$E^\circ_{Cell} = 1.1\ V$ for Daniel cell. Which of the following expressions are correct description of state of equilibrium in this cell? (Two or more than two options may be correct.) (i) $1.1 = K_c$ (ii) $\frac{2.303RT}{2F} \log K_c = 1.1$ (iii) $\log K_c = \frac{2.2}{0.059}$ (iv) $\log K_c = 1.1$

E^°Cell = 1.1V for Daniel cell. Which of the following expressions are correct description of state of equilib

NCERT Exemplar · Q45

E^\circ_{Cell} = 1.1\ V

(i)

1.1 = K_c

(ii)

\frac{2.303RT}{2F} \log K_c = 1.1

(iii)

\log K_c = \frac{2.2}{0.059}

(iv)

\log K_c = 1.1

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Concept understanding — Cell Representation Nernst Equation

Cell Representation and the Nernst Equation: From Intuition to Precision

Imagine you have a Daniell cell — a zinc rod in zinc sulphate solution connected by a salt bridge to a copper rod in copper sulphate solution. You know it produces a voltage. But what happens if you dilute the copper sulphate solution? Or if you change the temperature? The voltage changes. The Nernst equation is the tool that tells you exactly how much it changes.

The Intuition First

A battery works because the two half-cells "want" to react — zinc wants to lose electrons, copper ions want to gain them. This "want" is measured as a tendency, or potential. But the strength of that tendency depends on how crowded the ions are.

Think of it like this: If you have a room full of people who all want to leave (like zinc ions wanting to form), the push to get out is stronger when the room is packed. If the room is nearly empty, the push is weaker. Similarly, for copper ions wanting to enter the metal (gain electrons), the pull is stronger when there are many copper ions around, and weaker when there are few.

The Nernst equation quantifies this: the actual cell potential depends on the concentrations (or activities) of the ions involved.

The Precise Statement

For a general cell reaction:

aA + bB \rightarrow cC + dD

The cell potential $E$ under non-standard conditions is given by:

E = E^\circ - \frac{RT}{nF} \ln Q

Where:

$E$ = cell potential under the given conditions (in volts)
$E^\circ$ = standard cell potential (when all reactants/products are at 1 M, 1 atm, 25°C)
$R$ = universal gas constant (8.314 J/mol·K)
$T$ = temperature in Kelvin
$n$ = number of moles of electrons transferred in the balanced half-reactions
$F$ = Faraday constant (96,485 C/mol)
$Q$ = reaction quotient = $\frac{[C]^c [D]^d}{[A]^a [B]^b}$ (using concentrations for dilute solutions)

At 25°C (298 K), the equation simplifies to a very practical form:

E = E^\circ - \frac{0.0591}{n} \log_{10} Q

The 0.0591 comes from $\frac{2.303RT}{F}$ at 298 K. Notice it uses $\log_{10}$ (common log), not natural log.

Cell Representation: How We Write It

In electrochemistry, we represent a cell with a shorthand notation. For the Daniell cell:

\text{Zn(s)} \mid \text{Zn}^{2+}(\text{aq}) \parallel \text{Cu}^{2+}(\text{aq}) \mid \text{Cu(s)}

The single vertical line $\mid$ represents a phase boundary (solid electrode | solution). The double line $\parallel$ represents the salt bridge.

The anode (oxidation) is written on the left, the cathode (reduction) on the right. Electrons flow from left to right in the external circuit.

Applying the Nernst Equation to a Cell Representation

For the Daniell cell, the half-reactions are:

Anode (oxidation): $\text{Zn(s)} \rightarrow \text{Zn}^{2+}(\text{aq}) + 2e^-$
Cathode (reduction): $\text{Cu}^{2+}(\text{aq}) + 2e^- \rightarrow \text{Cu(s)}$

Overall: $\text{Zn(s)} + \text{Cu}^{2+}(\text{aq}) \rightarrow \text{Zn}^{2+}(\text{aq}) + \text{Cu(s)}$

Here $n = 2$ (two electrons transferred). The reaction quotient is:

Q = \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]}

So the Nernst equation becomes:

E = E^\circ - \frac{0.0591}{2} \log_{10} \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]}

Note

Solids (Zn, Cu) do not appear in $Q$ because their concentrations are constant (activity = 1).

A Worked Example

Suppose you have a Daniell cell where $[\text{Zn}^{2+}] = 0.1$ M and $[\text{Cu}^{2+}] = 1.0$ M at 25°C. $E^\circ$ for the cell is 1.10 V.

$E = 1.10 - \frac{0.0591}{2} \log_{10} \frac{0.1}{1.0}$ …

Why this formula?

Cell Representation & the Nernst Equation: Why It Works

The Core Question

Why does a cell's voltage change when concentrations change? The Nernst equation answers this — but the reason lies in the link between chemical free energy and electrical work.

1. The Fundamental Link: Gibbs Free Energy & Cell Potential

A galvanic cell does electrical work. The maximum useful work a cell can do equals the change in Gibbs free energy ( $\Delta G$ ):

\Delta G = -nFE_{\text{cell}}

Where:

$n$ = moles of electrons transferred
$F$ = Faraday constant ( $96,485 \, \text{C mol}^{-1}$ )
$E_{\text{cell}}$ = cell potential (volts)

Why negative? A spontaneous reaction has $\Delta G < 0$ and $E_{\text{cell}} > 0$ — the negative sign makes this consistent.

2. The Chemical Side: $\Delta G$ Depends on Concentration

For a general redox reaction:

aA + bB \rightarrow cC + dD

The Gibbs free energy under non-standard conditions is:

\Delta G = \Delta G^\circ + RT \ln Q

Where $Q$ is the reaction quotient:

Q = \frac{[C]^c [D]^d}{[A]^a [B]^b}

Why this form? It comes from the relationship between chemical potential and concentration — the entropy of mixing drives concentration dependence.

3. Combining Both Sides: The Derivation

Set the electrical work equal to the chemical free energy change:

-nFE_{\text{cell}} = -nFE^\circ_{\text{cell}} + RT \ln Q

Divide both sides by $-nF$ :

E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{RT}{nF} \ln Q

This is the Nernst equation.

4. The "Why" in Plain Terms

Concept	Physical Meaning
$E^\circ_{\text{cell}}$	Voltage when all species are at 1 M (standard state)
$-\frac{RT}{nF} \ln Q$	Correction factor — adjusts voltage for real concentrations
$Q$	Tells you how far the reaction is from equilibrium

Key insight: When $Q = K$ (equilibrium), $E_{\text{cell}} = 0$ — the battery is dead because no net reaction occurs.

5. The Common Form (log base 10)

At $25^\circ \text{C}$ ( $298 \, \text{K}$ ):

\frac{RT}{F} \ln 10 \approx 0.0592 \, \text{V}

So:

E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0592}{n} \log_{10} Q

Why convert to log? Exam convenience — most concentration values are powers of 10.

--- …

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At equilibrium, the cell potential is zero, so the Nernst equation gives $E^\circ_{Cell} = \frac{0.059}{n} \log K_c$ at 298 K. For the Daniel cell ( $n=2$ , $E^\circ=1.1\ \text{V}$ ), this yields $\log K_c = \frac{2.2}{0.059}$ , making options (ii) and (iii) correct.

The key is to remember what "equilibrium" means for an electrochemical cell. A Daniel cell (Zn-Cu) has a standard potential of 1.1 V when all species are at unit activity. But as the cell discharges, concentrations change, and the potential drops. Eventually, the cell reaches a state where no further net reaction occurs — that's equilibrium. At that point, the cell potential is exactly zero.

The Nernst equation connects the cell potential to the reaction quotient $Q$ :

E_{Cell} = E^\circ_{Cell} - \frac{RT}{nF} \ln Q

At equilibrium, $E_{Cell} = 0$ and $Q = K_c$ (the equilibrium constant). So:

0 = E^\circ_{Cell} - \frac{RT}{nF} \ln K_c

Rearranging:

E^\circ_{Cell} = \frac{RT}{nF} \ln K_c

This is the fundamental relation. For the Daniel cell, $n = 2$ (two electrons transferred: $\text{Zn} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cu}$ ) and $E^\circ_{Cell} = 1.1\ \text{V}$ .

Now let's examine each option.

Option (i): $1.1 = K_c$

This would mean the equilibrium constant equals 1.1. But $K_c$ is a dimensionless equilibrium constant, typically enormous for a spontaneous cell (around $10^{37}$ for Daniel cell). A potential of 1.1 V corresponds to a huge $K_c$ , not a number near 1. So this is clearly wrong — it confuses potential with the constant itself.
Option (ii): $\frac{2.303RT}{2F} \log K_c = 1.1$

This is just the natural log converted to base 10. Since $\ln K_c = 2.303 \log K_c$ , and $n=2$ , we have:

E^\circ_{Cell} = \frac{RT}{2F} \cdot 2.303 \log K_c = \frac{2.303RT}{2F} \log K_c

Setting this equal to 1.1 V is exactly the equilibrium condition. So (ii) is correct.

Option (iii): $\log K_c = \frac{2.2}{0.059}$ At 298 K, the factor $\frac{2.303RT}{F}$ equals 0.059 V. So: $\frac{2.303RT}{2F} = \frac{0.059}{2} = 0.0295$ …

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Common Mistakes in Nernst Equation & Equilibrium (Daniel Cell)

Students often lose marks here due to conceptual confusion between cell potential at equilibrium and standard cell potential. Let's break down the pitfalls.

🔴 Mistake 1: Confusing $E_{Cell}$ with $E^\circ_{Cell}$ at equilibrium

The error:

Thinking that at equilibrium, $E_{Cell} = E^\circ_{Cell} = 1.1\ V$ .

Why it's wrong:

At equilibrium, the cell reaction has stopped — no net driving force.

The Nernst equation at equilibrium becomes:

E_{Cell} = E^\circ_{Cell} - \frac{0.059}{n} \log K_c = 0

So $E_{Cell} = 0$ , not $1.1\ V$ .

How to avoid:

Remember:

$E^\circ_{Cell}$ is for standard conditions (1 M, 1 atm, 298 K).
At equilibrium, $E_{Cell} = 0$ and $Q = K_c$ .

🔴 Mistake 2: Misplacing the formula for $\log K_c$

The error:

Writing $\log K_c = \frac{n E^\circ_{Cell}}{0.059}$ but forgetting the sign or factor of 2.

Correct derivation:

0 = E^\circ_{Cell} - \frac{0.059}{n} \log K_c

\Rightarrow \log K_c = \frac{n E^\circ_{Cell}}{0.059}

For Daniel cell, $n = 2$ and $E^\circ_{Cell} = 1.1\ V$ :

\log K_c = \frac{2 \times 1.1}{0.059} = \frac{2.2}{0.059}

How to avoid:

Always write the Nernst equation fully before substituting:

E_{Cell} = E^\circ_{Cell} - \frac{0.059}{n} \log Q

Then set $E_{Cell} = 0$ and $Q = K_c$ .

🔴 Mistake 3: Forgetting the $2.303RT/F$ form

The error:

Using $0.059$ when the question gives $2.303RT/F$ — or vice versa.

Key fact:

At 298 K:

\frac{2.303RT}{F} = 0.059\ V

So option (ii) is correct because:

\frac{2.303RT}{2F} \log K_c = \frac{0.059}{2} \log K_c = 1.1

This is just a rearranged form of the correct relation.

How to avoid:

If the question uses $2.303RT/F$ , do not replace it with $0.059$ unless asked.
Check if the expression is algebraically equivalent to the standard formula. …

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Q. $E^\circ_{Cell} = 1.1\ V$ for Daniel cell. Which of the following expressions are correct description of state of equilibrium in this cell? (Two or more than two options may be correct.)

Concept understanding — Cell Representation Nernst Equation

Cell Representation and the Nernst Equation: From Intuition to Precision

The Intuition First

The Precise Statement

Cell Representation: How We Write It

Applying the Nernst Equation to a Cell Representation

A Worked Example

Why this formula?

Cell Representation & the Nernst Equation: Why It Works

The Core Question

1. The Fundamental Link: Gibbs Free Energy & Cell Potential

2. The Chemical Side: $\Delta G$ Depends on Concentration

3. Combining Both Sides: The Derivation

4. The "Why" in Plain Terms

5. The Common Form (log base 10)

Method: Nernst Equation at Equilibrium

Step-by-step reasoning

Common Mistakes in Nernst Equation & Equilibrium (Daniel Cell)

🔴 Mistake 1: Confusing $E_{Cell}$ with $E^\circ_{Cell}$ at equilibrium

🔴 Mistake 2: Misplacing the formula for $\log K_c$

🔴 Mistake 3: Forgetting the $2.303RT/F$ form

Q.ECell∘=1.1 VE^\circ_{Cell} = 1.1\ VECell∘​=1.1 V for Daniel cell. Which of the following expressions are correct description of state of equilibrium in this cell? (Two or more than two options may be correct.)

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Q. $E^\circ_{Cell} = 1.1\ V$ for Daniel cell. Which of the following expressions are correct description of state of equilibrium in this cell? (Two or more than two options may be correct.)

Unlock everything free for 14 days