For the given cell, $Mg \mid Mg^{2+} \parallel Cu^{2+} \mid Cu$ (Two or more than two options may be correct.) (i) $Mg$ is cathode (ii) $Cu$ is cathode (iii) The cell reaction is $Mg + Cu^{2+} \rightarrow Mg^{2+} + Cu$ (iv) $Cu$ is the oxidising agent

> [!TLDR] > In a galvanic cell, oxidation occurs at the anode (negative) and reduction at the cathode (positive). For the cell $Mg \mid Mg^{2+} \parallel Cu^{2+} \mid Cu$, magnesium is the anode (oxidised) and copper is the cathode (reduced). The cell reaction is $Mg + Cu^{2+} \rightarrow Mg^{2+} + Cu$, and $Cu^{2+}$ is the oxidising agent. Therefore, options (ii) and (iii) are correct. The key to this question lies in understanding the standard cell representation and the Nernst equation's conceptual foundation — but here, we don't even need numbers. The cell diagram itself tells us everything. In a galvanic (voltaic) cell, the anode is written on the left and the cathode on the right. The single vertical line `|` represents a phase boundary, and the double vertical line `||` represents the salt bridge. So the cell $Mg \mid Mg^{2+} \parallel Cu^{2+} \mid Cu$ tells us: - Left side: $Mg$ electrode in contact with $Mg^{2+}$ ions — this is the **anode** (oxidation occurs here). - Right side: $Cu$ electrode in contact with $Cu^{2+}$ ions — this is the **cathode** (reduction occurs here). Now let's go through each option step by step. 1. **Option (i): $Mg$ is cathode** This is false. In the cell diagram, magnesium is on the left, which is the anode. At the anode, oxidation happens: $Mg \rightarrow Mg^{2+} + 2e^-$. The anode is the negative electrode in a galvanic cell, not the cathode. 2. **Option (ii): $Cu$ is cathode** This is true. Copper is on the right side of the diagram, which is the cathode. At the cathode, reduction occurs: $Cu^{2+} + 2e^- \rightarrow Cu$. The cathode is the positive electrode. 3. **Option (iii): The cell reaction is $Mg + Cu^{2+} \rightarrow Mg^{2+} + Cu$** This is true. Combine the half-reactions: - Anode (oxidation): $Mg \rightarrow Mg^{2+} + 2e^-$ - Cathode (reduction): $Cu^{2+} + 2e^- \rightarrow Cu$ Adding them gives the overall cell reaction: $Mg + Cu^{2+} \rightarrow Mg^{2+} + Cu$. The electrons cance

For the given cell, Mg mid Mg²⁺ parallel Cu²⁺ mid Cu (Two or more than two options may be correct.) (i) Mg is

NCERT Exemplar · Q52

Mg \mid Mg^{2+} \parallel Cu^{2+} \mid Cu

(i)

Mg

is cathode

(ii)

Cu

is cathode

(iii) The cell reaction is

Mg + Cu^{2+} \rightarrow Mg^{2+} + Cu

(iv)

Cu

is the oxidising agent

Karnataka PUCMCQ· 1mImportance★★★★★

83% · 95/115 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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.

--- …

Start your 14-day free trial to unlock the full solution →

In a galvanic cell, oxidation occurs at the anode (negative) and reduction at the cathode (positive). For the cell $Mg \mid Mg^{2+} \parallel Cu^{2+} \mid Cu$ , magnesium is the anode (oxidised) and copper is the cathode (reduced). The cell reaction is $Mg + Cu^{2+} \rightarrow Mg^{2+} + Cu$ , and $Cu^{2+}$ is the oxidising agent. Therefore, options (ii) and (iii) are correct.

The key to this question lies in understanding the standard cell representation and the Nernst equation's conceptual foundation — but here, we don't even need numbers. The cell diagram itself tells us everything.

In a galvanic (voltaic) cell, the anode is written on the left and the cathode on the right. The single vertical line | represents a phase boundary, and the double vertical line || represents the salt bridge. So the cell $Mg \mid Mg^{2+} \parallel Cu^{2+} \mid Cu$ tells us:

Left side: $Mg$ electrode in contact with $Mg^{2+}$ ions — this is the anode (oxidation occurs here).
Right side: $Cu$ electrode in contact with $Cu^{2+}$ ions — this is the cathode (reduction occurs here).

Now let's go through each option step by step.

Option (i): $Mg$ is cathode

This is false. In the cell diagram, magnesium is on the left, which is the anode. At the anode, oxidation happens: $Mg \rightarrow Mg^{2+} + 2e^-$ . The anode is the negative electrode in a galvanic cell, not the cathode.
Option (ii): $Cu$ is cathode

This is true. Copper is on the right side of the diagram, which is the cathode. At the cathode, reduction occurs: $Cu^{2+} + 2e^- \rightarrow Cu$ . The cathode is the positive electrode.
Option (iii): The cell reaction is $Mg + Cu^{2+} \rightarrow Mg^{2+} + Cu$

This is true. Combine the half-reactions:
- Anode (oxidation): $Mg \rightarrow Mg^{2+} + 2e^-$
- Cathode (reduction): $Cu^{2+} + 2e^- \rightarrow Cu$ Adding them gives the overall cell reaction: $Mg + Cu^{2+} \rightarrow Mg^{2+} + Cu$ . The electrons cancel out. …

Start your 14-day free trial to unlock the full solution →

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). …
Start your 14-day free trial to see the full solution →
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). …
Start your 14-day free trial to see the full solution →
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: …
Start your 14-day free trial to see the full solution →
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). …
Start your 14-day free trial to see the full solution →
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. …
Start your 14-day free trial to see the full solution →

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	$-3$	$0.80 - 0.177 = 0.623\ \mathrm{V}$

Start your 14-day free trial to see the full solution →

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: …
Start your 14-day free trial to see the full solution →
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 …
Start your 14-day free trial to see the full solution →
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: …
Start your 14-day free trial to see the full solution →
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. …
Start your 14-day free trial to see the full solution →

←Previous questionQ51 — Molar conductivity of ionic solution depends on ______. (Two or more t…→Next questionQ53 — Match the terms given in Column I with the units given in Column II. C…

Q.For the given cell, $Mg \mid Mg^{2+} \parallel Cu^{2+} \mid Cu$ (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: Standard Cell Notation Interpretation

Step 1: Understand the cell notation

Step 2: Identify the electrodes

Step 3: Write the overall cell reaction

✗ Mistake 1: Confusing anode and cathode

✗ Mistake 2: Writing the cell reaction backwards

✗ Mistake 3: Saying the metal itself is the oxidising agent

1. The Nernst equation for this electrode

2. The qualitative rule that follows

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

Q.For the given cell, Mg∣Mg2+∥Cu2+∣CuMg \mid Mg^{2+} \parallel Cu^{2+} \mid CuMg∣Mg2+∥Cu2+∣Cu (Two or more than two options may be correct.)

🎓Unlock everything free for 14 days

Q.For the given cell, $Mg \mid Mg^{2+} \parallel Cu^{2+} \mid Cu$ (Two or more than two options may be correct.)

Unlock everything free for 14 days