Match List-I with List-II.
List-I (Parameter)
A. Cell constant
B. Molar conductivity
C. Conductivity

Q: Why is the unit of molar conductivity $\text{S cm}^2 \text{ mol}^{-1}$ instead of $\text{S m}^2 \text{ mol}^{-1}$?

Both are valid units! However, in laboratory settings, cell dimensions are typically measured in centimeters, making $\text{S cm}^2 \text{ mol}^{-1}$ the more commonly used practical unit.

Q: What is the difference between conductance and conductivity?

Conductance ($G$) is the ease with which current flows through a specific object, measured in Siemens ($\text{S}$). Conductivity ($\kappa$) is an intrinsic property of the material, measured in $\text{S m}^{-1}$ or $\Omega^{-1} \text{m}^{-1}$.

Q: Why is the degree of dissociation dimensionless?

It is defined as the number of moles dissociated divided by the total number of moles initially taken. Since it's a ratio of the same physical quantity (moles), the units cancel out.

Q: Can cell constant be negative?

No, the cell constant is the ratio of the distance between electrodes ($l$) to their cross-sectional area ($A$). Since both are physical dimensions, they are always positive.

Q: How do I quickly solve such matrix match questions in exams?

Use the elimination method! Find the easiest parameter you are 100% sure about (like degree of dissociation being dimensionless). Match it, and check the options. Often, matching just one or two parameters gives you the correct answer.

Question

Match List-I with List-II.

$\begin{array}{ll} 	ext{List-I (Parameter)} & 	ext{List-II (Unit)} \ 	ext{A. Cell constant} & 	ext{1. } 	ext{S cm}^2 	ext{ mol}^{-1} \ 	ext{B. Molar conductivity} & 	ext{2. Dimensionless} \ 	ext{C. Conductivity} & 	ext{3. } 	ext{m}^{-1} \ 	ext{D. Degree of dissociation of electrolyte} & 	ext{4. } \Omega^{-1}	ext{m}^{-1} \end{array}$

Choose the most appropriate answer from the options given below.

Options:

(A) A $ightarrow$ 3, B $ightarrow$ 1, C $ightarrow$ 4, D $ightarrow$ 2

(B) A $ightarrow$ 3, B $ightarrow$ 1, C $ightarrow$ 2, D $ightarrow$ 4

(C) A $ightarrow$ 1, B $ightarrow$ 4, C $ightarrow$ 3, D $ightarrow$ 2

(D) A $ightarrow$ 2, B $ightarrow$ 1, C $ightarrow$ 3, D $ightarrow$ 4

Sigma Percentile · Accepted Answer

### The Importance of Units in Electrochemistry

When diving into the world of electrochemistry, it is easy to get lost in a sea of formulas and concepts. However, the secret to mastering this chapter lies in understanding the **units** of various parameters.

Units are not just arbitrary letters; they tell a story about the physical reality of the system. Let's break down the parameters given in this problem one by one.

### Analyzing the Cell Constant

Imagine a conductivity cell where two electrodes are placed parallel to each other. The **cell constant** (denoted by $G^*$) is a geometric parameter of this cell.

It is defined as the ratio of the distance between the electrodes ($l$) to their cross-sectional area ($A$).

Mathematically, we write this as:

$$G^* = \frac{l}{A}$$

If we measure the length in meters ($	ext{m}$) and the area in square meters ($	ext{m}^2$), the unit becomes:

$$	ext{Unit} = \frac{	ext{m}}{	ext{m}^2} = 	ext{m}^{-1}$$

Thus, the cell constant perfectly matches with $	ext{m}^{-1}$.

### Decoding Molar Conductivity

Next, we have **molar conductivity** ($\Lambda_m$). This parameter is incredibly powerful because it normalizes the conductivity of a solution for one mole of the electrolyte.

It tells us the conducting power of all the ions produced by dissolving one mole of an electrolyte in a solution. The formula is:

$$\Lambda_m = \frac{\kappa 	imes 1000}{M}$$

Here, $\kappa$ is the conductivity and $M$ is the molarity. When we plug in the standard units, the unit for molar conductivity emerges as:

$$	ext{S cm}^2 	ext{ mol}^{-1}$$

This matches our first option in the second list.

### Understanding Conductivity

**Conductivity** ($\kappa$), also known as specific conductance, is the reciprocal of resistivity ($ho$).

It represents the ease with which electric current flows through a unit volume of the solution. We can derive its unit from the resistance formula:

$$R = ho \frac{l}{A}$$

Rearranging for conductivity, we get:

$$\kappa = \frac{1}{ho} = \frac{1}{R} 	imes \frac{l}{A}$$

Since resistance is measured in Ohms ($\Omega$) and the cell constant in $	ext{m}^{-1}$, the unit of conductivity is:

$$\Omega^{-1} 	ext{m}^{-1}$$

This perfectly aligns with the fourth option in the list.

### The Simplicity of Degree of Dissociation

Finally, we look at the **degree of dissociation** ($\alpha$).

This is a fundamental concept in ionic equilibrium and electrochemistry. It is defined as the fraction of the total number of moles of an electrolyte that dissociates into ions.

$$\alpha = \frac{	ext{Moles dissociated}}{	ext{Total moles}}$$

Because it is a ratio of two identical physical quantities (moles divided by moles), the units completely cancel out.

Therefore, the degree of dissociation is a **dimensionless** quantity.

### Final Conclusion

By systematically analyzing each parameter, we have successfully decoded their units.

- **Cell constant** matches with $	ext{m}^{-1}$.
- **Molar conductivity** matches with $	ext{S cm}^2 	ext{ mol}^{-1}$.
- **Conductivity** matches with $\Omega^{-1} 	ext{m}^{-1}$.
- **Degree of dissociation** is **dimensionless**.

This leads us to the correct matching sequence, proving that a solid grasp of basic definitions is all you need to conquer such problems!

Select Answer:

Visualized Solution

$Electrochemical Parameters$

$Cell Constant (G^{*})$

$Molar Conductivity (Λ_{m})$

$Conductivity (κ)$

$Degree of Dissociation (α)$

$Final Answer$

$The Way Forward$

The Sigma Insight: Electrolytic Conduction

The Importance of Units in Electrochemistry

Analyzing the Cell Constant

Decoding Molar Conductivity

Understanding Conductivity

The Simplicity of Degree of Dissociation

Final Conclusion

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