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

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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.

KEAM 2026Set eng-2026-04214 marksMCQ
$\Delta_rG^\circ$
›Reveal solutionSolution

$E^\circ_{cell}=0.34\,\text{V}$ , $n=6$ , so $\Delta_rG^\circ=-nFE^\circ=-196.86\,\text{kJ mol}^{-1}$ .

In the reaction $2Cr_{(s)}+3Cd^{2+}_{(aq)}\rightarrow 2Cr^{3+}_{(aq)}+3Cd_{(s)}$ , chromium is oxidised (anode) and $Cd^{2+}$ is reduced (cathode). The standard cell potential is
$E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode}=E^\circ_{Cd^{2+}/Cd}-E^\circ_{Cr^{3+}/Cr}=(-0.40)-(-0.74)=+0.34\,\text{V}.$
The number of electrons transferred: each Cr loses $3\,e^-$ ( $\times 2=6$ ) and each $Cd^{2+}$ gains $2\,e^-$ ( $\times 3=6$ ), so $n=6$ .

Using $\Delta_rG^\circ=-nFE^\circ_{cell}$ with $F=96500\,\text{C mol}^{-1}$ :
$\Delta_rG^\circ=-(6)(96500)(0.34)=-196{,}860\,\text{J mol}^{-1}=-196.86\,\text{kJ mol}^{-1}.$
✓Final answer
$The correct option is (A) .$
KEAM 2024Set eng-2024-06074 marksMCQ
$^0$
›Reveal solutionSolution

In a Daniell cell, oxidation occurs at the anode (Zn) and reduction at the cathode (Cu), the electrolytes are ZnSO $_4$ and CuSO $_4$ , and it converts chemical energy into electrical energy. Only the value of the standard cell potential, $1.1$ V, is stated correctly.

Evaluating each option: (A) is false because oxidation occurs at the anode, not the cathode. (B) is false because reduction occurs at the cathode, not the anode. (D) is false — a galvanic cell converts chemical energy into electrical energy (the reverse describes electrolysis). (E) is false because the electrolytes are ZnSO $_4$ and CuSO $_4$ , not FeSO $_4$ .

The standard electrode potentials are $E^0(\mathrm{Cu}^{2+}/\mathrm{Cu})=+0.34$ V and $E^0(\mathrm{Zn}^{2+}/\mathrm{Zn})=-0.76$ V, so
$E^0_{cell}=E^0_{cathode}-E^0_{anode}=0.34-(-0.76)=1.10\ \text{V}.$
✓Final answer
$The correct option is (C) .$
KEAM 2024Set eng-2024-06094 marksMCQ
$NH_4Cl$
›Reveal solutionSolution
In a mercury cell the anode is zinc amalgamated with mercury (Zn–Hg), and the cathode is HgO mixed with carbon.

The mercury (button) cell electrodes:

Anode: zinc–mercury amalgam, Zn(Hg) — oxidised: $Zn + 2OH^- \to ZnO + H_2O + 2e^-$ .

Cathode: paste of HgO with carbon — reduced: $HgO + H_2O + 2e^- \to Hg + 2OH^-$ .

Electrolyte: paste of KOH/ZnO.

The $NH_4Cl/ZnCl_2$ paste and $MnO_2$ –carbon belong to the dry (Leclanché) cell, not the mercury cell. Hence the anode is the zinc–mercury amalgam.

✓Final answer
$The correct option is (E) .$

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