Redox Reactions & Electrochemistry — Previous Year Questions (JEE Main 2026)
Solved JEE Main 2026 previous year questions on Redox Reactions and Electrochemistry — redox titrations, cell EMF, Nernst equation, molar conductivity and Kohlrausch law — with step-by-step solutions.
Solved JEE Main 2026 questions from the Redox Reactions and Electrochemistry chapter, covering redox titrations, electrochemical cells, the Nernst equation, fuel-cell work, molar conductivity, and Kohlrausch’s law, each with a concise step-by-step solution.
Solutions are AI-generated and pending review.
Solution
Each $\text{MnO}_4^-$ accepts $5$ electrons in acidic medium. Count the electrons the oxidisable species must give up.
Per mole of each compound:
- $\text{FeC}_2\text{O}_4$: $\text{Fe}^{2+}\!\to\!\text{Fe}^{3+}$ ($1e^-$) $+\ \text{C}_2\text{O}_4^{2-}\!\to\!2\text{CO}_2$ ($2e^-$) $= 3e^-$
- $\text{Fe}_2(\text{C}_2\text{O}_4)_3$: Fe already $+3$; $3$ oxalates $\to 3\times 2 = 6e^-$
- $\text{FeSO}_4$: $\text{Fe}^{2+}\!\to\!\text{Fe}^{3+} = 1e^-$
- $\text{Fe}_2(\text{SO}_4)_3$: nothing oxidisable $= 0e^-$
Total electrons released:
$$3 + 6 + 1 + 0 = 10\ e^-$$Moles of $\text{KMnO}_4$:
$$n = \frac{10}{5} = 2$$Answer: B (2)
Solution
Kohlrausch’s law builds $\text{BaSO}_4$ from the given electrolytes:
$$\Lambda^\circ_m(\text{BaSO}_4) = \Lambda^\circ_m(\text{BaCl}_2) + \Lambda^\circ_m(\text{H}_2\text{SO}_4) - 2\,\Lambda^\circ_m(\text{HCl}) = x_1 + x_2 - 2x_3$$(The $2\,\text{Cl}^-$ and $2\,\text{H}^+$ cancel, leaving $\text{Ba}^{2+} + \text{SO}_4^{2-}$.)
Molar conductivity links conductivity $x$ (S cm$^{-1}$) to solubility $S$ (mol L$^{-1}$). The factor $1000$ converts litres to cm$^3$:
$$\Lambda_m = \frac{1000\,x}{S} \implies S = \frac{1000\,x}{x_1 + x_2 - 2x_3}\ \text{mol L}^{-1}$$Solubility product ($1{:}1$ salt, $S = [\text{Ba}^{2+}] = [\text{SO}_4^{2-}]$):
$$K_{sp} = [\text{Ba}^{2+}][\text{SO}_4^{2-}] = S^2 = \frac{(1000\,x)^2}{(x_1 + x_2 - 2x_3)^2} = \frac{10^6\,x^2}{(x_1 + x_2 - 2x_3)^2}$$This is the expression in option A. (Option B drops the $10^6$ litre-to-cm$^3$ conversion; option C is its reciprocal $1/K_{sp}$; option D uses the wrong Kohlrausch combination $x_1 + x_2 + 2x_3$.)
Answer: A
Solution
For a spontaneous cell, the half-cell with the higher $E^0$ is the cathode (reduction).
- Cathode (higher $E^0$): $\text{AgBr}/\text{Ag}$, $E^0 = +0.07$ V
- Anode (lower $E^0$): $\text{Fe(OH)}_2/\text{Fe}$, $E^0 = -0.88$ V
Half-reactions:
$$\text{Anode (oxidation): } \text{Fe}(s) + 2\text{OH}^- \to \text{Fe(OH)}_2(s) + 2e^-$$$$\text{Cathode (reduction): } 2\text{AgBr}(s) + 2e^- \to 2\text{Ag}(s) + 2\text{Br}^-$$Overall reaction:
$$\text{Fe}(s) + 2\text{OH}^-(aq) + 2\text{AgBr}(s) \rightleftharpoons \text{Fe(OH)}_2(s) + 2\text{Ag}(s) + 2\text{Br}^-(aq)$$This matches option A. (Fe is oxidised, not reduced; $E^0_{cell} = +0.95$ V, not $-0.95$; $E^0_{cell}$ is intensive.)
Answer: A
Solution
$\text{K}_2\text{Cr}_2\text{O}_7$ (orange, $\text{Cr}^{+6}$) is a strong oxidising agent; it changes colour ($\to$ green $\text{Cr}^{3+}$) only when it meets a reducing agent. Every species in the correct set must be able to be oxidised further.
Check each option:
- A: $\text{Fe}^{2+}\!\to\!\text{Fe}^{3+}$, $\text{Sn}^{2+}\!\to\!\text{Sn}^{4+}$, $I^-\!\to\!I_2$, $S^{2-}\!\to\!S$ — all are reducing agents. ✓
- B: $\text{Fe}^{3+}$ is already oxidised (not a reductant). ✗
- C: $\text{Sn}^{4+}$ is the highest common state (not a reductant). ✗
- D: $\text{Fe}^{3+}$, $\text{SO}_4^{2-}$, $\text{Sn}^{4+}$ are all in their highest states (not reductants). ✗
Answer: A
Solution
$\text{M}^{x+}/\text{M}$ has the higher $E^\ominus$, so it is the cathode; $\text{Fe}^{3+}/\text{Fe}$ is the anode.
$$E^\ominus_{cell} = 0.15 - (-0.036) = 0.186\ \text{V}$$Nernst equation with $Q = 10^{-2}$ ($\log_{10}Q = -2$):
$$E_{cell} = E^\ominus_{cell} - \frac{0.059}{n}\log_{10}Q$$$$0.2057 = 0.186 - \frac{0.059}{n}(-2) = 0.186 + \frac{0.118}{n}$$$$\frac{0.118}{n} = 0.0197 \implies n = \frac{0.118}{0.0197} \approx 6$$Balance electrons in the overall reaction:
- Anode: $\text{Fe} \to \text{Fe}^{3+} + 3e^-$
- Cathode: $\text{M}^{x+} + x\,e^- \to \text{M}$
Total transferred $n = 3x$ (LCM of the two half-reactions), so:
$$3x = 6 \implies x = 2$$(A $+2$ p-block cation, e.g. $\text{Sn}^{2+}$/$\text{Pb}^{2+}$, is consistent.)
Answer: 2
Solution
In the fuel cell, $\text{CH}_3\text{OH}$ is the fuel (oxidised at the anode); $\text{O}_2$ is reduced at the cathode.
$$E^\circ_{cell} = E^\circ_{cath} - E^\circ_{an} = 1.23 - 0.02 = 1.21\ \text{V}$$Oxidation of $1$ mol $\text{CH}_3\text{OH}\to\text{CO}_2$ transfers $n = 6$ electrons.
Maximum electrical work:
$$W_{max} = nFE^\circ_{cell} = 6 \times 96500 \times 1.21 = 700{,}590\ \text{J}$$Useful work at 80% efficiency:
$$W = 0.80 \times 700{,}590 = 560{,}472\ \text{J}$$Isothermal compression against constant $P = 1\ \text{kPa} = 1000\ \text{Pa}$:
$$W = P\,|\Delta V| \implies |\Delta V| = \frac{560{,}472}{1000} = 560.47\ \text{m}^3$$Answer: 560
Solution
Evaluate each statement.
- Option 1 (correct): Cross-multiplying to balance electrons ($n_1 n_2$ each) gives $n_2\text{Red}_1+n_1\text{Ox}_2 \rightleftharpoons n_2\text{Ox}_1^{n_1+}+n_1\text{Red}_2^{n_2-}$. ✓
- Option 2 (correct): Electrons produced at the anode are consumed at the cathode, so they cancel in the overall equation. ✓
- Option 3 (correct): From the Nernst equation $E = E^\circ - \dfrac{RT}{nF}\ln Q$, rearranging gives $\dfrac{E-E^\circ}{RT/F} = -\dfrac{2.303}{n}\log_{10}Q$, so the slope magnitude $\propto \dfrac{1}{n}$. ✓
- Option 4 (INCORRECT): Electrical work $= \text{charge}\times\text{potential difference}$ (a product), i.e. $W = qE$, not the ratio $q/E$. ✗
Answer: D
Solution
Moles of $\text{MnO}_4^-$: $0.500 \times 0.2 = 0.1$ mol.
In basic (neutral) medium, $\text{MnO}_4^-$ is reduced to $\text{MnO}_2$, gaining $3\,e^-$ each:
$$\text{electrons available} = 0.1 \times 3 = 0.3\ \text{mol}\,e^-$$Iodide is oxidised $2I^- \to I_2 + 2e^-$ ($1\,e^-$ per $I^-$), so $0.3$ mol $I^-$ is oxidised:
$$n_{I_2} = \frac{0.3}{2} = 0.15\ \text{mol}$$(KI available $= 0.75$ mol, so $\text{MnO}_4^-$ is limiting — consistent.)
Thiosulphate titration: $I_2 + 2\text{S}_2\text{O}_3^{2-} \to 2I^- + \text{S}_4\text{O}_6^{2-}$
$$n_{\text{thio}} = 2 \times 0.15 = 0.30\ \text{mol}$$$$x = \frac{n_{\text{thio}}}{V} = \frac{0.30}{0.300} = 1\ \text{M}$$Answer: 1
Solution
Ag is the cathode, Zn the anode:
$$E^\circ_{cell} = 0.80 - (-0.76) = 1.56\ \text{V}$$Overall: $\text{Zn} + 2\text{Ag}^+ \to \text{Zn}^{2+} + 2\text{Ag}$, with $n = 2$ and
$$Q = \frac{[\text{Zn}^{2+}]}{[\text{Ag}^+]^2} = \frac{1}{x^2}$$Nernst equation:
$$E_{cell} = E^\circ_{cell} - \frac{0.059}{2}\log_{10}\frac{1}{x^2} = 1.56 - \frac{0.059}{2}\,(-2\log x) = 1.56 + 0.059\log x$$$$1.60 = 1.56 + 0.059\log x \implies 0.04 = 0.059\log x$$$$\log x = \frac{0.04}{0.059} = \frac{4}{5.9}$$Answer: B
Solution
Conductivity $\kappa = G \times \text{(cell constant)}$:
$$\kappa = (1.9\times10^{-3}) \times 1.3 = 2.47\times10^{-3}\ \text{S cm}^{-1}$$Concentration from $\Lambda_m = \dfrac{1000\,\kappa}{C}$:
$$C = \frac{1000\,\kappa}{\Lambda_m} = \frac{1000 \times 2.47\times10^{-3}}{123.5} = 0.02\ \text{mol L}^{-1}$$Mass of MX per litre: $0.02 \times 75 = 1.5$ g. Mass of solution per litre (density $1.0$ g mL$^{-1}$) $= 1000$ g.
Percentage (w/w):
$$x\% = \frac{1.5}{1000}\times100 = 0.15\% = 15\times10^{-2}\%$$Answer: 15
Solution
Potentials are not additive, but Gibbs energies ($\Delta G^\circ = -nFE^\circ$) are. Combine the half-reactions:
$$\text{Fe}^{2+} + 2e^- \to \text{Fe} \quad (n=2):\ \Delta G_1 = -2FX$$$$\text{Fe}^{3+} + 3e^- \to \text{Fe} \quad (n=3):\ \Delta G_2 = -3FY$$Target: $\text{Fe}^{3+} + e^- \to \text{Fe}^{2+}\ (n=1)$, obtained as $\Delta G_2 - \Delta G_1$:
$$\Delta G_3 = \Delta G_2 - \Delta G_1 = -3FY - (-2FX) = -3FY + 2FX$$Since $\Delta G_3 = -1\cdot F\cdot E^\ominus_{Fe^{3+}/Fe^{2+}}$:
$$-F\,E^\ominus = -3FY + 2FX \implies E^\ominus_{Fe^{3+}/Fe^{2+}} = 3Y - 2X$$Answer: B