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Chemistry · chapter PYQs

Electrochemistry, on the exam screen.

6 questions from NEET 2026, Re-NEET 2026, NEET 2020, 6 minutes, marked +4 / −1. See them with solutions.

Electrochemistry · PYQs

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Q1/6

  1. Question 1 · NEET 2026 · Q73 · +4 / −1

    A solution of copper sulphate is electrolysed for 10 minutes with a current of 1.5 amperes. The mass of copper deposited at cathode is: (Given: Molar mass of Cu = 63 g mol−1^{-1}; 1 F = 96487 C mol−1^{-1})

    Answer and explanation

    Answer: (2)

    Charge passed Q=i t=1.5×(10×60)=900Q = i\,t = 1.5 \times (10\times 60) = 900 C. Deposition follows Cu2++2e−→Cu\mathrm{Cu}^{2+} + 2\mathrm{e}^- \rightarrow \mathrm{Cu}, so mass =E i tF=63×1.5×6002×96487≈0.294= \frac{E\,i\,t}{F} = \frac{63 \times 1.5 \times 600}{2 \times 96487} \approx 0.294 g, i.e. 0.2938 g. The other values arise from using the wrong n-factor or the wrong time.

  2. Question 2 · NEET 2026 · Q79 · +4 / −1

    Calculate emf of the half cell given below:
    Pt(s)∣H2(g,2atm)∣HCl(aq,0.02M)\mathrm{Pt}(\text{s}) | \mathrm{H}_{2}(\mathrm{g}, 2 \mathrm{atm}) | \mathrm{HCl}(\mathrm{aq}, 0.02 \mathrm{M})
    EH2/H+∘=0E^{\circ}_{H_2/H^+} = 0 V
    (Given: 2.303 RTF=0.059\frac{2.303\,RT}{F} = 0.059, log⁡2=0.3010\log 2 = 0.3010)

    Answer and explanation

    Answer: (4)

    For H2→2H++2e−\mathrm{H}_{2} \rightarrow 2\mathrm{H}^{+} + 2\mathrm{e}^{-}, the Nernst equation gives E=E∘−0.0592log⁡[H+]2pH2E = E^{\circ} - \frac{0.059}{2}\log\frac{[H^+]^2}{p_{H_2}}, with [H+^{+}] = 0.02 M and p(H2_{2}) = 2 atm. So E=0−0.0295log⁡(0.02)22=−0.0295log⁡(2×10−4)=−0.0295(0.301−4)≈+0.109E = 0 - 0.0295\log\frac{(0.02)^2}{2} = -0.0295\log(2\times10^{-4}) = -0.0295(0.301 - 4) \approx +0.109 V.

  3. Question 3 · Re-NEET 2026 · Q56 · +4 / −1

    The standard electrode potential (E°) for the half-cell reaction Fe3++e−→Fe2+\mathrm{Fe}^{3+} + \mathrm{e}^- \rightarrow \mathrm{Fe}^{2+} at 298 K is (Given : E°(Fe3+/Fe\mathrm{Fe}^{3+}/\mathrm{Fe}) = -0.04 V and E°(Fe2+/Fe\mathrm{Fe}^{2+}/\mathrm{Fe}) = -0.44 V at 298 K)

    Answer and explanation

    Answer: (2)

    Electrode potentials are NOT additive when the electron counts differ; the additive quantity is ΔG∘=−nFE∘\Delta G^\circ = -nFE^\circ. For Fe3++3e−→Fe\mathrm{Fe}^{3+} + 3\mathrm{e}^- \rightarrow \mathrm{Fe}, ΔG1∘=−3F(−0.04)=0.12F\Delta G_1^\circ = -3F(-0.04) = 0.12F; for Fe2++2e−→Fe\mathrm{Fe}^{2+} + 2\mathrm{e}^- \rightarrow \mathrm{Fe}, ΔG2∘=−2F(−0.44)=0.88F\Delta G_2^\circ = -2F(-0.44) = 0.88F. Subtracting, ΔG3∘=0.12F−0.88F=−0.76F\Delta G_3^\circ = 0.12F - 0.88F = -0.76F, and since n=1n = 1, E∘=+0.76 VE^\circ = +0.76\ \text{V}. NCERT Class 12, Electrochemistry (electrode potential and Gibbs energy).

  4. Question 4 · Re-NEET 2026 · Q62 · +4 / −1

    For a salt XY\mathbf{XY}, which is a strong electrolyte, the plot of Λm\Lambda_m versus c\sqrt{c} has a slope of -90.0 S cm2mol−3/2L1/2\text{cm}^2\text{mol}^{-3/2}\text{L}^{1/2} at 298 K. At 0.01 M concentration of XY\mathbf{XY}, the value of Λm\Lambda_m is 145.0 S cm2mol−1\text{cm}^2\text{mol}^{-1}. The limiting molar conductivity of Y−\mathbf{Y^-} ion (λY−0\lambda^0_{Y^-}, in S cm2mol−1\text{cm}^2\text{mol}^{-1}) at 298 K will be (Given : λX+0\lambda^0_{X^+} = 74.0 S cm2mol−1\text{cm}^2\text{mol}^{-1})

    Answer and explanation

    Answer: (1)

    For a strong electrolyte the Debye-Huckel-Onsager relation is Λm=Λm0−Ac\Lambda_m = \Lambda_m^0 - A\sqrt{c}, where the plot's slope is −A-A. At c=0.01 Mc = 0.01\ \text{M}, c=0.1\sqrt{c} = 0.1, so Λm0=145.0+(90.0)(0.1)=154.0\Lambda_m^0 = 145.0 + (90.0)(0.1) = 154.0. Kohlrausch's law then gives λY−0=154.0−74.0=80.0 S cm2mol−1\lambda^0_{Y^-} = 154.0 - 74.0 = 80.0\ \text{S cm}^2\text{mol}^{-1}. NCERT Class 12, Electrochemistry (variation of conductivity with concentration).

  5. Question 5 · NEET 2020 · Q93 · +4 / −1

    The number of Faradays (F) required to produce 20 g of calcium from molten CaCl2_{2} (atomic mass of Ca = 40 g mol−1^{-1}) is

    Answer and explanation

    Answer: (4)

    The reaction for the production of calcium is Ca2++2e−→Ca\mathrm{Ca}^{2+} + 2\mathrm{e}^- → \mathrm{Ca}. The number of moles of calcium produced is 20 g40 gmol−1=0.5 mol\frac{20\ g}{40\ g mol^{-1}} = 0.5\ mol. Since 2 moles of electrons are required to produce 1 mole of calcium, the number of Faradays required is 0.5×2=1 F0.5 \times 2 = 1\ F. Chapter: Electrochemistry.

  6. Question 6 · NEET 2020 · Q130 · +4 / −1

    On electrolysis of dil. sulphuric acid using Platinum (Pt) electrode, the product obtained at anode will be

    Answer and explanation

    Answer: (1)

    On electrolysis of dilute sulfuric acid with platinum electrodes, oxygen gas is evolved at the anode due to the oxidation of water. The reaction is 2H2O→O2+4H++4e−2\mathrm{H}_{2}\mathrm{O} → \mathrm{O}_{2} + 4\mathrm{H}^{+} + 4\mathrm{e}^{-}. Chapter: Electrochemistry.