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

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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 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. …

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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

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.)

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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