S + 2 H^{+}$

The left side is neutral. Add two electrons on the right to make that side neutral too:

H_{2} S \to S + 2 H^{+} + 2 e^{-}

How do you equalise electrons and combine the half-reactions?

Multiply the reduction half-reaction by 2 and the oxidation half-reaction by 5. The least common multiple of the electron counts, 5 and 2, is 10, so these multipliers make the electrons accepted equal the electrons released. Then add the reactions and cancel matching terms.

The scaled reduction is:

2 M n O_{4}^{-} + 16 H^{+} + 10 e^{-} \to 2 M n^{2 +} + 8 H_{2} O

The scaled oxidation is:

5 H_{2} S \to 5 S + 10 H^{+} + 10 e^{-}

Add the equations and cancel the matching ten-electron terms:

2 M n O_{4}^{-} + 16 H^{+} + 5 H_{2} S \to 2 M n^{2 +} + 5 S + 10 H^{+} + 8 H_{2} O

Now cancel ten hydrogen ions from each side. Six hydrogen ions remain on the reactant side: $16 - 10 = 6$

The final net ionic equation is:

2 M n O_{4}^{-} + 6 H^{+} + 5 H_{2} S \to 2 M n^{2 +} + 5 S + 8 H_{2} O

This equation gives net consumption and production. Keep the scaled half-reactions for the electron-transfer count.

How do you calculate the answer and check that 18 is correct?

The given 5 mol of hydrogen sulfide matches its coefficient in the net equation, so no further scaling is needed. Read water production from the net equation and electron transfer from either scaled half-reaction.

The water formed is:

x = 5 \times \frac{8}{5} = 8 m o l

Each mole of hydrogen sulfide loses two moles of electrons: $y = 5 \times 2 = 10 m o l$

Equivalently, two moles of permanganate each accept five moles of electrons: $y = 2 \times 5 = 10 m o l$

The numerical entry is 18:

x + y = 8 + 10 = 18

Check atom conservation on both sides of the net equation:

Manganese: 2 atoms on each side.
Sulfur: 5 atoms on each side.
Oxygen: 8 atoms on each side.
Hydrogen: 16 atoms on each side, including hydrogen from both the acid and hydrogen sulfide.

Check charge independently:

Q_{l e f t} = 2 (- 1) + 6 (+ 1) = + 4

Q_{r i g h t} = 2 (+ 2) = + 4

Electrons cancel because they are transferred internally, not because zero electrons are involved. The donor loses exactly what the acceptor gains.

Why does counting electrons twice give 28 instead of 18?

The illustrative wrong numerical result 28 comes from adding electrons released to electrons accepted. It is not an actual answer option. The faulty calculation counts the same transferred electrons twice:

y_{w r o n g} = 10 + 10 = 20 m o l

(x + y)_{w r o n g} = 8 + 20 = 28

The ten moles released by hydrogen sulfide are the same ten moles accepted by permanganate. Correct accounting is:

electrons lost = electrons gained = electrons transferred = 10 m o l

Use either scaled half-reaction to calculate transferred electrons, never the sum of both counts.

How can you apply this method to two related acidic-medium questions?

Scale the balanced reaction to the stated amount, then count electrons from one half-reaction. Both questions below are original practice, not additional verified JEE PYQs.

What changes if only 2.5 mol of hydrogen sulfide reacts?

Original practice 1: For 2.5 mol of hydrogen sulfide undergoing the same reaction with the same products, find the water amount, electron-transfer amount and their sum. Every amount is halved because the sample is half the original amount:

scale factor = \frac{2.5}{5} = \frac{1}{2}

x = 8 \times \frac{1}{2} = 4 m o l

y = 10 \times \frac{1}{2} = 5 m o l

x + y = 9

Each mole of hydrogen sulfide still releases two moles of electrons. Changing the sample size changes the amounts, not the reaction ratios.

What happens when 1 mol of dichromate oxidises iron(II)?

Original practice 2: In acidic solution, 1 mol of dichromate is completely reduced to chromium(III) by iron(II), which becomes iron(III). Find the moles of iron(II) consumed, water produced and electrons transferred.

The answers are:

Iron(II) consumed: 6 mol.
Water produced: 7 mol.
Electrons transferred: 6 mol.

Check these using the reduction half-reaction:

C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 C r^{3 +} + 7 H_{2} O

Write the oxidation half-reaction and multiply it by 6:

F e^{2 +} \to F e^{3 +} + e^{-}

6 F e^{2 +} \to 6 F e^{3 +} + 6 e^{-}

Adding and cancelling electrons gives:

C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 F e^{2 +} \to 2 C r^{3 +} + 7 H_{2} O + 6 F e^{3 +}

The six iron(II) ions each lose one electron, matching the six accepted by dichromate. In both practice questions, apply the scale factor to every stoichiometric quantity, but count the electron transfer only once.

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Frequently asked questions

What is the answer to the JEE Advanced 2023 H2S and KMnO4 numerical?

The answer is 18. When 5 mol of H2S is oxidised to elemental sulfur by acidic permanganate, 8 mol of water forms and 10 mol of electrons is transferred. The required sum is 8 + 10 = 18.

How do you balance the H2S and permanganate reaction in acidic medium?

Write separate half-reactions for permanganate reduction to Mn²⁺ and H2S oxidation to sulfur. Balance oxygen with water, hydrogen with H⁺, and charge with electrons, then multiply the reduction by 2 and the oxidation by 5. Adding and cancelling gives 2MnO₄⁻ + 6H⁺ + 5H₂S → 2Mn²⁺ + 5S + 8H₂O.

Why is the electron transfer 10 mol and not 20 mol?

The 10 mol of electrons released by H2S is the same 10 mol accepted by permanganate. Adding electrons lost and gained counts the transfer twice. Use either half-reaction to find the transferred amount, not the sum of both.

What is the answer if 2.5 mol of H2S reacts instead of 5 mol?

For the same reaction and products, halve every stoichiometric amount. The water formed is 4 mol and the electron transfer is 5 mol, so their sum is 9. This is an original practice variation, not an additional verified JEE previous-year question.