Redox Reactions and Electrode Processes redox reactions,: NEET MCQs with Answers

15 free questions on Redox Reactions and Electrode Processes redox reactions, for NEET (UG), starting with previous-year NEET and AIPMT questions. Try each one, then tap Show answer for the answer and a short explanation.

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Question 1

Previous-year question
Which option correctly describes a redox couple? [NEET 2023 (Manipur)]
  1. Both are oxidised forms involving the same element.
  2. Both are reduced forms involving the same element.
  3. Both the reduced and oxidised forms involving the same element.
  4. Cathode and anode together.
Show answer
Answer: (C) Both the reduced and oxidised forms involving the same element.
Explanation: A redox couple consists of the oxidised and reduced forms of the same substance taking part in an oxidation or reduction half-reaction, e.g. Zn²⁺/Zn or Cu²⁺/Cu (written oxidised form/reduced form). It does not contain two oxidised or two reduced forms, and the cathode–anode combination is a cell, not a couple.

Question 2

Which statement correctly describes the reaction NaH(s) + H₂O(l) → NaOH(aq) + H₂(g)?
  1. Hydride hydrogen is oxidised from –1 to 0 and one hydrogen of water is reduced from +1 to 0, both ending up in H₂
  2. Water is the reducing agent, because its hydrogen goes to a lower oxidation number
  3. It is not a redox reaction, because the oxidation numbers of sodium and oxygen do not change
  4. NaH is the oxidising agent, because hydride hydrogen is reduced from –1 to 0
Show answer
Answer: (A) Hydride hydrogen is oxidised from –1 to 0 and one hydrogen of water is reduced from +1 to 0, both ending up in H₂
Explanation: Hydrogen in NaH is –1 and hydrogen in water is +1, and each H₂ molecule is formed from one atom of each kind. Hydride loses an electron and is oxidised, so NaH is the reducing agent. A hydrogen of water gains that electron and is reduced, so water is the oxidising agent, even though sodium and oxygen do not change.

Question 3

Which metal does not react with steam but still displaces hydrogen from dilute hydrochloric acid?
  1. Sodium
  2. Iron
  3. Silver
  4. Tin
Show answer
Answer: (D) Tin
Explanation: Hydrogen is displaced from acids more easily than from water, so some metals that do not react with steam can still release it from acids; tin and cadmium are examples. Sodium reacts even with cold water, and iron reacts with steam. Silver lies below hydrogen in the activity series and does not displace hydrogen from dilute acid.

Question 4

Assertion: Hydrogen chloride can be prepared by heating sodium chloride with concentrated H₂SO₄, but pure hydrogen iodide cannot be prepared by heating sodium iodide with it. Reason: HI is a much stronger reducing agent than HCl, so it reduces concentrated H₂SO₄ and is itself oxidised to I₂.
  1. Assertion is true, Reason is false
  2. Assertion is true, Reason is true, but Reason does not explain Assertion
  3. Assertion is true, Reason is true, and Reason correctly explains Assertion
  4. Assertion is false, Reason is true
Show answer
Answer: (C) Assertion is true, Reason is true, and Reason correctly explains Assertion
Explanation: With NaCl, concentrated H₂SO₄ acts only as a non-volatile acid, and sulphur keeps its +6 state. HI is a strong reductant, so the HI first formed reduces the acid and is oxidised to iodine. Reducing strength of the hydrogen halides rises from HCl to HI, which explains the difference.

Question 5

In the titration of liberated iodine against sodium thiosulphate solution, starch is used as the indicator. Which procedure is correct?
  1. Starch is added when the iodine colour has faded to pale yellow, and the end point is the disappearance of the blue colour
  2. Starch is added at the very start, and the end point is the change from colourless to pale yellow
  3. Starch is added at the very start, and the end point is the first appearance of a blue colour
  4. Starch is added after the end point, to confirm that a blue colour remains
Show answer
Answer: (A) Starch is added when the iodine colour has faded to pale yellow, and the end point is the disappearance of the blue colour
Explanation: Starch forms an intense blue complex with iodine, and the colour vanishes when thiosulphate has reduced the last of the iodine to iodide. It is added near the end, when the solution is pale yellow, because a large amount of iodine binds strongly to starch and is released only slowly. Early addition would therefore blur the end point.

Question 6

In which of the following reactions does concentrated H₂SO₄ act only as an acid, and not as an oxidising agent?
  1. Zn + 2H₂SO₄ → ZnSO₄ + SO₂ + 2H₂O
  2. C + 2H₂SO₄ → CO₂ + 2SO₂ + 2H₂O
  3. 2HBr + H₂SO₄ → Br₂ + SO₂ + 2H₂O
  4. 2NaCl + H₂SO₄ → Na₂SO₄ + 2HCl
Show answer
Answer: (D) 2NaCl + H₂SO₄ → Na₂SO₄ + 2HCl
Explanation: Sulphuric acid acts as an oxidising agent only when sulphur is reduced from +6, as shown here by the formation of SO₂. With sodium chloride, sulphur stays at +6 in Na₂SO₄ and no element changes its oxidation number. The acid simply supplies protons and drives off volatile HCl.

Question 7

A copper rod is placed in colourless AgNO₃ solution (beaker X), and a silver rod is placed in blue Cu(NO₃)₂ solution (beaker Y). Given E°(Ag⁺/Ag) = +0.80 V and E°(Cu²⁺/Cu) = +0.34 V, what is observed?
  1. Both solutions show a change, because each metal displaces the other from solution
  2. Neither solution shows a change, because both metals lie below hydrogen in the series
  3. X stays colourless, while the blue colour of Y fades and copper deposits on the silver rod
  4. X gradually turns blue and silver deposits on the copper rod, while Y shows no change
Show answer
Answer: (D) X gradually turns blue and silver deposits on the copper rod, while Y shows no change
Explanation: Ag⁺ has the higher reduction potential, so copper reduces Ag⁺ with E°cell = +0.46 V. In X, copper dissolves as blue Cu²⁺ and silver metal deposits, so the blue colour grows with time. The reverse change in Y has a negative E°cell, so nothing happens.
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Question 8

Assertion: The compound AgF₂, if formed, acts as a very strong oxidising agent. Reason: Silver is in the unusual +2 state in AgF₂ and readily gains an electron to reach the more stable +1 state.
  1. Assertion and Reason are both true, and Reason correctly explains Assertion
  2. Assertion is false but Reason is true
  3. Assertion and Reason are both true, but Reason does not explain Assertion
  4. Assertion is true but Reason is false
Show answer
Answer: (A) Assertion and Reason are both true, and Reason correctly explains Assertion
Explanation: Silver commonly shows the +1 state, and +2 is unstable for it. Fluorine, the strongest oxidising element, is needed to reach this state. Ag²⁺ therefore takes up an electron very readily to form Ag⁺, oxidising other species in the process, so the Reason explains the Assertion.

Question 9

One mole each of CuSO₄, K₂Cr₂O₇, H₂O₂ and O₃ is treated separately with excess KI in acidic medium. Which two of them liberate the same number of moles of I₂?
  1. CuSO₄ and O₃
  2. H₂O₂ and O₃
  3. K₂Cr₂O₇ and O₃
  4. CuSO₄ and H₂O₂
Show answer
Answer: (B) H₂O₂ and O₃
Explanation: H₂O₂ + 2I⁻ + 2H⁺ → 2H₂O + I₂ and O₃ + 2I⁻ + 2H⁺ → O₂ + I₂ + H₂O each give 1 mol of I₂. In 2Cu²⁺ + 4I⁻ → 2CuI + I₂, one mole of Cu²⁺ gives only 0.5 mol of I₂. One mole of dichromate takes up 6 electrons and gives 3 mol of I₂.

Question 10

In the reaction PbO₂ + 4HCl → PbCl₂ + Cl₂ + 2H₂O, what fraction of the HCl taken is actually oxidised?
  1. One-half
  2. One-fourth
  3. Three-fourths
  4. All of it
Show answer
Answer: (A) One-half
Explanation: Of the four chloride ions, two become Cl₂ and rise from –1 to 0, so they are oxidised. The other two stay at –1 in PbCl₂ and act only as an acid and a source of chloride. Hence 2 out of 4 HCl units, or one-half, are oxidised, matching the 2 electrons gained by Pb(+4) going to Pb(+2).

Question 11

Assertion: Fluorine is not used in aqueous solution to displace chlorine from chloride ions. Reason: Fluorine oxidises water to oxygen in preference to reacting with the chloride ions.
  1. Assertion is false, Reason is true
  2. Assertion is true, Reason is true, but Reason does not explain Assertion
  3. Assertion is true, Reason is false
  4. Assertion is true, Reason is true, and Reason correctly explains Assertion
Show answer
Answer: (D) Assertion is true, Reason is true, and Reason correctly explains Assertion
Explanation: Fluorine is so strong an oxidant that it attacks water itself: 2F₂ + 2H₂O → 4H⁺ + 4F⁻ + O₂. Because this reaction takes over in water, displacement of Cl₂, Br₂ or I₂ by F₂ is generally not carried out in aqueous solution. The Reason therefore explains the Assertion directly.

Question 12

Given E°: Fe³⁺/Fe²⁺ = +0.77 V, Sn⁴⁺/Sn²⁺ = +0.15 V, I₂/I⁻ = +0.54 V and Br₂/Br⁻ = +1.09 V. Which pair of species can remain together in aqueous solution without reacting under standard conditions?
  1. Br₂ and Fe²⁺
  2. Fe³⁺ and Br⁻
  3. Fe³⁺ and Sn²⁺
  4. Fe³⁺ and I⁻
Show answer
Answer: (B) Fe³⁺ and Br⁻
Explanation: A redox reaction is feasible when the oxidant couple has a higher E° than the reductant couple. Fe³⁺ (+0.77 V) can oxidise Sn²⁺ (+0.15 V) and I⁻ (+0.54 V), and Br₂ (+1.09 V) can oxidise Fe²⁺. Fe³⁺ cannot oxidise Br⁻ because +0.77 V is below +1.09 V, so these two can coexist.

Question 13

The standard reduction potentials are Al³⁺/Al = –1.66 V, Zn²⁺/Zn = –0.76 V, Fe²⁺/Fe = –0.44 V, Ni²⁺/Ni = –0.25 V and Cu²⁺/Cu = +0.34 V. Judging from these values alone, a NiSO₄ solution can be stored without loss of Ni²⁺ in a vessel made of
  1. copper
  2. aluminium
  3. iron
  4. zinc
Show answer
Answer: (A) copper
Explanation: A metal with a lower reduction potential than Ni²⁺/Ni reduces Ni²⁺ and goes into solution, depositing nickel. Zinc, iron and aluminium all lie below nickel, so each would lower the Ni²⁺ concentration. Copper has a higher potential and cannot reduce Ni²⁺.

Question 14

In the titration of Fe²⁺ in acidic solution, no separate indicator is added when KMnO₄ is the titrant, but an indicator such as diphenylamine is added when K₂Cr₂O₇ is the titrant. Which is the best reason for this difference?
  1. K₂Cr₂O₇ does not oxidise Fe²⁺ without an indicator acting as a catalyst
  2. KMnO₄ reacts faster with Fe²⁺ than K₂Cr₂O₇ does, so the end point is reached before any colour can form
  3. MnO₄⁻ is intensely coloured and Mn²⁺ is almost colourless, so the first excess drop gives a lasting pink; the orange-to-green change of dichromate is not sharp enough to see
  4. Dichromate is colourless in acidic solution, whereas permanganate is coloured in acidic solution
Show answer
Answer: (C) MnO₄⁻ is intensely coloured and Mn²⁺ is almost colourless, so the first excess drop gives a lasting pink; the orange-to-green change of dichromate is not sharp enough to see
Explanation: Permanganate acts as its own indicator: MnO₄⁻ is deep purple while Mn²⁺ is nearly colourless, so the first slight excess leaves a permanent pink tint. Dichromate is orange and is reduced to green Cr³⁺, and this gradual colour change does not mark the end point clearly. A redox indicator such as diphenylamine is therefore added to the solution when dichromate is used.

Question 15

Which reaction is wrongly matched with its type?
  1. Ca(OH)₂ → CaO + H₂O : redox decomposition reaction
  2. Zn + H₂SO₄ → ZnSO₄ + H₂ : non-metal displacement redox reaction
  3. Fe + CuSO₄ → FeSO₄ + Cu : metal displacement redox reaction
  4. C + O₂ → CO₂ : redox combination reaction
Show answer
Answer: (A) Ca(OH)₂ → CaO + H₂O : redox decomposition reaction
Explanation: In the decomposition of Ca(OH)₂ the oxidation numbers of Ca (+2), O (−2) and H (+1) do not change, so it is not a redox reaction. Not every decomposition reaction is redox. The other three each involve a change in oxidation number: Zn and Fe are oxidised in displacement reactions, and carbon is oxidised in the combination.

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