Nernst Equation and Concentration Cells

Introduction

Your smartphone battery voltage drops from 4.2V when fully charged to 3.0V when depleted - but the chemistry inside hasn’t changed! What’s happening? The Nernst equation explains how cell potential varies with ion concentrations. This fundamental equation connects electrochemistry with thermodynamics and is one of the most important formulas in JEE Chemistry!

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Interactive: Nernst Equation Demonstration

Watch how EMF changes as you vary ion concentrations:

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Derivation of Nernst Equation

From Thermodynamics

For a general cell reaction at temperature T:

$$\Delta G = \Delta G° + RT \ln Q$$

where Q = reaction quotient

Connecting to Electrochemistry

We know:

$$\Delta G = -nFE_{\text{cell}}$$ $$\Delta G° = -nFE°_{\text{cell}}$$

Substituting:

$$-nFE_{\text{cell}} = -nFE°_{\text{cell}} + RT \ln Q$$

Dividing by -nF:

$$E_{\text{cell}} = E°_{\text{cell}} - \frac{RT}{nF} \ln Q$$

Converting to log₁₀

$$\ln Q = 2.303 \log Q$$ $$\boxed{E_{\text{cell}} = E°_{\text{cell}} - \frac{2.303RT}{nF} \log Q}$$

At 298 K (25°C)

Substituting values:

• R = 8.314 J/(mol·K)
• T = 298 K
• F = 96,500 C/mol

$$\frac{2.303RT}{F} = \frac{2.303 \times 8.314 \times 298}{96500} = 0.0591 \approx 0.059$$ $$\boxed{E_{\text{cell}} = E°_{\text{cell}} - \frac{0.059}{n} \log Q}$$

This is the Nernst equation at 25°C - memorize this form for JEE!

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Understanding the Reaction Quotient (Q)

For General Reaction

$$aA + bB \rightarrow cC + dD$$ $$Q = \frac{[C]^c[D]^d}{[A]^a[B]^b}$$

Important Rules for Q

1. Solids and pure liquids: Activity = 1 (excluded from Q)
2. Solutions: Use molar concentration [M]
3. Gases: Use partial pressure in bar or atm
4. Electrodes: Pure metals/solids omitted

Example: Daniell Cell

$$\text{Zn}_{(s)} + \text{Cu}^{2+}_{(aq)} \rightarrow \text{Zn}^{2+}_{(aq)} + \text{Cu}_{(s)}$$ $$Q = \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]}$$

Note: Solid Zn and Cu are omitted!

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Applying the Nernst Equation

Example 1: Simple Metal-Ion Cell

Problem: Calculate EMF of the cell at 25°C:

$$\text{Zn} | \text{Zn}^{2+}(0.1M) || \text{Cu}^{2+}(0.01M) | \text{Cu}$$

Given: E°(Zn²⁺/Zn) = -0.76 V, E°(Cu²⁺/Cu) = +0.34 V

Solution:

Step 1: Calculate E°cell

$$E°_{\text{cell}} = E°_{\text{cathode}} - E°_{\text{anode}} = 0.34 - (-0.76) = 1.10 \text{ V}$$

Step 2: Write cell reaction

$$\text{Zn} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cu}$$

Step 3: Find Q

$$Q = \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} = \frac{0.1}{0.01} = 10$$

Step 4: Apply Nernst equation (n = 2)

$$E_{\text{cell}} = 1.10 - \frac{0.059}{2} \log 10$$ $$E_{\text{cell}} = 1.10 - 0.0295 \times 1$$ $$\boxed{E_{\text{cell}} = 1.0705 \text{ V}}$$

Example 2: Gas Electrode

Problem: Calculate EMF at 25°C:

$$\text{Pt} | \text{H}_2(2 \text{ atm}) | \text{H}^+(0.1M) || \text{Ag}^+(0.01M) | \text{Ag}$$

Given: E°(Ag⁺/Ag) = +0.80 V, E°(H⁺/H₂) = 0.00 V

Solution:

Cell reaction:

$$\text{H}_2 + 2\text{Ag}^+ \rightarrow 2\text{H}^+ + 2\text{Ag}$$ $$E°_{\text{cell}} = 0.80 - 0 = 0.80 \text{ V}$$ $$Q = \frac{[\text{H}^+]^2}{P_{\text{H}_2}[\text{Ag}^+]^2} = \frac{(0.1)^2}{2 \times (0.01)^2} = \frac{0.01}{2 \times 0.0001} = 50$$ $$E_{\text{cell}} = 0.80 - \frac{0.059}{2} \log 50$$ $$E_{\text{cell}} = 0.80 - 0.0295 \times 1.699$$ $$\boxed{E_{\text{cell}} = 0.75 \text{ V}}$$

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Nernst Equation for Half-Cells

For Single Electrode

For a half-reaction: $\text{M}^{n+} + ne^- \rightarrow \text{M}$

$$\boxed{E = E° - \frac{0.059}{n} \log \frac{1}{[\text{M}^{n+}]}}$$ $$\boxed{E = E° + \frac{0.059}{n} \log [\text{M}^{n+}]}$$

Example: Cu²⁺/Cu Electrode

For Cu²⁺(0.01M) at 25°C:

$$E = 0.34 + \frac{0.059}{2} \log(0.01)$$ $$E = 0.34 + 0.0295 \times (-2)$$ $$E = 0.34 - 0.059 = 0.281 \text{ V}$$

Lower concentration → Lower electrode potential!

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Concentration Cells

A concentration cell is a galvanic cell where both electrodes are made of the same material but are immersed in solutions of different concentrations.

Principle

• EMF is generated solely due to concentration difference
• Ions flow from higher to lower concentration
• Electrons flow to equalize concentrations

General Setup

$$\text{M} | \text{M}^{n+}(C_1) || \text{M}^{n+}(C_2) | \text{M}$$

where C₁ < C₂ (C₁ is dilute, C₂ is concentrated)

EMF of Concentration Cell

E°cell = 0 (same electrodes!)

Cell reaction:

$$\text{M}^{n+}(C_2) \rightarrow \text{M}^{n+}(C_1)$$

(Ions move from concentrated to dilute)

$$Q = \frac{C_1}{C_2}$$ $$\boxed{E_{\text{cell}} = 0 - \frac{0.059}{n} \log \frac{C_1}{C_2} = \frac{0.059}{n} \log \frac{C_2}{C_1}}$$

Since C₂ > C₁, the log term is positive → Positive EMF!

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Types of Concentration Cells

1. Electrolyte Concentration Cell

Same electrode, different electrolyte concentrations

Example:

$$\text{Zn} | \text{Zn}^{2+}(0.01M) || \text{Zn}^{2+}(1M) | \text{Zn}$$ $$E_{\text{cell}} = \frac{0.059}{2} \log \frac{1}{0.01} = \frac{0.059}{2} \times 2 = 0.059 \text{ V}$$

At anode (dilute side): $\text{Zn} \rightarrow \text{Zn}^{2+}(0.01M) + 2e^-$

At cathode (concentrated side): $\text{Zn}^{2+}(1M) + 2e^- \rightarrow \text{Zn}$

Net: Zn²⁺ moves from concentrated to dilute until equilibrium!

2. Electrode Concentration Cell (Gas)

Different gas pressures at the same electrode

Example: Hydrogen concentration cell

$$\text{Pt} | \text{H}_2(P_1) | \text{H}^+(1M) || \text{H}^+(1M) | \text{H}_2(P_2) | \text{Pt}$$

where P₁ < P₂

$$E_{\text{cell}} = \frac{0.059}{2} \log \frac{P_2}{P_1}$$

3. Amalgam Concentration Cell

Different concentrations of metal in mercury amalgam

Example:

$$\text{Zn-Hg}(C_1) | \text{Zn}^{2+} || \text{Zn}^{2+} | \text{Zn-Hg}(C_2)$$

Same principle applies!

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Equilibrium and the Nernst Equation

At Equilibrium

When a cell reaches equilibrium:

• Ecell = 0 (no driving force)
• Q = K (reaction quotient = equilibrium constant)

Substituting into Nernst equation:

$$0 = E°_{\text{cell}} - \frac{0.059}{n} \log K$$ $$\boxed{E°_{\text{cell}} = \frac{0.059}{n} \log K}$$

This connects electrode potential to equilibrium constant!

Example: Finding Equilibrium Constant

For the reaction: $\text{Zn} + \text{Cu}^{2+} \rightleftharpoons \text{Zn}^{2+} + \text{Cu}$

E°cell = 1.10 V, n = 2

$$1.10 = \frac{0.059}{2} \log K$$ $$\log K = \frac{2 \times 1.10}{0.059} = 37.29$$ $$\boxed{K = 10^{37.29} \approx 2 \times 10^{37}}$$

Huge K! This reaction goes essentially to completion.

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Relationship Summary

Connecting Everything

$$\boxed{\Delta G° = -nFE°_{\text{cell}} = -RT \ln K = -2.303RT \log K}$$

At 298 K:

$$\boxed{E°_{\text{cell}} = \frac{0.059}{n} \log K}$$ $$\boxed{\Delta G° = -2.303 \times 8.314 \times 298 \times \log K = -5.706 \log K \text{ kJ/mol}}$$

E°cell	ΔG°	K	Reaction
Positive	Negative	> 1	Spontaneous (products favored)
Zero	Zero	= 1	Equilibrium
Negative	Positive	< 1	Non-spontaneous (reactants favored)

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Practice Problems

Level 1: JEE Main

Q1. Calculate EMF of the cell at 25°C:

$$\text{Zn} | \text{Zn}^{2+}(0.01M) || \text{Cu}^{2+}(0.1M) | \text{Cu}$$

Given: E°cell = 1.10 V

Q2. For a concentration cell:

$$\text{Ag} | \text{Ag}^+(0.001M) || \text{Ag}^+(0.1M) | \text{Ag}$$

Calculate EMF at 25°C.

Q3. At what concentration of Ag⁺ will the electrode potential be 0.80 V at 25°C? Given: E°(Ag⁺/Ag) = +0.80 V

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Level 2: JEE Main/Advanced

Q4. Calculate the equilibrium constant for the reaction:

$$\text{Fe}^{2+} + \text{Ag}^+ \rightarrow \text{Fe}^{3+} + \text{Ag}$$

Given: E°(Ag⁺/Ag) = +0.80 V, E°(Fe³⁺/Fe²⁺) = +0.77 V

Q5. A hydrogen electrode is immersed in a solution of pH = 3 at 25°C. Calculate its electrode potential.

Q6. For the cell:

$$\text{Pt} | \text{H}_2(1 \text{ atm}) | \text{H}^+(xM) || \text{H}^+(0.1M) | \text{H}_2(1 \text{ atm}) | \text{Pt}$$

If EMF = 0.0295 V, find x.

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Level 3: JEE Advanced

Q7. Calculate ΔG° for the reaction at 25°C:

$$2\text{Al} + 3\text{Cu}^{2+} \rightarrow 2\text{Al}^{3+} + 3\text{Cu}$$

Given: E°(Cu²⁺/Cu) = +0.34 V, E°(Al³⁺/Al) = -1.66 V

Q8. A cell has E°cell = 1.5 V. At what ratio of [products]/[reactants] will the cell EMF become zero at 25°C? (Assume n = 2)

Q9. For the reaction:

$$\text{Zn} + 2\text{Ag}^+ \rightleftharpoons \text{Zn}^{2+} + 2\text{Ag}$$

If initial [Ag⁺] = 1M and [Zn²⁺] = 0.01M, what will be the EMF when [Ag⁺] drops to 0.5M?

Given: E°cell = 1.56 V

Q10. The EMF of the cell:

$$\text{Zn} | \text{Zn}^{2+}(C_1) || \text{Zn}^{2+}(C_2) | \text{Zn}$$

is 0.0295 V at 25°C. Find the ratio C₂/C₁.

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Solutions to Practice Problems

A1.

$$Q = \frac{0.01}{0.1} = 0.1$$ $$E = 1.10 - \frac{0.059}{2} \log(0.1) = 1.10 - \frac{0.059}{2}(-1) = 1.10 + 0.0295 = \boxed{1.1295 \text{ V}}$$

A2.

$$E = \frac{0.059}{1} \log \frac{0.1}{0.001} = 0.059 \times 2 = \boxed{0.118 \text{ V}}$$

A3.

$$0.80 = 0.80 + \frac{0.059}{1} \log[\text{Ag}^+]$$ $$\log[\text{Ag}^+] = 0$$ $$\boxed{[\text{Ag}^+] = 1 \text{ M}}$$

(At standard concentration!)

A4.

$$E°_{\text{cell}} = 0.80 - 0.77 = 0.03 \text{ V}$$ $$0.03 = \frac{0.059}{1} \log K$$ $$K = 10^{0.03/0.059} = 10^{0.508} \approx \boxed{3.2}$$

A5. pH = 3 → [H⁺] = 10⁻³ M

$$E = 0 + \frac{0.059}{2} \log(10^{-3})^2 = \frac{0.059}{2} \times (-6) = \boxed{-0.177 \text{ V}}$$

A6.

$$0.0295 = \frac{0.059}{2} \log \frac{0.1}{x}$$ $$1 = \log \frac{0.1}{x}$$ $$\frac{0.1}{x} = 10$$ $$\boxed{x = 0.01 \text{ M}}$$

A7.

$$E°_{\text{cell}} = 0.34 - (-1.66) = 2.00 \text{ V}$$ $$\Delta G° = -nFE° = -6 \times 96500 \times 2.00 = \boxed{-1158 \text{ kJ}}$$

A8.

$$0 = 1.5 - \frac{0.059}{2} \log Q$$ $$\log Q = \frac{2 \times 1.5}{0.059} = 50.85$$ $$\boxed{Q = 10^{50.85} \approx 7.1 \times 10^{50}}$$

A9. When [Ag⁺] = 0.5M, Zn reacted = 0.5/2 = 0.25 mol So [Zn²⁺] = 0.01 + 0.25 = 0.26M (if initial Zn was sufficient)

$$Q = \frac{0.26}{(0.5)^2} = \frac{0.26}{0.25} = 1.04$$ $$E = 1.56 - \frac{0.059}{2} \log(1.04) = 1.56 - 0.0005 \approx \boxed{1.559 \text{ V}}$$

A10.

$$0.0295 = \frac{0.059}{2} \log \frac{C_2}{C_1}$$ $$\log \frac{C_2}{C_1} = 1$$ $$\boxed{\frac{C_2}{C_1} = 10}$$

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Common Mistakes to Avoid

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Advanced Applications

pH Measurement

A pH meter uses the Nernst equation!

For hydrogen electrode in solution of unknown pH:

$$E = 0 - \frac{0.059}{2} \log \frac{(P_{\text{H}_2})}{[\text{H}^+]^2}$$

At 1 atm H₂:

$$E = -0.059 \log \frac{1}{[\text{H}^+]} = 0.059 \log[\text{H}^+]$$ $$E = -0.059 \text{ pH}$$ $$\boxed{\text{pH} = -\frac{E}{0.059}}$$

Solubility Product from EMF

For AgCl electrode:

$$\text{AgCl} + e^- \rightarrow \text{Ag} + \text{Cl}^-$$ $$E = E° - 0.059 \log[\text{Cl}^-]$$

Can be used to find Ksp!

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Key Formulas Summary

Must Memorize for JEE

Nernst equation (298 K):

$$\boxed{E = E° - \frac{0.059}{n} \log Q}$$

Concentration cell:

$$\boxed{E = \frac{0.059}{n} \log \frac{C_{\text{concentrated}}}{C_{\text{dilute}}}}$$

Equilibrium connection:

$$\boxed{E° = \frac{0.059}{n} \log K}$$

Thermodynamic relations:

$$\boxed{\Delta G° = -nFE°}$$ $$\boxed{\Delta G° = -2.303RT \log K}$$

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Related Topics

Within Electrochemistry

• Electrochemical Cells - Foundation of cell EMF
• Electrode Potentials - Standard E° values and electrochemical series
• Oxidation-Reduction - Redox fundamentals
• Batteries - Practical voltage calculations in primary/secondary cells
• Electrolysis - Reverse EMF considerations
• Corrosion - Concentration cells in rusting

Equilibrium and Thermodynamics

• Chemical Equilibrium - Relationship between K and E°
• Gibbs Energy - ΔG° = -nFE° derivation
• Entropy - Spontaneity and equilibrium
• Ionic Equilibrium - pH calculations with Nernst
• Solubility Product - Ksp from EMF measurements

Solutions and Concentrations

• Concentration Methods - Molarity in Nernst calculations
• Colligative Properties - Activity vs concentration

Physics Connections

• Current Electricity - Voltage and potential difference
• Thermodynamics - Energy conversions
• Electrostatics - Electric Potential - Potential concepts

Mathematics Connections

• Logarithms - Log calculations in Nernst equation

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