Redox Reactions and Electrochemistry — NEET Chemistry MCQs with answers

Redox Reactions and Electrochemistry is a Class 12 Chemistry chapter in the NEET (UG) syllabus. NEET720 has 1,051 reviewed practice questions on it, each with a quick answer and a step-by-step explanation. The 8 questions below are free and fixed, so you can bookmark this page; the full chapter, plus mistake tracking and spaced revision, is in the app.

Topics covered

Oxidation number and balancing · Galvanic cells, EMF and Nernst equation · Electrochemical cells · Electrolysis and Faraday laws · Conductance and Kohlrausch law · Batteries, fuel cells and corrosion · Batteries and Fuel Cells · Oxidation number rules · Equivalent concept · Standard electrode potentials · Nernst equation · Batteries and Corrosion · Nernst equation graph · Balancing half-reactions · Standard reduction potentials · Concentration cell · Galvanic vs electrolytic cells · Faraday's law of electrolysis · Molar conductivity vs concentration · Kohlrausch's law · Gibbs free energy and EMF · Electrode reactions · Equivalent conductivity · Molar conductivity vs sqrt(C) · Electrode sign convention · Faraday's laws · Oxidizing power of halogens · Standard cell EMF · Salt bridge function · Equilibrium and cell EMF · Equivalent weight from electrolysis · Types of electrodes · Conductometric titration · Nernst equation for concentration cell · Oxidation number · Oxidation number in polyatomic ion · Types of redox reactions · Cell EMF calculation · Nernst equation · Faraday's laws of electrolysis

8 free Redox Reactions and Electrochemistry practice questions with answers

Choose an answer in your head before opening it. Each explanation says why the correct option is right and, where relevant, why the tempting wrong option is wrong.

1. Question 1 · easy · Batteries, fuel cells and corrosion

   Which of the following statements correctly distinguishes a secondary cell from a primary cell?

   • A.A secondary cell can be recharged by passing current in the opposite direction, while a primary cell cannot be reused once discharged
   • B.A secondary cell produces a higher EMF than any primary cell
   • C.A primary cell uses a paste electrolyte while a secondary cell always uses a liquid electrolyte
   • D.A primary cell requires an external power source to operate, while a secondary cell operates spontaneously
   Show answer and explanation

   Answer: A. A secondary cell can be recharged by passing current in the opposite direction, while a primary cell cannot be reused once discharged

   Secondary cells (e.g., lead storage, Ni-Cd) are reversible: after discharge, passing current in the reverse direction regenerates the original reactants, so the cell can be reused. Primary cells (e.g., dry cell, mercury cell) undergo an irreversible reaction and are discarded after use.

   In a primary cell, the electrode reactions proceed only once in the forward (discharge) direction because the products cannot be efficiently converted back to reactants — the cell reaction is effectively irreversible under practical conditions. In a secondary (storage) cell, the discharge reaction is chemically reversible: applying an external EMF greater than the cell's own EMF in the opposite sense drives the reaction backward, regenerating the original electrode materials (e.g., in lead storage battery, PbSO4 formed on both electrodes is converted back to Pb and PbO2 during charging). This reversibility is the defining criterion, not EMF magnitude or electrolyte state.

   Common mistake: Assuming the electrolyte's physical state (paste vs liquid) determines rechargeability rather than the reversibility of the electrode reaction

2. Question 2 · easy · Batteries, fuel cells and corrosion

   In the lead storage battery, what serves as the electrolyte?

   • A.Aqueous NH4Cl and ZnCl2 paste
   • B.38% aqueous sulphuric acid (density about 1.28 g/mL)
   • C.Potassium hydroxide solution
   • D.Molten cryolite-alumina mixture
   Show answer and explanation

   Answer: B. 38% aqueous sulphuric acid (density about 1.28 g/mL)

   The lead storage battery uses about 38% aqueous H2SO4 as its electrolyte, with Pb as anode and PbO2 as cathode; during discharge both electrodes convert to PbSO4 and the H2SO4 concentration decreases.

   Lead storage battery: anode = spongy Pb, cathode = PbO2 (on Pb grid), electrolyte = 38% H2SO4. Discharge reactions: Anode: Pb + SO4^2- → PbSO4 + 2e-. Cathode: PbO2 + SO4^2- + 4H+ + 2e- → PbSO4 + 2H2O. Overall: Pb + PbO2 + 2H2SO4 → 2PbSO4 + 2H2O. As H2SO4 is consumed and water is produced, the electrolyte's density falls, which is used to monitor charge state. On recharging, the reverse reaction regenerates Pb, PbO2 and H2SO4.

   Common mistake: Mixing up electrolytes of different named batteries (dry cell vs lead storage vs Ni-Cd)

3. Question 3 · medium · Batteries, fuel cells and corrosion

   In the H2-O2 fuel cell using porous carbon electrodes and aqueous KOH electrolyte, what happens at the cathode during operation?

   • A.O2 is reduced: O2 + 2H2O + 4e- → 4OH-
   • B.H2 is oxidized: H2 + 2OH- → 2H2O + 2e-
   • C.O2 is oxidized to O3 releasing electrons
   • D.Water is oxidized to O2 and H+ ions
   Show answer and explanation

   Answer: A. O2 is reduced: O2 + 2H2O + 4e- → 4OH-

   In the H2-O2 fuel cell, H2 is oxidized at the anode and O2 is reduced at the cathode. The cathode half-reaction in alkaline medium is O2 + 2H2O + 4e- → 4OH-.

   The H2-O2 fuel cell converts chemical energy directly to electrical energy. At the anode: 2H2 + 4OH- → 4H2O + 4e- (oxidation of H2). At the cathode: O2 + 2H2O + 4e- → 4OH- (reduction of O2). Overall: 2H2 + O2 → 2H2O. Electrons flow from anode to cathode through the external circuit, and OH- ions migrate through the electrolyte to maintain charge balance. Only pure water is the product, making this an environmentally clean power source used in spacecraft.

   Common mistake: Swapping anode and cathode half-reactions, or believing O2 is oxidized instead of reduced

4. Question 4 · medium · Batteries, fuel cells and corrosion

   A mercury cell delivers a nearly constant voltage of 1.35 V throughout its useful life. If such a cell supplies a steady current of 50 mA for 2 hours, calculate the total electrical work done by the cell (in joules). (Assume ideal constant-voltage operation.)

   • A.486 J
   • B.162 J
   • C.6.75 J
   • D.972 J
   Show answer and explanation

   Answer: A. 486 J

   Electrical work W = V x I x t. Convert current to amperes (0.05 A) and time to seconds (7200 s): W = 1.35 x 0.05 x 7200 = 486 J.

   Given: V = 1.35 V, I = 50 mA = 0.05 A, t = 2 h = 2 x 3600 s = 7200 s. Electrical work (energy delivered) W = V x I x t = 1.35 x 0.05 x 7200. Step 1: 1.35 x 0.05 = 0.0675. Step 2: 0.0675 x 7200 = 486 J. So the cell does 486 J of electrical work.

   Common mistake: Failing to convert time to seconds or current to amperes before substituting into W = VIt

5. Question 5 · hard · Batteries, fuel cells and corrosion

   During recharging of a lead storage battery, a current of 10 A is passed for 965 seconds through the cell, driving the reverse of 2PbSO4 + 2H2O → Pb + PbO2 + 2H2SO4. Calculate the total mass of PbSO4 (molar mass 303 g/mol) consumed at the two electrodes combined. (F = 96500 C/mol; the overall reaction involves a 2-electron transfer per 2 mol PbSO4, i.e., 1 mol electrons per 1 mol PbSO4 consumed.)

   • A.30.3 g
   • B.60.6 g
   • C.15.15 g
   • D.3.03 g
   Show answer and explanation

   Answer: A. 30.3 g

   Q = It = 10 x 965 = 9650 C. Moles of electrons = 9650/96500 = 0.1 mol. Since the electron:PbSO4 ratio is 1:1, moles PbSO4 = 0.1 mol, mass = 0.1 x 303 = 30.3 g.

   Step 1: Charge Q = I x t = 10 x 965 = 9650 C. Step 2: Moles of electrons n(e-) = Q/F = 9650/96500 = 0.1 mol. Step 3: In the reaction 2PbSO4 + 2H2O -> Pb + PbO2 + 2H2SO4, 2 mol electrons are transferred through the external circuit while 2 mol PbSO4 (1 mol at each electrode) are converted — a 1:1 ratio of mol electrons to mol PbSO4 consumed overall. Step 4: moles PbSO4 = n(e-) = 0.1 mol. Step 5: mass = 0.1 x 303 = 30.3 g.

   Common mistake: Using wrong electron:PbSO4 stoichiometric ratio (2:1 or 1:2 instead of 1:1)

6. Question 6 · medium · Batteries, fuel cells and corrosion

   A nickel-cadmium (Ni-Cd) cell discharges according to Cd + 2NiO(OH) + 2H2O → Cd(OH)2 + 2Ni(OH)2. If the cell delivers a steady current of 2 A for 965 seconds, calculate the mass of cadmium consumed (Cd = 112 g/mol).

   • A.2.24 g
   • B.5.6 g
   • C.1.12 g
   • D.0.56 g
   Show answer and explanation

   Answer: C. 1.12 g

   Q = It = 2 x 965 = 1930 C. n(e-) = 1930/96500 = 0.02 mol. Since Cd loses 2 electrons per atom (Cd → Cd2+ + 2e-), moles Cd = 0.01 mol, mass = 0.01 x 112 = 1.12 g.

   Step 1: Q = It = 2 x 965 = 1930 C. Step 2: n(e-) = Q/F = 1930/96500 = 0.02 mol. Step 3: At the Cd anode, Cd → Cd2+ + 2e-, a 2-electron oxidation, so moles Cd = n(e-)/2 = 0.02/2 = 0.01 mol. Step 4: mass = 0.01 x 112 = 1.12 g.

   Common mistake: Using the wrong electron:metal stoichiometric ratio for a 2-electron oxidation

7. Question 7 · medium · Batteries, fuel cells and corrosion

   In a dry (Leclanché) cell, the cathode reaction is MnO2 + NH4+ + e- → MnO(OH) + NH3, a 1-electron reduction per MnO2. If a current of 965 mA flows for 1000 seconds, calculate the mass of MnO2 consumed (MnO2 = 87 g/mol).

   • A.1.74 g
   • B.0.435 g
   • C.8.7 g
   • D.0.87 g
   Show answer and explanation

   Answer: D. 0.87 g

   Q = It = 0.965 x 1000 = 965 C. n(e-) = 965/96500 = 0.01 mol. Since the reduction is 1 electron per MnO2, moles MnO2 = 0.01 mol, mass = 0.01 x 87 = 0.87 g.

   Step 1: I = 965 mA = 0.965 A. Step 2: Q = It = 0.965 x 1000 = 965 C. Step 3: n(e-) = Q/F = 965/96500 = 0.01 mol. Step 4: The cathode half-reaction MnO2 + NH4+ + e- → MnO(OH) + NH3 shows a 1:1 ratio of electrons to MnO2, so moles MnO2 = 0.01 mol. Step 5: mass = 0.01 x 87 = 0.87 g.

   Common mistake: Assuming a 2-electron reduction for MnO2 by analogy with other metal oxide reductions

8. Question 8 · medium · Batteries, fuel cells and corrosion

   In the H2-O2 fuel cell, the overall reaction is 2H2 + O2 → 2H2O, requiring 2 mol electrons per mol of water formed. If a total charge of 1.93 x 10^5 C is passed through the cell, calculate the mass of water produced (H2O = 18 g/mol).

   • A.36 g
   • B.18 g
   • C.9 g
   • D.4.5 g
   Show answer and explanation

   Answer: B. 18 g

   n(e-) = Q/F = 1.93x10^5/96500 = 2 mol. Since 2 mol electrons form 1 mol H2O (overall reaction: 4e- form 2 mol H2O), moles H2O = 1 mol, mass = 18 g.

   Step 1: n(e-) = Q/F = 193000/96500 = 2 mol. Step 2: Overall cell reaction 2H2 + O2 → 2H2O transfers 4 mol electrons for every 2 mol H2O formed, i.e., a 2:1 ratio of electrons to water. Step 3: moles H2O = n(e-)/2 = 2/2 = 1 mol. Step 4: mass H2O = 1 x 18 = 18 g.

   Common mistake: Using the wrong electron:water stoichiometric ratio from the overall balanced equation

Questions about Redox Reactions and Electrochemistry for NEET

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