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MHT-CET Chemistry · Electrochemistry

Cell Constant and Conductivity Measurements

Conductance is the reciprocal of resistance; conductivity is conductance scaled by the cell's geometry, the cell constant l/a — so κ = (cell constant)/R, and the cell constant itself is found once with a standard KCl solution.

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Why this matters

23 PYQs, one HARD (a lost exponent in the stem). Three question shapes: multiply κ by R to get the cell constant, divide the cell constant by R to get κ, and pick the correct or incorrect relation among k = 1/ρ, k = G·(l/a), k = Λ·c. The rest is units: siemens, S cm⁻¹, and what 1 S is not.

Concept 1 of 3

Conductance, Conductivity and Their Units

Intuition

Resistance R measures how hard current finds it to flow; conductance G = 1/R measures how easy. Both depend on the shape of the sample. Conductivity κ removes the shape: κ = G × (l/a), the conductance of a 1 cm cube — so it is a property of the solution alone.

Definition

  • G=1RG = \dfrac{1}{R}, unit siemens (S) =Ω−1=A V−1=C V−1s−1= \Omega^{-1} = \text{A V}^{-1} = \text{C V}^{-1}\text{s}^{-1}. The ohm itself is NOT a unit of conductance.
  • κ=1ρ=G⋅la\kappa = \dfrac{1}{\rho} = G \cdot \dfrac{l}{a}, unit S cm−1\text{S cm}^{-1} (SI: S m−1\text{S m}^{-1}). Molar conductivity has S cm2 mol−1\text{S cm}^2\,\text{mol}^{-1} — the extra cm² is how you tell them apart.
  • κ=Λ⋅c1000\kappa = \dfrac{\Lambda \cdot c}{1000} (c in mol L⁻¹) is also correct; k=1R⋅alk = \dfrac{1}{R}\cdot\dfrac{a}{l} is NOT — it inverts the cell constant.
  • Who conducts: molten NaCl and salt solutions (mobile ions); NOT crystalline NaCl (ions fixed), diamond or sulphur. Urea in water gives no ions, so its conductivity is that of distilled water.

Conductivity

κ=1ρ=G⋅la=1R⋅la\kappa = \frac{1}{\rho} = G\cdot\frac{l}{a} = \frac{1}{R}\cdot\frac{l}{a}

Worked example

A solution in a cell with electrodes 1.5 cm apart and 0.5 cm² in area has resistance 200 Ω. Find G and κ.
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1ElectrochemistryMODERATE
Which of the following expressions for conductivity of solution of an electrolyte is NOT correct?

[Q94 · 4th May Shift 2 · 2023]

Reading S cm² mol⁻¹ as the unit of conductivity

That is MOLAR conductivity. Conductivity is S cm⁻¹ (or S m⁻¹). The options always offer both; the one with mol⁻¹ in it belongs to Λ.
Drill 5 more on conductance, conductivity and their units

Concept 2 of 3

Cell Constant: l/a = κ × R

Intuition

The distance between the electrodes divided by their area is fixed for a given cell. It is never measured with a ruler; the cell is filled with KCl of known conductivity, the resistance is read, and cell constant = κ × R. Once known, it converts every later resistance into a conductivity.

Definition

  • Cell constant =la= \dfrac{l}{a}, unit cm−1\text{cm}^{-1}. From geometry: electrodes 0.92 cm apart, area 1.2 cm² → 0.767 cm−10.767\ \text{cm}^{-1}.
  • Determination: cell constant=κKCl×RKCl\text{cell constant} = \kappa_{\text{KCl}} \times R_{\text{KCl}}. 0.1 M KCl, κ=1.70×10−4\kappa = 1.70 \times 10^{-4}, R=100 ΩR = 100\ \Omega → 0.017 cm−10.017\ \text{cm}^{-1}.
  • Standard solutions: 1 M, 0.1 M or 0.01 M KCl, whose conductivities are tabulated. NOT saturated KCl — its concentration changes with temperature.
  • la=k⋅R\dfrac{l}{a} = k \cdot R is the formula; la=kR\dfrac{l}{a} = \dfrac{k}{R} and Rk\dfrac{R}{k} are the distractors.

Cell constant

la=κ×R\frac{l}{a} = \kappa \times R

Worked example

0.01 M KCl (κ=1.41×10−3 S cm−1\kappa = 1.41 \times 10^{-3}\ \text{S cm}^{-1}) shows a resistance of 850 Ω in a cell. Find the cell constant.
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2ElectrochemistryEASY
The resistance of conductivity cell filled with 0.1 M KCl solution is 100 ohm and conductivity is 1.70×10−41.70 \times 10^{-4} S cm−1^{-1}. What is the cell constant of the cell?

[Q83 · 11th May Shift 1 · 2023]

Dividing when the cell constant is asked

κ=(l/a)/R\kappa = (l/a)/R, so l/a=κRl/a = \kappa R: MULTIPLY. Dividing gives κ/R\kappa/R, a number four orders too small — and it is offered as an option.
Drill 8 more on cell constant: l/a = κ × r

Concept 3 of 3

Conductivity From the Cell Constant and Resistance

Intuition

The reverse move. With the cell constant known, fill the cell with the unknown solution, read R, and κ = (cell constant)/R. Dilute solutions have large R (thousands of ohms) and κ of order 10⁻⁴ S cm⁻¹.

Definition

  • κ=l/aR\kappa = \dfrac{l/a}{R}. Cell constant 0.9 cm⁻¹, R=6530 ΩR = 6530\ \Omega → κ=1.38×10−4 Ω−1cm−1\kappa = 1.38 \times 10^{-4}\ \Omega^{-1}\text{cm}^{-1}.
  • 0.84 cm⁻¹ and 14000 Ω (5×10−45 \times 10^{-4} M NaCl) → 6.0×10−56.0 \times 10^{-5}; 1.32 cm⁻¹ and 528 Ω → 0.00250.0025.
  • Order-of-magnitude check: a 0.1 M salt is about 10−210^{-2} S cm⁻¹; a 10−310^{-3} M one about 10−410^{-4}. A result of 10210^{2} means you multiplied.
  • Units on the answer: Ω−1cm−1\Omega^{-1}\text{cm}^{-1}, never Ω cm−1\Omega\,\text{cm}^{-1} — the paper plants that too.

Conductivity of the unknown

κ=cell constantR\kappa = \frac{\text{cell constant}}{R}

Worked example

A cell of constant 0.653 cm⁻¹ shows 6530 Ω with 0.001 M AgNO₃. Find κ.
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 3ElectrochemistryEASY
Cell constant of a conductivity cell is 0.9 cm−1^{-1} and resistance shown by AgNO3_3 solution is 6530 ohm. What is the conductivity of AgNO3_3 solution?

[Q75 · 11th May Shift 2 · 2023]

Trusting a printed exponent over the order of magnitude

One 2023 paper printed κ of 0.1 M KCl as 1.90×10−61.90 \times 10^{-6} and keyed a cell constant of 218.5 cm⁻¹; the working needs κ = 1.90. When the arithmetic lands on no option, match the mantissa and let the sanity range decide the exponent.
Drill 7 more on conductivity from the cell constant and resistance

Summary — formulas & gotchas at a glance

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Formulas (3)

Watch out for (3)

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