Consider a cell given below: $Cu \mid Cu^{2+} \parallel Cl^- \mid Cl_2, Pt$ Write the reactions that occur at anode and cathode.

Consider a cell given below: Cu mid Cu²⁺ parallel Cl^- mid Cl₂, Pt Write the reactions that occur at anode and

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Cu \mid Cu^{2+} \parallel Cl^- \mid Cl_2, Pt

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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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In this electrochemical cell, the anode is where oxidation occurs (Cu → Cu²⁺ + 2e⁻) and the cathode is where reduction occurs (Cl₂ + 2e⁻ → 2Cl⁻). The cell notation tells us the left side is the anode and the right side is the cathode.

Let's understand what this cell notation actually means before jumping into the reactions.

The Language of Cell Notation

The notation $Cu \mid Cu^{2+} \parallel Cl^- \mid Cl_2, Pt$ follows a standard convention. The single vertical line $\mid$ represents a phase boundary (solid electrode | solution). The double vertical line $\parallel$ represents the salt bridge that connects the two half-cells.

By convention, the anode (where oxidation happens) is written on the left, and the cathode (where reduction happens) is written on the right. This is a critical rule to remember.

Watch out

A common mistake is to reverse the electrodes. Remember: Left = Anode (oxidation), Right = Cathode (reduction) in standard cell notation.

Step-by-Step Breakdown

1. Identify the two half-cells

The left half-cell is $Cu \mid Cu^{2+}$ . This means a copper metal electrode is in contact with a solution containing Cu²⁺ ions.

The right half-cell is $Cl^- \mid Cl_2, Pt$ . Here, a platinum electrode (inert, written last) is in contact with a solution containing Cl⁻ ions and chlorine gas (Cl₂). Platinum is used because it doesn't participate chemically — it just conducts electrons.

2. Determine the reaction at the anode (left side)

At the anode, oxidation occurs — the species loses electrons. Looking at the left half-cell, copper metal (Cu) can lose two electrons to become Cu²⁺ ions:

Cu(s) \rightarrow Cu^{2+}(aq) + 2e^-

This is oxidation because the oxidation state of copper increases from 0 to +2.

Tip

A quick way to confirm: if the electrode is a metal (like Cu) and it's on the left, it almost always undergoes oxidation. The metal dissolves into the solution.

3. Determine the reaction at the cathode (right side)

At the cathode, reduction occurs — the species gains electrons. Looking at the right half-cell, chlorine gas (Cl₂) can gain two electrons to become two chloride ions (Cl⁻):

Cl_2(g) + 2e^- \rightarrow 2Cl^-(aq)

This is reduction because the oxidation state of chlorine decreases from 0 to -1.

4. Verify the overall cell reaction (optional but helpful) …

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Here are the most common mistakes students make with this specific electrochemical cell setup, along with how to avoid each.

Mistake 1: Misidentifying the Anode and Cathode

The Error:

Students often assume the left side is always the anode and the right side is always the cathode. In this cell, they might write the oxidation of $Cu$ at the left electrode (which is correct) but then incorrectly write the reduction of $Cl_2$ at the right electrode (which is also correct, but for the wrong reason).

Why it happens:

They memorize "anode on left, cathode on right" without understanding the underlying chemistry. The cell notation $Cu \mid Cu^{2+} \parallel Cl^- \mid Cl_2, Pt$ tells us the anode is on the left and the cathode is on the right only if the cell is spontaneous. Here, it is.

How to avoid it:

Always determine the direction of electron flow based on the standard reduction potentials ( $E^\circ$ ).

Step 1: Write the two half-reactions.
- Left half-cell: $Cu^{2+} + 2e^- \rightarrow Cu$ (Reduction potential $E^\circ = +0.34$ V)
- Right half-cell: $Cl_2 + 2e^- \rightarrow 2Cl^-$ (Reduction potential $E^\circ = +1.36$ V)
Step 2: The half-cell with the higher reduction potential (more positive) will undergo reduction (gain electrons). Here, $Cl_2/Cl^-$ has $+1.36$ V > $+0.34$ V, so $Cl_2$ is reduced at the cathode.
Step 3: The other half-cell (with lower $E^\circ$ ) will undergo oxidation (lose electrons). Here, $Cu$ is oxidized at the anode.

Result: Anode = Left (Cu electrode), Cathode = Right (Pt electrode).

Mistake 2: Writing the Wrong Half-Reaction at the Anode

The Error:

Students write the reduction of $Cu^{2+}$ at the anode (e.g., $Cu^{2+} + 2e^- \rightarrow Cu$ ) instead of the oxidation of $Cu$ .

Why it happens:

They confuse the species present. The anode is where oxidation occurs (loss of electrons). The cell notation shows $Cu$ (solid) in contact with $Cu^{2+}$ (aqueous). The only species that can be oxidized is the solid $Cu$ metal.

How to avoid it:

Remember the mnemonic: "An Ox, Red Cat" (Anode = Oxidation, Cathode = Reduction).

At the anode, look for a species that can lose electrons (increase in oxidation state).
- $Cu(s) \rightarrow Cu^{2+}(aq) + 2e^-$ (Oxidation: $Cu$ goes from 0 to +2)
At the cathode, look for a species that can gain electrons (decrease in oxidation state).
- $Cl_2(g) + 2e^- \rightarrow 2Cl^-(aq)$ (Reduction: $Cl$ goes from 0 to -1)

Correct Anode Reaction:

Cu(s) \rightarrow Cu^{2+}(aq) + 2e^-

Mistake 3: Forgetting the Inert Electrode (Pt) in the Cathode Reaction

The Error:

Students write the cathode reaction as $Pt + Cl_2 \rightarrow ...$ or simply $Cl_2 + 2e^- \rightarrow 2Cl^-$ but then forget to mention that Pt is just an inert conductor.

Why it happens:

They see $Pt$ in the cell notation and think it participates chemically. In reality, $Pt$ is inert (does not react). It only provides a surface for the $Cl_2$ gas to interact with the $Cl^-$ solution.

How to avoid it: …

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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.Consider a cell given below:
$Cu \mid Cu^{2+} \parallel Cl^- \mid Cl_2, Pt$
Write the reactions that occur at anode and cathode.

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)

The Language of Cell Notation

Step-by-Step Breakdown

Method: Electrode Identification & Half-Reaction Writing

Steps

Mistake 1: Misidentifying the Anode and Cathode

Mistake 2: Writing the Wrong Half-Reaction at the Anode

Mistake 3: Forgetting the Inert Electrode (Pt) in the Cathode Reaction

1. The Nernst equation for this electrode

2. The qualitative rule that follows

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

Q.Consider a cell given below: Cu∣Cu2+∥Cl−∣Cl2,PtCu \mid Cu^{2+} \parallel Cl^- \mid Cl_2, PtCu∣Cu2+∥Cl−∣Cl2​,Pt Write the reactions that occur at anode and cathode.

The Language of Cell Notation

Step-by-Step Breakdown

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Q.Consider a cell given below:
$Cu \mid Cu^{2+} \parallel Cl^- \mid Cl_2, Pt$
Write the reactions that occur at anode and cathode.

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