I. For aqueous NaCl at 25 °C the molar conductivities at two concentrations are measured as: _m(0.010\ M)=123.9\ S\ cm²\ mol^-1 _m(0.040\ M)=120.7\ S\ cm²\ mol^-1 Molar conductivity for strong electrolytes can be represented as _m = °_m - A√(c). Determine ° (molar conductivity at infinite dilution) and the constant A for NaCl solution. (3) II. Write the electrode reactions that occur in a hydrogen–oxygen fuel cell. Also, state two advantages of fuel cells compared to conventional batteries. (2)

I. A = 32\ S cm²mol^-1M^-1/2, °_m = 127.1\ S cm²mol^-1; II. Anode: 2H_2 + 4OH^- → 4H_2O + 4e^-, Cathode: O_2 + 2H_2O + 4e^- → 4OH^-; advantages: high efficiency and continuous operation without recharging (pollution-free).

ChemistryCBSE Class 12 · Sample paper, session 2026-27 · 5 marks · Long answer

CBSE Class 12 Chemistry sample paper 2026-27: question 33(B)

From the official CBSE sample question paper for 2026-27, section E, question 33(B), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q33(B) · 5 marks · Long answerElectrochemistry

I. For aqueous NaCl at 25 °C the molar conductivities at two concentrations are measured as:
$Λ_{m} (0.010 M) = 123.9 S c m^{2} m o l^{- 1}$
$Λ_{m} (0.040 M) = 120.7 S c m^{2} m o l^{- 1}$
Molar conductivity for strong electrolytes can be represented as $Λ_{m} = Λ_{m}^{\circ} - A c$ . Determine $Λ^{\circ}$ (molar conductivity at infinite dilution) and the constant $A$ for NaCl solution. (3)

II. Write the electrode reactions that occur in a hydrogen–oxygen fuel cell. Also, state two advantages of fuel cells compared to conventional batteries. (2)

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2026-27.

Hints

Open one hint at a time and try again before the next.

Hint 1

Write the Kohlrausch equation once for each concentration, using $c$ .

Hint 2

Subtracting the two equations removes $Λ_{m}^{\circ}$ and leaves A.

Hint 3

For the fuel cell, H₂ is oxidised at the anode and O₂ is reduced at the cathode; think about efficiency and the fuel supply for the advantages.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

I. Use $Λ_{m} = Λ_{m}^{\circ} - A c$ .
For $c_{1} = 0.010$ M, $c_{1} = 0.100$ :
$123.9 = Λ_{m}^{\circ} - 0.100 A ... (1)$
½ mark
For $c_{2} = 0.040$ M, $c_{2} = 0.200$ :
$120.7 = Λ_{m}^{\circ} - 0.200 A ... (2)$
½ mark
Subtract (2) from (1):
$123.9 - 120.7 = - 0.100 A + 0.200 A$
$3.2 = 0.100 A$
½ mark
$A = 32 S c m^{2} m o l^{- 1} M^{- 1/2}$
½ mark
Substitute $A$ in (1):
$Λ_{m}^{\circ} = 123.9 + 32 \times 0.100 = 123.9 + 3.2$
½ mark
$Λ_{m}^{\circ} = 127.1 S c m^{2} m o l^{- 1}$
½ mark
II. Electrode reactions (alkaline electrolyte, as in NCERT):
- Anode: $2 H_{2} (g) + 4 O H^{-} (a q) \to 4 H_{2} O (l) + 4 e^{-}$
- Cathode: $O_{2} (g) + 2 H_{2} O (l) + 4 e^{-} \to 4 O H^{-} (a q)$
- Overall: $2 H_{2} (g) + O_{2} (g) \to 2 H_{2} O (l)$
(In an acidic medium: anode $2 H_{2} \to 4 H^{+} + 4 e^{-}$ ; cathode $O_{2} + 4 H^{+} + 4 e^{-} \to 2 H_{2} O$ .)
1 mark
Advantages over conventional batteries:
1. They have higher efficiency (about 70%) in converting chemical energy into electrical energy.
½ mark
2. They work continuously as long as the reactants H $_{2}$ and O $_{2}$ are supplied, so they need no recharging. They are also pollution-free, since the only product is water.
½ mark

Answer: I. $A = 32 S c m^{2} m o l^{- 1} M^{- 1/2}$ , $Λ_{m}^{\circ} = 127.1 S c m^{2} m o l^{- 1}$ ; II. Anode: $2 H_{2} + 4 O H^{-} \to 4 H_{2} O + 4 e^{-}$ , Cathode: $O_{2} + 2 H_{2} O + 4 e^{-} \to 4 O H^{-}$ ; advantages: high efficiency and continuous operation without recharging (pollution-free).

Where the marks are

Sets up equation (1) using $0.010 = 0.100$ (½ mark)
Sets up equation (2) using $0.040 = 0.200$ (½ mark)
Subtracts the equations to get $3.2 = 0.100 A$ (½ mark)
Finds $A = 32$ with unit (½ mark)
Substitutes A back into equation (1) (½ mark)
Finds $Λ_{m}^{\circ} = 127.1 S c m^{2} m o l^{- 1}$ (½ mark)
Writes correct anode and cathode reactions (½ + ½ mark)
States two valid advantages of fuel cells (½ + ½ mark)

Revise the chapter

Chapter: Electrochemistry (notes, formulas, where marks are lost and every past question)
Topic: Conductivity, Molar Conductivity and Kohlrausch's Law

Practise this chapterBoard-style questions from Electrochemistry, one at a time, marked against the answer key and marking points.Back to the full paperSample paper 2026-27: every question, by section.

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CBSE Class 12 Chemistry sample paper 2026-27: question 33(B)

Hints

Step-by-step solution

Revise the chapter

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