Redox Reactions NEET PYQReal NEET & AIPMT 2005–2021 questions, year-tagged, with solutions
By the RankUp team, founded by Dr. Deepak Sahu (MBBS Gold Medalist, NEET PG AIR 1731) · Updated
Every Redox Reactions question from the NEET & AIPMT papers on this site (2005–2021), newest first. Each is tagged with its paper. Tap an option to check your answer and read the explanation.
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Redox Reactions previous year questions
12 questions. Tap an option to check your answer.
Q1NEET 2021
Which of the following reactions is the metal displacement reaction ? Choose the right option.
Not quite — the answer is B.
In Cr₂O₃ + 2Al → Al₂O₃ + 2Cr, aluminium displaces chromium metal from its oxide, a metal displacement reaction. Fe + 2HCl displaces hydrogen, not a metal, and the other two are decomposition reactions.
Q2NEET 2020 · Q125
What is the change in oxidation number of carbon in the following reaction ? CH₄(g) + 4Cl₂(g) → CCl₄(l) + 4HCl(g)
Not quite — the answer is C.
In CH₄, carbon is bonded to four less electronegative H atoms, so x + 4(+1) = 0 gives −4; in CCl₄, x + 4(−1) = 0 gives +4. Carbon is oxidised from −4 to +4.
Q3NEET 2018 · Q46
The correct order of N-compounds in its decreasing order of oxidation states is
Not quite — the answer is C.
Oxidation states of N: HNO₃ +5, NO +2, N₂ 0, NH₄Cl −3. So the decreasing order is HNO₃ > NO > N₂ > NH₄Cl; the reverse sequence is increasing order.
Q4NEET 2018 · Q82
For the redox reaction MnO₄⁻ + C₂O₄²⁻ + H⁺ → Mn²⁺ + CO₂ + H₂O, the correct coefficients of the reactants for the balanced equation are (MnO₄⁻, C₂O₄²⁻, H⁺)
Not quite — the answer is B.
Mn goes from +7 to +2 (gain 5e⁻) and each C₂O₄²⁻ loses 2e⁻. Equalising electrons: 2MnO₄⁻ + 5C₂O₄²⁻; balancing O and H gives 2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O.
Q5AIPMT 2015
Which of the following processes does not involve oxidation of iron?
Not quite — the answer is D.
In Fe(CO)₅ iron is in the zero oxidation state, as in the free metal, so no oxidation occurs. Iron is oxidised in rusting, in liberating H₂ from steam (to Fe₃O₄) and in displacing Cu from CuSO₄ (to Fe²⁺).
Q6AIPMT 2014
The pair of compounds that can exist together is:
Not quite — the answer is B.
FeCl₂ and SnCl₂ are both reducing agents, so they do not react with each other. SnCl₂ reduces HgCl₂ and FeCl₃, and FeCl₃ oxidises I⁻ to I₂, so those pairs cannot coexist.
Q7AIPMT 2012 Prelims
In which of the following compounds, nitrogen exhibits highest oxidation state?
Not quite — the answer is D.
Oxidation states of N: NH₂OH −1, N₂H₄ −2, NH₃ −3, and in N₃H (hydrazoic acid) 3x + 1 = 0, so x = −1/3. The highest (least negative) is −1/3 in N₃H. NH₂OH at −1 is the nearest trap.
Q8AIPMT 2010 Prelims · Q77
Oxidation states of P in H₄P₂O₅, H₄P₂O₆, H₄P₂O₇, are respectively
Not quite — the answer is A.
With H = +1 and O = −2: H₄P₂O₅: 4 + 2x − 10 = 0, x = +3; H₄P₂O₆: x = +4; H₄P₂O₇: x = +5. So +3, +4, +5.
Q9AIPMT 2009 Prelims · Q152
Oxidation numbers of P in PO₄³⁻, of S in SO₄²⁻ and that of Cr in Cr₂O₇²⁻, are respectively :
Not quite — the answer is D.
PO₄³⁻: x − 8 = −3, x = +5. SO₄²⁻: x − 8 = −2, x = +6. Cr₂O₇²⁻: 2x − 14 = −2, x = +6. So +5, +6 and +6.
Q10AIPMT 2008 Prelims · Q75
Number of moles of MnO₄⁻ required to oxidize one mole of ferrous oxalate completely in acidic medium will be
Not quite — the answer is B.
In FeC₂O₄ both parts are oxidised: Fe²⁺ → Fe³⁺ (1 e⁻) and C₂O₄²⁻ → 2CO₂ (2 e⁻), so 3 electrons per mole. Each MnO₄⁻ → Mn²⁺ takes 5 electrons, so 3/5 = 0.6 mol MnO₄⁻ is needed. 0.4 mol counts only the oxalate electrons.
Q11AIPMT 2007 Prelims · Q99
The number of moles of KMnO₄ that will be needed to react with one mole of sulphite ion in acidic solution is
Not quite — the answer is B.
SO₃²⁻ → SO₄²⁻ loses 2 electrons, while MnO₄⁻ → Mn²⁺ gains 5 in acidic medium. So 1 mol sulphite needs 2/5 mol KMnO₄. 4/5 counts the sulphite change twice.
Q12AIPMT 2005 Prelims · Q61
The number of moles of KMnO₄ reduced by one mole of KI in alkaline medium is :
Not quite — the answer is D.
In alkaline medium MnO₄⁻ → MnO₂ (gain 3e⁻) and I⁻ → IO₃⁻ (loss 6e⁻): 2MnO₄⁻ + I⁻ + H₂O → 2MnO₂ + IO₃⁻ + 2OH⁻. So one mole KI reduces two moles KMnO₄. 'One fifth' wrongly assumes MnO₄⁻ goes to Mn²⁺ (5 electrons) with I⁻ giving only 1 electron.
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