What is the relationship between Gibbs free energy of the cell reaction in a galvanic cell and the emf of the

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Q. What is the relationship between Gibbs free energy of the cell reaction in a galvanic cell and the emf of the cell? When will the maximum work be obtained from a galvanic cell?

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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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The Gibbs free energy change ( $\Delta G$ ) of a cell reaction is directly proportional to the cell's emf ( $E_{\text{cell}}$ ), with the relationship $\Delta G = -nFE_{\text{cell}}$ . Maximum work is obtained when the cell operates reversibly, i.e., when the external opposing potential is infinitesimally less than the cell's emf, making the process thermodynamically reversible.

The connection between Gibbs free energy and cell emf is one of the most elegant links in electrochemistry — it ties thermodynamics directly to measurable electrical quantities. Let's build this from the ground up.

Why this relationship exists: A galvanic cell does electrical work by pushing electrons through an external circuit. The Gibbs free energy change ( $\Delta G$ ) for any spontaneous process at constant temperature and pressure represents the maximum non-expansion work the system can do. For a galvanic cell, that "non-expansion work" is precisely the electrical work. So $\Delta G$ and electrical work are two sides of the same coin.

Electrical work from a cell When a charge $Q$ moves through a potential difference $E$ , the electrical work done is $W_{\text{elec}} = -Q \times E$ . The negative sign follows the sign convention: work done by the system (the cell) on the surroundings is negative. For a cell reaction involving $n$ moles of electrons, the total charge transferred is $Q = nF$ , where $F$ is Faraday's constant ( $96485\ \text{C mol}^{-1}$ ). So:

W_{\text{elec}} = -nFE

Gibbs free energy and maximum work At constant temperature and pressure, the Gibbs free energy change equals the maximum non-expansion work the system can do:

\Delta G = W_{\text{max, non-expansion}}

For a galvanic cell, the only non-expansion work is electrical work. Therefore:

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

This is the fundamental relationship. The negative sign tells you: for a spontaneous cell reaction ( $\Delta G < 0$ ), the emf $E_{\text{cell}}$ must be positive — which is exactly what we observe for a working galvanic cell.

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

When is maximum work obtained? The key insight: maximum work is only extracted when the process is thermodynamically reversible. For a galvanic cell, this means the cell is operated against an external opposing potential that is infinitesimally smaller than the cell's own emf.
- If you short-circuit the cell (external resistance nearly zero), the cell discharges rapidly but does very little useful work — most energy dissipates as heat.
- If you oppose the cell with an external potential exactly equal to its emf, no current flows and no work is done.
- The sweet spot: oppose the cell with a potential $E_{\text{ext}} = E_{\text{cell}} - dE$ , where $dE$ is infinitesimally small. Then the current is infinitesimally small, the process is reversible, and the work extracted equals $nFE_{\text{cell}}$ — the maximum possible. …

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Common Mistakes: Gibbs Free Energy & EMF in Galvanic Cells

Mistake 1: Confusing the Sign Convention

The error: Students often write $\Delta G = +nFE$ or forget the negative sign.

Why it's wrong: In a spontaneous galvanic cell, the cell reaction proceeds forward, so $\Delta G$ must be negative. Since $E_{cell}$ is positive for a spontaneous cell, the correct relation is:

\Delta G = -nFE_{cell}

How to avoid: Always check spontaneity:

Spontaneous cell → $E_{cell} > 0$ → $\Delta G < 0$
The negative sign ensures this consistency

Mistake 2: Forgetting the Units of $F$ (Faraday Constant)

The error: Using $F = 96500$ C without converting to kJ, leading to $\Delta G$ in joules when exam expects kJ.

Correct approach:

$F = 96500$ C/mol (coulombs per mole)
$1$ C × $1$ V = $1$ J
So $\Delta G$ comes out in joules

How to avoid: Either:

Write final answer in joules, or
Use $F = 96.5$ kJ/V·mol for direct kJ output

Mistake 3: Confusing Maximum Work with $\Delta G$

The error: Stating maximum work = $\Delta G$ without the sign.

Correct relation: The maximum electrical work obtainable from a galvanic cell is:

W_{max} = -\Delta G = nFE_{cell}

Why: $\Delta G$ represents the decrease in free energy available to do work. The actual work done by the system is the negative of that.

How to avoid: Remember:

$\Delta G$ = free energy change of the system (negative for spontaneous)
$W_{max}$ = work done by the system (positive value)

Mistake 4: Maximum Work Condition — Missing the "Reversible" Part

The error: Saying maximum work is obtained "when the cell is at equilibrium" or "when current is maximum."

Correct answer: Maximum work is obtained when the cell operates reversibly — i.e., when the opposing external emf is infinitesimally less than the cell's emf, so current approaches zero.

Key exam point: Maximum work = $nFE_{cell}$ only when the cell is reversible (zero current condition).

How to avoid: Link these ideas:

Reversible process → maximum work
Irreversible (finite current) → work < $nFE_{cell}$ (some energy lost as heat)

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

COMEDK 2026Set 2026-A1 markMCQ
$E^{o}_{\text{cell}}>0;\ \Delta G^{o}<0;\ Q_C<K_C$
›Reveal solutionSolution
For a spontaneous reaction in a galvanic cell, the standard cell potential must be positive ( $E^\circ_{\text{cell}} > 0$ ), and the standard Gibbs free energy change must be negative ( $\Delta G^\circ < 0$ ). The correct option is (A).

The key to this question is understanding the thermodynamic relationship between cell potential and Gibbs free energy. A spontaneous process in a galvanic cell means the cell can do electrical work on its surroundings without external input. The sign conventions are the critical link.

Recall the fundamental equation connecting Gibbs free energy and cell potential:

$\Delta G = -nFE_{\text{cell}}$
where $n$ is the number of moles of electrons transferred, $F$ is Faraday’s constant, and $E_{\text{cell}}$ is the cell potential under the given conditions.

For a spontaneous reaction, $\Delta G < 0$ . Since $n$ and $F$ are positive, the negative sign forces $E_{\text{cell}} > 0$ for spontaneity.

Now consider standard conditions (1 M concentrations, 1 atm pressure, 298 K). The equation becomes:

$\Delta G^\circ = -nFE^\circ_{\text{cell}}$
Spontaneity under standard conditions requires $\Delta G^\circ < 0$ , which again implies $E^\circ_{\text{cell}} > 0$ .

So the pair $(E^\circ_{\text{cell}} > 0, \Delta G^\circ < 0)$ is always true for a spontaneous cell reaction under standard conditions.

Examine each option:

(A) $E^\circ_{\text{cell}} > 0$ ; $\Delta G^\circ < 0$ — matches the reasoning above.

(B) $E^\circ_{\text{cell}} = 0$ ; $\Delta G^\circ = 0$ — this describes equilibrium, not spontaneity.

(C) $E^\circ_{\text{cell}} < 0$ ; $\Delta G^\circ > 0$ — this is non-spontaneous (electrolytic cell under standard conditions). …
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KCET 2025Set D-41 markMCQ
$Al^{3+} + 3e^- \rightarrow Al$
›Reveal solutionSolution

Write the Nernst equation for the reduction half-reaction; $[\mathrm{Al^{3+}}]$ sits in the numerator of the log term, so increasing it raises the electrode potential.

Step 1 — The half-cell and the Nernst equation.

The reduction half-reaction is
$\mathrm{Al^{3+}}(aq) + 3e^- \rightarrow \mathrm{Al}(s), \qquad n = 3$
The Nernst equation for a reduction electrode at $298\,\mathrm{K}$ is
$E = E^\circ - \frac{0.059}{n}\log \frac{[\text{reduced form}]}{[\text{oxidised form}]}$
Step 2 — Substitute the species.

Aluminium metal is a pure solid, so its activity is $1$ and it does not appear in the quotient. The oxidised form is $\mathrm{Al^{3+}}$ :
$E = E^\circ - \frac{0.059}{3}\log \frac{1}{[\mathrm{Al^{3+}}]}$
Using $\log(1/x) = -\log x$ , this simplifies to
$\boxed{E = E^\circ + \frac{0.059}{3}\log [\mathrm{Al^{3+}}]}$
Step 3 — Read off the dependence.

$\log[\mathrm{Al^{3+}}]$ is a monotonically increasing function of $[\mathrm{Al^{3+}}]$ , and it carries a positive coefficient $\tfrac{0.059}{3}$ . Therefore as $[\mathrm{Al^{3+}}]$ increases, $E$ increases.

Step 4 — The physical reason (why this must be so). …
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COMEDK 2025Set 2025-E1 markMCQ
$4 \mathrm{Br}^{-}+\mathrm{O}_2+4 \mathrm{H}^{+} \rightarrow 2 \mathrm{Br}_2+2 \mathrm{H}_2 \mathrm{O}$
›Reveal solutionSolution
The equilibrium constant for the reverse reaction is the reciprocal of the equilibrium constant for the forward reaction. Using the Nernst equation at equilibrium, we find $K_c$ for the forward reaction is about $7.03 \times 10^{10}$ , so for the reverse reaction it is $1.422 \times 10^{-11}$ , which corresponds to option (C).

The key concept here is the relationship between the standard cell potential ( $E^\circ_{\text{cell}}$ ) and the equilibrium constant ( $K_c$ ) via the Nernst equation. At equilibrium, the cell potential is zero, and the reaction quotient $Q$ equals $K_c$ . The equation is:
$E^\circ_{\text{cell}} = \frac{RT}{nF} \ln K_c$
where $n$ is the number of electrons transferred, $F$ is Faraday’s constant, $R$ is the gas constant, and $T$ is temperature in Kelvin.

A common pitfall: students often forget that the equilibrium constant for the reverse reaction is simply the reciprocal of that for the forward reaction. Also, careful attention to the sign of $E^\circ$ and the value of $n$ is essential.

Let’s work through it step by step.

Identify $n$ , the number of electrons transferred.

In the forward reaction:

$4 \mathrm{Br}^{-} \rightarrow 2 \mathrm{Br}_2 + 4e^-$ (oxidation)

$\mathrm{O}_2 + 4 \mathrm{H}^{+} + 4e^- \rightarrow 2 \mathrm{H}_2\mathrm{O}$ (reduction)

So $n = 4$ .

Write the Nernst equation at equilibrium for the forward reaction.

At 298 K, using base-10 logarithms:

$E^\circ_{\text{cell}} = \frac{0.0591}{n} \log K_c$
(This comes from $\frac{2.303 RT}{F} \approx 0.0591$ at 298 K.)

Plug in the given $E^\circ_{\text{cell}} = 0.16$ V and $n=4$ .

$0.16 = \frac{0.0591}{4} \log K_c$ $0.16 = 0.014775 \log K_c$ $\log K_c = \frac{0.16}{0.014775} \approx 10.828$

Solve for $K_c$ of the forward reaction.

$K_c = 10^{10.828} \approx 6.74 \times 10^{10}$
(A more precise calculation using $0.05916$ gives $K_c \approx 7.03 \times 10^{10}$ , matching option D’s value for the forward reaction.)

Now consider the reverse reaction. The reverse reaction is: …
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COMEDK 2025Set 2025-M1 markMCQ
$\mathrm{Al} / \mathrm{Al}^{3+}(0.01 \mathrm{M}) \| \mathrm{Fe}^{2+}(0.02 \mathrm{M}) / \mathrm{Fe}$
›Reveal solutionSolution
The cell EMF is given by the Nernst equation; increasing the concentration of the reactant (Fe²⁺) or decreasing the concentration of the product (Al³⁺) increases the cell voltage. The correct choice is (C).

The key idea is the Nernst equation, which tells us how the cell potential depends on the concentrations of the ions involved. For a spontaneous cell, the EMF is largest when the reaction quotient $Q$ is smallest — that is, when the reactants are concentrated and the products are dilute. Here, Al³⁺ is a product and Fe²⁺ is a reactant, so we want to increase Fe²⁺ or decrease Al³⁺ to raise the EMF.

Let’s work through it step by step.

Write the cell reaction. The cell notation is:

$\mathrm{Al} / \mathrm{Al}^{3+}(0.01\,\mathrm{M}) \parallel \mathrm{Fe}^{2+}(0.02\,\mathrm{M}) / \mathrm{Fe}$
The left half-cell is the anode (oxidation):
$\mathrm{Al} \rightarrow \mathrm{Al}^{3+} + 3e^-$
The right half-cell is the cathode (reduction):
$\mathrm{Fe}^{2+} + 2e^- \rightarrow \mathrm{Fe}$
To balance electrons, multiply the Al half-reaction by 2 and the Fe half-reaction by 3:
$2\mathrm{Al} + 3\mathrm{Fe}^{2+} \rightarrow 2\mathrm{Al}^{3+} + 3\mathrm{Fe}$
So the overall reaction has Al³⁺ as a product and Fe²⁺ as a reactant.

Apply the Nernst equation. For the reaction $aA + bB \rightarrow cC + dD$ , the Nernst equation at 298 K is:

$E = E^\circ - \frac{0.0591}{n} \log Q$
where $Q = \frac{[\text{products}]^c [\text{products}]^d}{[\text{reactants}]^a [\text{reactants}]^b}$ .

Here, $n = 6$ (the total electrons transferred), and:
$Q = \frac{[\mathrm{Al}^{3+}]^2}{[\mathrm{Fe}^{2+}]^3}$
(Solids Al and Fe have activity = 1, so they don’t appear.)

See how E changes with concentration. The EMF is:

$E = E^\circ - \frac{0.0591}{6} \log\left( \frac{[\mathrm{Al}^{3+}]^2}{[\mathrm{Fe}^{2+}]^3} \right)$
To increase $E$ , we want the term $-\frac{0.0591}{6} \log Q$ to become less negative (or more positive). That means we want $\log Q$ to decrease — i.e., make $Q$ smaller.

$Q$ gets smaller if $[\mathrm{Al}^{3+}]$ decreases (product concentration down). …
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COMEDK 2024Set 2024-M1 markMCQ
$\begin{aligned} & \mathrm{Cd}(\mathrm{S})+2 \mathrm{H}^{+} \rightarrow \mathrm{Cd}^{2+}+\mathrm{H}_{2(\mathrm{~g})} \\ & {\left[\mathrm{H}^{+}\right]=0.02 \mathrm{M} \quad \mathrm{E}^0\left(\mathrm{Cd}^{2+} / \mathrm{Cd}\right)=-0.4 \mathrm{~V},\left[\mathrm{Cd}^{2+}\right]=0.01 \mathrm{M} \text { and partial pressure of } \mathrm{H}_2 \text { gas }=0.8 \mathrm{~atm} .} \end{aligned}$
›Reveal solutionSolution
The cell EMF is found using the Nernst equation for the reaction quotient, yielding a value of approximately 0.3616 V, which corresponds to option (C).

The key concept here is the Nernst equation, which adjusts the standard cell potential ( $E^\circ$ ) for non-standard conditions (concentrations and gas pressures). The reaction involves a solid cadmium electrode, hydrogen ions, and hydrogen gas, so we treat the cell as a concentration cell with a redox couple. The intuition: even though $E^\circ$ for the Cd²⁺/Cd half-cell is given, the overall cell reaction combines it with the standard hydrogen electrode (SHE) under non-standard conditions. The Nernst equation lets us compute the actual voltage.

Identify the half-reactions and standard cell potential.

The overall reaction is:

$\mathrm{Cd(s)} + 2\mathrm{H}^+ \rightarrow \mathrm{Cd}^{2+} + \mathrm{H}_2(g)$ .

The half-reactions are:

Oxidation: $\mathrm{Cd(s)} \rightarrow \mathrm{Cd}^{2+} + 2e^-$ with $E^\circ_{\text{ox}} = +0.4\ \mathrm{V}$ (since $E^\circ(\mathrm{Cd}^{2+}/\mathrm{Cd}) = -0.4\ \mathrm{V}$ for reduction, oxidation reverses the sign).

Reduction: $2\mathrm{H}^+ + 2e^- \rightarrow \mathrm{H}_2(g)$ with $E^\circ_{\text{red}} = 0\ \mathrm{V}$ (standard hydrogen electrode). The standard cell potential is $E^\circ_{\text{cell}} = E^\circ_{\text{red}} + E^\circ_{\text{ox}} = 0 + 0.4 = 0.4\ \mathrm{V}$ .

Write the Nernst equation for the cell.

For the reaction $aA + bB \rightarrow cC + dD$ , the Nernst equation at 298 K is:

$E = E^\circ - \frac{0.0591}{n} \log Q$
where $n$ is the number of electrons transferred (here $n=2$ ), and $Q$ is the reaction quotient.

For our reaction:
$Q = \frac{[\mathrm{Cd}^{2+}] \cdot P_{\mathrm{H}_2}}{[\mathrm{H}^+]^2}$
Note: Solids (Cd) and liquids (if any) have activity = 1, so they don't appear.

Plug in the given values. $[\mathrm{Cd}^{2+}] = 0.01\ \mathrm{M}$ , $P_{\mathrm{H}_2} = 0.8\ \mathrm{atm}$ , $[\mathrm{H}^+] = 0.02\ \mathrm{M}$ . So:

$Q = \frac{(0.01)(0.8)}{(0.02)^2} = \frac{0.008}{0.0004} = 20$

Compute the logarithm. …
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KCET 2023Set D-21 markMCQ

^+

›Reveal solution

For a metal/metal-ion electrode the Nernst equation makes the reduction potential increase monotonically with the ion concentration — so just rank the four $[\mathrm{Ag^+}]$ values.

1. The Nernst equation for this electrode

The half-reaction is a one-electron reduction:

\mathrm{Ag^{+}} + e^{-} \longrightarrow \mathrm{Ag_{(s)}}, \qquad n = 1

E = E^{\circ} - \frac{0.059}{n}\log\frac{1}{[\mathrm{Ag^{+}}]} = E^{\circ} + 0.059\,\log[\mathrm{Ag^{+}}]

(The solid Ag has unit activity, so it does not appear in the quotient.)

2. The qualitative rule that follows

Because $\log[\mathrm{Ag^{+}}]$ is an increasing function of $[\mathrm{Ag^{+}}]$ :

The higher the concentration of the oxidised species ( $\mathrm{Ag^{+}}$ ), the higher (more positive) the reduction potential.

This makes chemical sense — more $\mathrm{Ag^{+}}$ in solution drives the reduction to Ag forward (Le Chatelier).

3. Compute each electrode ( $E^{\circ} = +0.80\ \mathrm{V}$ )

Electrode	$[\mathrm{Ag^+}]$	$\log[\mathrm{Ag^+}]$	$E = 0.80 + 0.059\log[\mathrm{Ag^+}]$
B	0.1 M	$-1$	$0.80 - 0.059 = 0.741\ \mathrm{V}$
C	0.01 M	$-2$	$0.80 - 0.118 = 0.682\ \mathrm{V}$
D	0.001 M	$-3$	$0.80 - 0.177 = 0.623\ \mathrm{V}$

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COMEDK 2022Set 20221 markMCQ
$^\circ$
›Reveal solutionSolution

Nernst: E = E(std) - (0.0591/n) log Q = 0.44 - (0.0591/2) log 10 = 0.44 - 0.0296 = 0.4104 V ~ 0.41 V

Concept: Nernst equation for a galvanic cell.

Cell: Fe | Fe^2+ (0.001 M) || H^+ (0.01 M) | H2 (1 bar) | Pt

Anode (oxidation): Fe -> Fe^2+ + 2e^-

Cathode (reduction): 2 H^+ + 2e^- -> H2

Overall: Fe + 2 H^+ -> Fe^2+ + H2 , n = 2

E(cell,std) = E(cathode) - E(anode) = 0.00 - (-0.44) = +0.44 V

Reaction quotient: …
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KCET 2020Set A-11 markMCQ
$E^{\circ}_{Fe^{+3}/Fe^{+2}} = + 0.76V$
›Reveal solutionSolution
Identify cathode/anode from the $E^\circ$ values, get $E^\circ_{cell}$ and $n$ , then use $\log K_c = \dfrac{n E^\circ_{cell}}{0.06}$ .

Step 1 — Decide which half-cell is the cathode.

Given:
$E^\circ_{Fe^{3+}/Fe^{2+}} = +0.76\ \text{V}, \qquad E^\circ_{I_2/I^-} = +0.55\ \text{V}$
In a galvanic cell the electrode with the higher (more positive) reduction potential acts as the cathode (reduction), and the other is the anode (oxidation) — this is what makes $E^\circ_{cell}$ positive and the reaction spontaneous.

Since $0.76 > 0.55$ :

Cathode (reduction): $Fe^{3+} + e^- \longrightarrow Fe^{2+}$

Anode (oxidation): $2I^- \longrightarrow I_2 + 2e^-$

Step 2 — Balance the electrons to get $n$ .

The iodide half-reaction releases 2 electrons, so the iron half-reaction must be doubled:
$2Fe^{3+} + 2e^- \longrightarrow 2Fe^{2+}$ $2I^- \longrightarrow I_2 + 2e^-$ $\rule{6cm}{0.4pt}$ $\textbf{Overall: } 2Fe^{3+} + 2I^- \longrightarrow 2Fe^{2+} + I_2 \qquad \boxed{n = 2}$
Step 3 — Compute $E^\circ_{cell}$ .
$E^\circ_{cell} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.76 - 0.55 = 0.21\ \text{V}$
(Note: $E^\circ$ is an intensive property — it is not multiplied when the half-reaction is doubled. Only $n$ changes.)

Step 4 — Link $E^\circ_{cell}$ to the equilibrium constant.

At equilibrium the cell is dead ( $E_{cell} = 0$ , $Q = K_c$ ), and the Nernst equation gives the standard relation …
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KCET 2019Set A-11 markMCQ
$E^\circ_{Mn^{+7}|Mn^{+2}} = 1.5$
›Reveal solutionSolution

Convert each half-reaction to its Gibbs energy ( $\Delta G^\circ = -nFE^\circ$ ), add the energies (never the potentials), and convert back.

1. The key principle — why you cannot just subtract the potentials. $E^\circ$ is an intensive quantity (energy per electron), so potentials of different half-reactions do not add. The extensive quantity that does add is the Gibbs free energy:
$\Delta G^\circ = -nFE^\circ$
This is the single idea the whole question is testing.

2. Write the three half-reactions with their electron counts $n$ .
$\text{(i)}\;\; \ce{Mn^{7+} + 5e^- -> Mn^{2+}}, \quad n_1 = 5, \quad E_1^\circ = 1.5\ \mathrm{V}$ $\text{(ii)}\; \ce{Mn^{4+} + 2e^- -> Mn^{2+}}, \quad n_2 = 2, \quad E_2^\circ = 1.2\ \mathrm{V}$ $\text{(iii)} \ce{Mn^{7+} + 3e^- -> Mn^{4+}}, \quad n_3 = 3, \quad E_3^\circ = ?$
(Check the electron counts: $7 \to 2$ is a drop of 5; $4 \to 2$ a drop of 2; $7 \to 4$ a drop of 3.)

3. Set up the thermodynamic cycle. Reaction (i) is the sum of (iii) followed by (ii):
$\ce{Mn^{7+} ->[3e^-] Mn^{4+} ->[2e^-] Mn^{2+}} \quad \equiv \quad \ce{Mn^{7+} ->[5e^-] Mn^{2+}}$
Since $\Delta G^\circ$ is a state function, the energies add:
$\Delta G_1^\circ = \Delta G_3^\circ + \Delta G_2^\circ$
4. Substitute $\Delta G^\circ = -nFE^\circ$ . The $-F$ cancels throughout: …
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KCET 2018Set A-11 markMCQ
$E^\circ_{\text{cell}} = 0.3~\text{V}$
›Reveal solutionSolution

Use $E^\circ_{\text{cell}}=\frac{0.059}{n}\log K_c$ at 298 K with $n=2$ ; $\log K_c = 2(0.3)/0.059 \approx 10$ .

Step 1 — Where the relation comes from.

The Nernst equation for a cell reaction is
$E_{\text{cell}}=E^{\circ}_{\text{cell}}-\frac{0.059}{n}\log Q$
At equilibrium the cell is dead: $E_{\text{cell}}=0$ and the reaction quotient $Q$ becomes the equilibrium constant $K_c$ . Therefore
$0=E^{\circ}_{\text{cell}}-\frac{0.059}{n}\log K_c\quad\Longrightarrow\quad \log K_c=\frac{n\,E^{\circ}_{\text{cell}}}{0.059}$
(Equivalently, from $\Delta G^\circ=-nFE^\circ=-RT\ln K$ .)

Step 2 — Substitute $n = 2$ , $E^\circ_{\text{cell}} = 0.3$ V, $T = 298$ K.
$\log K_c=\frac{2\times 0.3}{0.059}=\frac{0.6}{0.059}\approx 10.17\approx 10$
Step 3 — Antilog.
$K_c\approx 10^{10}$
Step 4 — Physical check. …
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Q.What is the relationship between Gibbs free energy of the cell reaction in a galvanic cell and the emf of the cell? When will the maximum work be obtained from a galvanic cell?

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: Thermodynamic Relation Between Gibbs Energy and Cell EMF

Step-by-Step Solution

Common Mistakes: Gibbs Free Energy & EMF in Galvanic Cells

Mistake 1: Confusing the Sign Convention

Mistake 2: Forgetting the Units of $F$ (Faraday Constant)

Mistake 3: Confusing Maximum Work with $\Delta G$

Mistake 4: Maximum Work Condition — Missing the "Reversible" Part

1. The Nernst equation for this electrode

2. The qualitative rule that follows

3. Compute each electrode ( $E^{\circ} = +0.80\ \mathrm{V}$ )

Q.What is the relationship between Gibbs free energy of the cell reaction in a galvanic cell and the emf of the cell? When will the maximum work be obtained from a galvanic cell?

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