Which cell will measure standard electrode potential of copper electrode? (i) Pt(s) mid H₂(g, 0.1bar) mid H^+(

NCERT Exemplar · Q1

Q. Which cell will measure standard electrode potential of copper electrode?

(i)

Pt(s) \mid H_2(g, 0.1\ \text{bar}) \mid H^+(aq., 1\ M) \parallel Cu^{2+}(aq., 1\ M) \mid Cu

(ii)

Pt(s) \mid H_2(g, 1\ \text{bar}) \mid H^+(aq., 1\ M) \parallel Cu^{2+}(aq., 2\ M) \mid Cu

(iii)

Pt(s) \mid H_2(g, 1\ \text{bar}) \mid H^+(aq., 1\ M) \parallel Cu^{2+}(aq., 1\ M) \mid Cu

(iv)

Pt(s) \mid H_2(g, 1\ \text{bar}) \mid H^+(aq., 0.1\ M) \parallel Cu^{2+}(aq., 1\ M) \mid Cu

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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}

E = 1.10 - \frac{0.0591}{2} \log_{10}(0.1)

\log_{10}(0.1) = -1

E = 1.10 - \frac{0.0591}{2}(-1) = 1.10 + 0.02955 = 1.1296 \text{ V}

The cell voltage is slightly higher than standard because the zinc ion concentration is lower (less "push" from the anode side, so the net driving force is larger).

Watch out

A common mistake is to forget that $n$ must match the balanced equation. If you write the half-reactions with different numbers of electrons, you'll get the wrong $n$ . Always check that the overall reaction is balanced.

The Key Insight

The Nernst equation is not just a formula — it's a statement that electrochemical potential is a logarithmic function of concentration. This means:

Diluting the reactant side (lowering $[ \text{Cu}^{2+} ]$ ) decreases $E$
Diluting the product side (lowering $[ \text{Zn}^{2+} ]$ ) increases $E$
At equilibrium, $E = 0$ and $Q = K$ (the equilibrium constant), giving $\ln K = \frac{nFE^\circ}{RT}$

This last point connects electrochemistry directly to thermodynamics — the Nernst equation is just the Gibbs free energy equation ( $\Delta G = -nFE$ ) written in terms of concentrations.

Cell representation and the Nernst equation together form a heavily tested pair within the NCERT/CBSE Class 12 Chemistry Electrochemistry chapter, and ‘how to write cell representation’ or ‘Nernst equation for cell reaction’ are common important-question searches for board exams, JEE Main and NEET. Being fluent in both the notation and the formula is essential for solving electrochemistry numericals quickly in competitive exams.

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.

6. Cell Representation: How to Write $Q$

For a cell written as:

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

The reaction is:

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

Solids are omitted from $Q$ (activity = 1):

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

Why omit solids? Their concentration doesn't change — they're pure phases with fixed chemical potential.

7. Exam-Ready Summary

Step	What to Do	Why
1	Write balanced half-reactions	Identify $n$ (electrons transferred)
2	Write overall reaction	Determine $Q$ form
3	Plug into Nernst	Corrects $E^\circ$ for real conditions
4	Use $\log_{10}$ at $25^\circ \text{C}$	$0.0592/n$ is exam standard

Final takeaway: The Nernst equation is thermodynamics in disguise — it's the Gibbs free energy equation rewritten in electrical units. Every time you use it, you're balancing chemical potential against electrical potential.

The standard electrode potential of copper is measured using a cell where all species are in their standard states: $H_2$ at 1 bar, $H^+$ at 1 M, and $Cu^{2+}$ at 1 M. Only option (iii) satisfies all three conditions.

To measure the standard electrode potential of a half-cell, we must construct a cell where that half-cell is combined with a standard hydrogen electrode (SHE) — and every species in the entire cell must be in its standard state.

The standard hydrogen electrode is defined as: $Pt(s) \mid H_2(g, 1\ \text{bar}) \mid H^+(aq., 1\ M)$ . Any deviation from 1 bar or 1 M means the cell no longer gives the standard potential — you'd get a non-standard cell potential instead.

For the copper electrode, the half-reaction is:

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

Its standard state requires $Cu^{2+}$ concentration = 1 M and solid copper (activity = 1, always true for a pure solid).

So the correct cell must have:

$H_2$ pressure = 1 bar
$H^+$ concentration = 1 M
$Cu^{2+}$ concentration = 1 M

Let's check each option.

Option (i): $Pt(s) \mid H_2(g, 0.1\ \text{bar}) \mid H^+(aq., 1\ M) \parallel Cu^{2+}(aq., 1\ M) \mid Cu$

Hydrogen pressure is 0.1 bar, not 1 bar. The SHE is not in its standard state. ✗
Option (ii): $Pt(s) \mid H_2(g, 1\ \text{bar}) \mid H^+(aq., 1\ M) \parallel Cu^{2+}(aq., 2\ M) \mid Cu$

Hydrogen side is standard, but $Cu^{2+}$ is 2 M, not 1 M. The copper half-cell is not in its standard state. ✗
Option (iii): $Pt(s) \mid H_2(g, 1\ \text{bar}) \mid H^+(aq., 1\ M) \parallel Cu^{2+}(aq., 1\ M) \mid Cu$

All three conditions are met: $H_2$ at 1 bar, $H^+$ at 1 M, $Cu^{2+}$ at 1 M. This is the correct standard cell. ✓
Option (iv): $Pt(s) \mid H_2(g, 1\ \text{bar}) \mid H^+(aq., 0.1\ M) \parallel Cu^{2+}(aq., 1\ M) \mid Cu$

Hydrogen side has $H^+$ at 0.1 M, not 1 M. SHE is not standard. ✗

Watch out

A common mistake is to think that only the half-cell being measured needs to be in its standard state. In fact, both half-cells must be in their standard states to measure a standard electrode potential. The SHE is the reference, and it must itself be standard.

Tip

The Nernst equation tells us that if any concentration or pressure deviates from the standard value, the cell potential changes by $\frac{0.059}{n}\log Q$ . For the SHE, $Q$ involves $[H^+]$ and $P_{H_2}$ , so even a small deviation shifts the measured potential away from the true standard value.

✓Final answer

The correct option is (iii) .

Method: Standard Cell Condition Check (IUPAC Definition)

Why this method?

The standard electrode potential of an electrode is defined under standard conditions:

All solutes at 1 M concentration
All gases at 1 bar pressure
Temperature usually 298 K (implied)

For the copper electrode, we need a cell where the copper half-cell is exactly at standard state, and the reference hydrogen electrode is also at standard state.

Steps

Identify the half-cells
- Left: Hydrogen electrode (reference)
- Right: Copper electrode (test)
Check the hydrogen electrode
- Must have $H^+$ concentration = 1 M
- $H_2$ gas pressure = 1 bar
- Platinum is the inert conductor (always present)
Check the copper electrode
- Must have $Cu^{2+}$ concentration = 1 M
- Solid copper metal (always present)
Eliminate options that violate any standard condition

Applying to the options

Option	$H_2$ pressure	$[H^+]$	$[Cu^{2+}]$	Standard?
(i)	0.1 bar	1 M	1 M	✗ (gas pressure wrong)
(ii)	1 bar	1 M	2 M	✗ (copper ion wrong)
(iii)	1 bar	1 M	1 M	✓ All correct
(iv)	1 bar	0.1 M	1 M	✗ (acid concentration wrong)

Final Answer

Option (iii) is the correct cell to measure the standard electrode potential of copper.

Pt(s) \mid H_2(g, 1\ \text{bar}) \mid H^+(aq., 1\ M) \parallel Cu^{2+}(aq., 1\ M) \mid Cu

Key takeaway: For any standard electrode potential measurement, both half-cells must be at their standard states — solutes at 1 M, gases at 1 bar.

Common Mistakes & How to Avoid Them

Mistake 1: Ignoring the Definition of "Standard" Conditions

The Error:

Students often pick option (i) or (iv) because they focus only on the copper side ( $Cu^{2+}$ concentration) and forget that both half-cells must be at standard conditions for a standard electrode potential measurement.

Why It's Wrong:

Standard electrode potential ( $E^\ominus$ ) is defined when:

All solutes are at 1 M concentration
All gases are at 1 bar pressure
Temperature is 298 K (implied)

How to Avoid:

Always check every component in the cell representation:

$H^+$ must be $1\ M$ (eliminates D)
$H_2$ gas must be $1\ bar$ (eliminates A)
$Cu^{2+}$ must be $1\ M$ (eliminates B)

✓ Correct answer is (iii): $Pt(s) \mid H_2(g, 1\ \text{bar}) \mid H^+(aq., 1\ M) \parallel Cu^{2+}(aq., 1\ M) \mid Cu$

Mistake 2: Confusing "Standard" with "Any Reference"

The Error:

Some students think any SHE (Standard Hydrogen Electrode) works, even if its conditions are non-standard. They pick (i) because $H_2$ at $0.1\ bar$ still acts as a reference.

Why It's Wrong:

The SHE is only "standard" when $P_{H_2} = 1\ bar$ and $[H^+] = 1\ M$ . If either changes, the half-cell potential shifts according to the Nernst equation:

E_{H^+/H_2} = E^\ominus_{H^+/H_2} - \frac{0.059}{2} \log \frac{P_{H_2}}{[H^+]^2}

At $0.1\ bar$ and $1\ M\ H^+$ :

E = 0 - \frac{0.059}{2} \log \frac{0.1}{1^2} = +0.0295\ V

This is not zero — so you're not measuring a standard potential.

How to Avoid:

Remember: SHE = 1 bar H₂, 1 M H⁺. Any deviation changes the reference potential.

Mistake 3: Forgetting the Nernst Equation Applies to Both Half-Cells

The Error:

Students check only the copper side for standard conditions and assume the hydrogen side is automatically fine.

Why It's Wrong:

The measured cell potential is:

E_{cell} = E_{Cu} - E_{SHE}

If $E_{SHE} \neq 0$ , then $E_{cell} \neq E^\ominus_{Cu}$ even if $[Cu^{2+}] = 1\ M$ .

How to Avoid:

Treat each half-cell independently. For a standard measurement:

Left half-cell (SHE): must give $E = 0\ V$
Right half-cell (copper): must give $E = E^\ominus_{Cu^{2+}/Cu}$

Only option (iii) satisfies both.

Mistake 4: Misreading the Cell Diagram Notation

The Error:

Students confuse the single vertical line ( $|$ ) with the double line ( $\parallel$ ) and misidentify which side is anode/cathode.

Why It's Wrong:

Single line ( $|$ ): phase boundary
Double line ( $\parallel$ ): salt bridge (separates half-cells)
Left = anode (oxidation), Right = cathode (reduction)

In all options, SHE is on the left (anode) and copper on the right (cathode). The measured potential is the reduction potential of copper.

How to Avoid:

Practice reading cell diagrams left-to-right:

Anode | Anode solution || Cathode solution | Cathode

Quick Checklist for "Standard Electrode Potential" Questions

Component	Required Condition	Check in Options
$H_2$ pressure	$1\ bar$	Eliminates A
$[H^+]$	$1\ M$	Eliminates D
$[Cu^{2+}]$	$1\ M$	Eliminates B
All three	Must match	✓ Only C works

Final Tip: When you see "standard electrode potential," immediately think: 1 M, 1 bar, 298 K — for every species in the cell.

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

AP EAPCET 2026Set ap-2026-05-19-AN1 markMCQ
$Zn^{2+}$
›Reveal solutionSolution
The cell is a Daniell cell (Zn|Zn²⁺||Cu²⁺|Cu) with an external opposing voltage $E_{\text{ext}} < 1.1\ \text{V}$ . Since the opposing voltage is less than the cell’s standard emf (1.1 V), the cell still drives electrons from Zn (anode) to Cu (cathode) — so Statement I is correct. As the cell operates, Zn dissolves (weight decreases) and Cu deposits (weight increases) — so Statement II is also correct. Hence the correct option is (A).

Concept and Intuition

This is a Daniell cell — the classic Zn–Cu galvanic cell. Normally, without any external source, electrons flow spontaneously from the Zn electrode (anode, oxidation) to the Cu electrode (cathode, reduction). The standard cell potential is about 1.1 V.

Here, an external voltage source $E_{\text{ext}}$ is inserted in the external circuit, but it is less than 1.1 V. That means the external source is trying to oppose the cell’s natural electron flow, but it is too weak to reverse it. The cell still operates as a galvanic cell (not an electrolytic cell), because the net driving force remains from Zn to Cu.

The Nernst equation helps us understand that even if concentrations change slightly, the cell potential remains positive as long as the reaction quotient $Q$ is less than the equilibrium constant $K$ . Since $E_{\text{ext}} < E_{\text{cell}}$ , the cell continues to discharge.

Step-by-step reasoning

Identify the spontaneous direction In a Zn–Cu cell, the standard reduction potentials are:

$\begin{aligned} \text{Cu}^{2+} + 2e^- &\rightarrow \text{Cu} \quad E^\circ = +0.34\ \text{V} \\ \text{Zn}^{2+} + 2e^- &\rightarrow \text{Zn} \quad E^\circ = -0.76\ \text{V} \end{aligned}$
The cell reaction is:
$\text{Zn} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cu}$
with $E^\circ_{\text{cell}} = 0.34 - (-0.76) = 1.10\ \text{V}$ .

Electrons flow from Zn (anode) to Cu (cathode) in the external circuit.

Effect of the external source $E_{\text{ext}} < 1.1\ \text{V}$

The external source is connected opposing the cell’s natural polarity. However, because its voltage is less than the cell’s emf, the net potential difference still drives electrons from Zn to Cu.

Tip

Think of it like two batteries in series opposing: the larger one (the cell) wins, so current flows in its direction. Here the cell is the “larger” battery.

Evaluate Statement I: “Electrons flow from Cu electrode to Zn electrode.”

The statement claims the opposite of the spontaneous direction. But as argued, the external source is too weak to reverse the flow. Therefore electrons still flow from Zn to Cu, not from Cu to Zn.

Statement I is false as written — wait, careful: The statement says “from Cu electrode to Zn electrode”. That is the reverse of the actual direction. So Statement I is incorrect.

Watch out

A common mistake is to assume that any external voltage reverses the cell. But reversal only happens if $E_{\text{ext}} > E_{\text{cell}}$ . Here $E_{\text{ext}} < 1.1$ , so the cell still discharges normally.

Evaluate Statement II: “With increase in time, the weight of Zn electrode decreases and weight of Cu electrode increases.”

In the spontaneous reaction:

Zn anode: $\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^-$ (Zn dissolves, so its mass decreases).

Cu cathode: $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$ (Cu deposits, so its mass increases). This happens regardless of the external source as long as the cell is discharging. Since $E_{\text{ext}} < E_{\text{cell}}$ , the cell continues to discharge, so Zn loses mass and Cu gains mass. Statement II is correct.

Combine the two statements

Statement I: Incorrect (electrons flow Zn → Cu, not Cu → Zn).

Statement II: Correct. Therefore the correct choice is (D).

$E_{\text{ext}}$

✓Final answer
$The correct option is (D) . ANSWER: D$
AP EAPCET 2026Set ap-2026-05-20-AN1 markMCQ
$Zn^{2+}$
›Reveal solutionSolution
In a standard Zn–Cu galvanic cell, electrons flow from the Zn anode (oxidation) to the Cu cathode (reduction), so conventional current flows from Cu to Zn. The Zn electrode loses mass (Zn → Zn²⁺) and the Cu electrode gains mass (Cu²⁺ → Cu). Therefore Statement I is correct, Statement II is incorrect, and the answer is (C).

Concept and Intuition: The Cell Representation and the Nernst Equation

This question tests your understanding of the direction of current and the mass changes at the electrodes in a working galvanic cell. The key is to remember that in a galvanic (voltaic) cell, the spontaneous redox reaction drives electrons through the external circuit. By convention, current flows opposite to electron flow (from positive to negative). The electrode where oxidation occurs (anode) loses mass; the electrode where reduction occurs (cathode) gains mass. The external opposing voltage $E_{\text{ext}} < 1.1\ \text{V}$ tells us the cell is still operating spontaneously (the cell’s emf is about 1.1 V for Zn–Cu under standard conditions).

Step-by-step reasoning

Identify the spontaneous reaction In a Zn–Cu cell, zinc is more reactive (higher reduction potential for Zn²⁺/Zn is –0.76 V, for Cu²⁺/Cu is +0.34 V). The spontaneous reaction is:

$\text{Zn(s)} + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu(s)}$
Zinc is oxidized (loses electrons), copper ions are reduced (gain electrons).

Determine electron flow and conventional current

Oxidation at the Zn electrode (anode): $\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^-$ Electrons leave the Zn electrode and travel through the external circuit toward the Cu electrode.

Reduction at the Cu electrode (cathode): $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$ Electrons arrive at the Cu electrode.

Conventional current is defined as the flow of positive charge, opposite to electron flow. So electrons flow from Zn → Cu, meaning conventional current flows from Cu → Zn.

Statement I says: “Current flows from Cu electrode to Zn electrode.” This matches the conventional current direction. Statement I is correct.

Analyze mass changes at the electrodes

At the Zn anode: Zn metal is converted to Zn²⁺ ions, which go into solution. The solid Zn electrode loses mass.

At the Cu cathode: Cu²⁺ ions from solution gain electrons and deposit as solid Cu on the electrode. The Cu electrode gains mass.

Statement II says: “With increase in time the mass of Zn electrode increases and mass of Cu electrode decreases.” This is the exact opposite of what happens. Statement II is incorrect.

Check the role of the external voltage source

The diagram shows $E_{\text{ext}} < 1.1\ \text{V}$ . This is an opposing voltage, but because it is less than the cell’s emf, the cell still operates spontaneously (just with a reduced net voltage). The direction of electron flow and the mass changes remain the same as in an unopposed cell. So the presence of this small opposing voltage does not alter the correctness of the statements.

Watch out

A common mistake is to confuse electron flow with conventional current. Remember: electrons flow from anode to cathode (negative to positive inside the cell), but conventional current flows from cathode to anode (positive to negative). Also, do not assume that the external voltage source reverses the cell — it only opposes it, and since $E_{\text{ext}} < 1.1\ \text{V}$ , the cell still runs spontaneously.

Tip

A quick memory aid: Anode = Oxidation = Loss of mass (think “An Ox” — Anode Oxidation). Cathode = Reduction = Gain of mass (think “Red Cat” — Reduction at Cathode). For a Zn–Cu cell, Zn is the anode, so it loses mass; Cu is the cathode, so it gains mass.

✓Final answer
$The correct option is (C) . ANSWER: C$
AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
$\Delta G^\circ$
›Reveal solutionSolution
This applies $\Delta G^\circ = -nFE^\circ_{cell}$ to the Ca/Au3+ redox couple; the standard free energy change works out to $-2.53\times10^3$ kJ.

Concept and Intuition

The standard free energy change of a redox reaction is related to the cell's standard EMF by $\Delta G^\circ = -nFE^\circ_{cell}$ , where $n$ is the total number of electrons transferred as balanced by the overall equation, and $F$ is Faraday's constant. A positive $E^\circ_{cell}$ (spontaneous reaction) corresponds to a negative $\Delta G^\circ$ .

Step-by-Step Solution

Identify the half-reactions: reduction at cathode, $Au^{3+} + 3e^- \rightarrow Au$ , $E^\circ = +1.50$ V; oxidation at anode, $Ca \rightarrow Ca^{2+} + 2e^-$ , $E^\circ_{Ca^{2+}/Ca} = -2.87$ V.

Balance electrons: multiply the Au half-reaction by 2 and the Ca half-reaction by 3, giving $n = 6$ electrons transferred overall, matching the given equation $3Ca + 2Au^{3+} \rightarrow 3Ca^{2+} + 2Au$ .

Compute $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 1.50 - (-2.87) = 4.37$ V.

Compute $\Delta G^\circ = -nFE^\circ_{cell} = -(6)(96500\ \text{C/mol})(4.37\ \text{V})$ .

Calculate: $6 \times 96500 = 579000$ ; $579000 \times 4.37 = 2{,}530{,}230$ J $= 2530.23$ kJ. So $\Delta G^\circ = -2530.23$ kJ $\approx -2.53 \times 10^3$ kJ.

Common Mistakes

Using $n = 3$ or $n = 2$ (from the individual half-reactions) instead of the correctly balanced $n = 6$ for the overall reaction.

Sign errors — forgetting the negative sign in $\Delta G^\circ = -nFE^\circ$ , or mishandling the negative anode potential when computing $E^\circ_{cell}$ .

Misplacing the decimal/power of ten when converting J to kJ (giving $10^4$ instead of $10^3$ ).

✓Final answer
$-2.53 \times 10^3$

ANSWER: A
AP EAPCET 2022Set ap-2022-07-11-FN1 markMCQ
$\Delta G^0$
›Reveal solutionSolution
This tests the relation $\Delta G^\circ=-nFE^\circ_{cell}$ , with the key subtlety of finding the correct $n$ (total electrons transferred, found via LCM of the two half-cell electron counts).

Concept and Intuition

The standard Gibbs energy change of a cell reaction is directly proportional to the number of moles of electrons transferred in the balanced overall reaction and the cell potential. Since the anode gives 3 electrons per P atom and the cathode needs 2 electrons per Q $^{2+}$ ion, the reaction must be balanced so both half-reactions transfer the same total number of electrons — found via the LCM of 3 and 2, which is 6.

Step-by-Step Solution

Anode (oxidation): $P \to P^{3+} + 3e^-$ ; Cathode (reduction): $Q^{2+} + 2e^- \to Q$ .

Balance electrons: multiply the anode reaction by 2 and the cathode reaction by 3, so $n = 6$ electrons flow in the overall balanced cell reaction ( $2P + 3Q^{2+} \to 2P^{3+} + 3Q$ ).

Apply $\Delta G^\circ = -nFE^\circ_{cell} = -6\times96500\,\text{C/mol}\times0.2\,\text{V}$ .

Compute: $-6\times96500\times0.2 = -115800\,\text{J} = -115.8\,\text{kJ}$ .

Common Mistakes

Using $n=1$ , $2$ , or $3$ instead of the correctly balanced $n=6$ .

Forgetting to convert Joules to kilojoules.

✓Final answer
$The correct option is (A) — 115.8 kJ.$

ANSWER: A
AP EAPCET 2022Set eng-2022-07-07-FN1 markMCQ
$Q. The energy conversion involved in a galvanic cell is (A) Chemical energy to mechanical energy (B) Chemical energy to electrical energy (C) Electrical energy to chemical energy (D) Electrical energy to thermal energy$
›Reveal solutionSolution
A galvanic cell is the device that turns a spontaneous chemical (redox) reaction into usable electrical energy — the opposite of electrolysis.

Concept and Intuition

In a galvanic cell, a spontaneous ( $\Delta G < 0$ ) redox reaction is split into separate oxidation (anode) and reduction (cathode) half-reactions connected by an external circuit, so the electron transfer that would otherwise happen directly (releasing energy as heat) instead flows through a wire and does electrical work. This is the reverse energy conversion of an electrolytic cell, which consumes electrical energy to drive a non-spontaneous chemical reaction.

Step-by-Step Solution

Identify the device: a galvanic/voltaic cell (e.g. Daniell cell) generates electricity from a spontaneous redox reaction.

The reaction proceeds because it is thermodynamically favourable (releases chemical energy).

That released energy is captured as an electric current through the external circuit rather than dissipated as heat.

Hence: chemical energy $\to$ electrical energy.

Common Mistakes

Confusing a galvanic cell (chemical $\to$ electrical) with an electrolytic cell (electrical $\to$ chemical), which is the reverse process.

✓Final answer
$The correct option is (B) — Chemical energy to electrical energy.$

ANSWER: B
AP EAPCET 2021Set ap-2021-09-07-FN1 markMCQ
$Cr/Cr^{3+}\ //\ Cd^{2+}/Cd$
›Reveal solutionSolution
Compute $E^\circ_{cell}$ from the given half-cell potentials, balance electrons for the overall cell reaction to get $n$ , then use $\Delta G^\circ = -nFE^\circ_{cell}$ .

Concept and Intuition

In the cell notation $Cr/Cr^{3+}\,//\,Cd^{2+}/Cd$ , the left electrode (written first) is the anode (oxidation), and the right electrode is the cathode (reduction). The cell potential is always cathode minus anode: $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$ . Once $E^\circ_{cell}$ is known, $\Delta G^\circ$ follows from the fundamental relation between electrical work and free energy, $\Delta G^\circ = -nFE^\circ_{cell}$ , where $n$ is the total number of electrons transferred in the balanced overall cell reaction (found by matching the electron counts of the two half-reactions via their LCM).

Step-by-Step Solution

Identify electrodes: anode = $Cr^{3+}/Cr$ ( $E^\circ=-0.74\ V$ , oxidation: $Cr\to Cr^{3+}+3e^-$ ), cathode = $Cd^{2+}/Cd$ ( $E^\circ=-0.40\ V$ , reduction: $Cd^{2+}+2e^-\to Cd$ ).

Compute cell potential: $E^\circ_{cell} = E^\circ_{cathode}-E^\circ_{anode} = (-0.40)-(-0.74) = 0.34\ V$ (positive, confirming the reaction is spontaneous as written -- consistent with it being a genuine galvanic cell).

Balance the overall reaction by matching electrons: the anode releases 3 electrons per $Cr$ , the cathode consumes 2 electrons per $Cd^{2+}$ ; the LCM of 3 and 2 is 6, so multiply the anode reaction by 2 and the cathode reaction by 3: $2Cr \to 2Cr^{3+}+6e^-$ and $3Cd^{2+}+6e^- \to 3Cd$ , giving overall $2Cr+3Cd^{2+}\to 2Cr^{3+}+3Cd$ with $n=6$ .

Apply $\Delta G^\circ = -nFE^\circ_{cell} = -6\times96{,}500\times0.34$ .

Compute: $6\times96{,}500=579{,}000$ ; $579{,}000\times0.34=196{,}860\ J/mol \approx -196.8\ kJ/mol$ (negative, consistent with a spontaneous cell reaction).

Common Mistakes

Reversing anode and cathode (using $E^\circ_{anode}-E^\circ_{cathode}$ ), which flips the sign of $E^\circ_{cell}$ and hence of $\Delta G^\circ$ .

Using $n=1$ or $n=2$ (from an unbalanced half-reaction) instead of the correctly balanced $n=6$ for the overall reaction, which would give a $\Delta G^\circ$ several times too small in magnitude.

✓Final answer
$-196.8\ kJ/mol$

ANSWER: C
AP EAPCET 2021Set eng-2021-10-05-FN1 markMCQ
$X\,|\,X^+(x\ M)\,||\,Y^+(y\ M)\,|\,Y$
›Reveal solutionSolution
By the standard cell-notation convention (anode | ... || ... | cathode) and a positive measured emf, the spontaneous cell reaction is $X + Y^+ \to X^+ + Y$ .

Concept and Intuition

Electrochemical cell notation has a fixed convention: the electrode written on the left is the anode, where oxidation occurs, and the electrode on the right is the cathode, where reduction occurs — the double vertical line "||" represents the salt bridge separating the two half-cells. A positive emf value for the cell as written confirms the reaction is indeed spontaneous in the direction implied by this left-to-right convention (anode oxidation feeding electrons to cathode reduction).

Step-by-Step Solution

Left half-cell, $X\,|\,X^+(x\,M)$ : this is the anode — oxidation occurs here, $X \to X^+ + e^-$ .

Right half-cell, $Y^+(y\,M)\,|\,Y$ : this is the cathode — reduction occurs here, $Y^+ + e^- \to Y$ .

Combine the two half-reactions (balancing the single electron transferred in each): $X + Y^+ \to X^+ + Y$ .

The given emf is $+0.5\,V$ (positive), which confirms this reaction as written is indeed the spontaneous direction (a negative emf would instead indicate the reverse reaction is spontaneous).

Common Mistakes

Reversing which side is the anode/cathode — cell notation is standardized with anode always on the left, cathode on the right.

Writing the overall reaction with charges/species that don't correspond to the actual redox half-reactions implied by the cell diagram (e.g. inventing an $X^-$ or $Y^-$ species, as in option D, which isn't consistent with the $X^+$ / $Y^+$ notation given).

✓Final answer
$X + Y^+ \longrightarrow X^+ + Y$

ANSWER: C

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