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.

Showing the 12 most recent of 14 on this concept.

CBSE 2026Set ANNUAL1 mark
$Q. What is the potential difference between the two electrodes of the galvanic cell called?$
›Reveal solutionSolution

The potential difference between the two electrodes of a galvanic cell (measured when no current is drawn) is called the electromotive force (EMF) or cell potential, $E_{\text{cell}}$ .

Concept. In a galvanic (voltaic) cell, the two half-cells are at different electrode potentials. The difference between the cathode and anode potentials is what pushes electrons through the external circuit:

$E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}}$

When this potential difference is measured under zero-current conditions (using a potentiometer, so the cell reaction is effectively at equilibrium and no IR drop occurs), it is the maximum potential difference the cell can deliver and is termed the electromotive force (EMF) of the cell.

✓Final answer
$E_{\text{cell}}$
CBSE 2026Set ANNUAL1 markMCQ
$Q. Consider the following statements about a reaction at equilibrium: A(g) + B(g) \leftrightarrow C(g). Statement I: Adding an inert gas at constant volume will shift the equilibrium to the right. Statement II: A catalyst changes the position of equilibrium.$
(a) i) Both statement I and II are correct
(b) ii) Both statement I and II are incorrect
(c) iii) Statement I is correct and statement II is incorrect
(d) iv) Statement I is incorrect and statement II is correct
›Reveal solutionSolution
[!TLDR]

ii) Both statement I and II are incorrect

Why

Adding an inert gas at constant volume does not change partial pressures/concentrations of reacting species, so it does not shift equilibrium (Statement I false). A catalyst speeds up attainment of equilibrium equally in both directions and never shifts its position (Statement II false).

[!ANSWER]

ii) Both statement I and II are incorrect
CBSE 2025Set ANNUAL1 mark
$Zn(s)+Cu^{2+}(aq) \rightarrow Zn^{2+}(aq)+Cu(s)$
›Reveal solutionSolution
An external potential exactly equal and opposite to the cell's own EMF brings the system to balance, so no net current flows in either direction — this is the basis of potentiometric EMF measurement.

The Daniell-type cell $Zn(s)|Zn^{2+}(aq)\|Cu^{2+}(aq)|Cu(s)$ spontaneously drives current in the galvanic direction (electrons flow from Zn anode to Cu cathode through the external circuit) with an EMF of 1.1 V under standard conditions.

If an external opposing potential is applied, it works against this spontaneous cell reaction:

If the external potential is less than 1.1 V, the cell's own EMF still dominates, and current continues to flow in the original (galvanic) direction, though at a reduced magnitude.

If the external potential is greater than 1.1 V, it overpowers the cell's own EMF, and current is forced to flow in the reverse direction (the cell now behaves as an electrolytic cell, being charged/driven backward).

If the external potential exactly equals 1.1 V (as given here), the two opposing potentials exactly cancel, and no net current flows — the system is in electrochemical balance (equilibrium).

(This exact-balance condition is the working principle behind the potentiometric method of accurately measuring a cell's true EMF.)

✓Final answer
$Since the applied external potential (1.1 V) exactly equals and opposes the cell's own EMF, the two cancel and no net current flows in either direction.$
CBSE 2025Set ANNUAL1 markMCQ
$Q. The correct statement in a cell of zinc and copper is$
(a) zinc acts as cathode and copper as anode
(b) zinc acts as anode and copper as cathode
(c) the standard reduction potential of zinc is more than that of copper
(d) the flow of electrons is from copper to zinc
›Reveal solutionSolution

Zinc has a lower (more negative) standard reduction potential than copper, so it is oxidized (anode) while copper is reduced (cathode).

In a Daniell-type zinc–copper cell, E°(Zn²⁺/Zn) = −0.76 V is lower than E°(Cu²⁺/Cu) = +0.34 V. The electrode with the lower (more negative) reduction potential is oxidized — zinc loses electrons and acts as the anode (Zn → Zn²⁺ + 2e⁻) — while the electrode with the higher reduction potential is reduced — copper gains electrons and acts as the cathode (Cu²⁺ + 2e⁻ → Cu). Electrons flow through the external circuit from zinc to copper (not the reverse), and since Zn has the more negative (smaller) standard reduction potential, statement (c) is false too.

✓Final answer
$(b) zinc acts as anode and copper as cathode.$
CBSE 2024Set D1 markMCQ
$Q. The electromotive force of the cell Zn | ZnSO4 || CuSO4 | Cu is 1.1 volt. Its cathode is$
(a) Zn
(b) Cu
(c) ZnSO4
(d) CuSO4
›Reveal solutionSolution
Reduction happens at the cathode; Cu2+ is reduced to Cu, so Cu is the cathode.

In the Daniell cell Zn | ZnSO4 || CuSO4 | Cu:

Anode (oxidation, left): Zn -> Zn2+ + 2e-

Cathode (reduction, right): Cu2+ + 2e- -> Cu

By convention the electrode written on the right of a cell notation is the cathode where reduction occurs. The standard EMF = E(cathode) - E(anode) = 0.34 - (-0.76) = +1.10 V, matching the given 1.1 V. Because copper has the higher (more positive) reduction potential, Cu2+ is reduced and copper is the cathode.

✓Final answer
$(b) Cu — the copper electrode is the cathode (reduction of Cu2+).$
CBSE 2024Set ANNUAL1 markMCQ
$Q. An electrochemical cell can behave like an electrolytic cell when _______.$
(a) Ecell = 0
(b) Ecell > Eext
(c) Eext > Ecell
(d) Ecell = Eext
›Reveal solutionSolution
A galvanic (electrochemical) cell starts behaving like an electrolytic cell when an external potential greater than the cell's own emf is applied against it, reversing the direction of current flow.

Consider a Daniell cell: Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s), which normally works as a galvanic cell producing a cell potential Ecell, with electrons flowing from Zn (anode) to Cu (cathode) through the external circuit.

If an external opposing emf (Eext) is applied to this cell:

When Eext < Ecell, the cell continues to work as a galvanic cell, but the current decreases.

When Eext = Ecell, no current flows through the cell (this is used to measure the cell's emf accurately, e.g. using a potentiometer).

When Eext > Ecell, the direction of current flow reverses. Electrons are now forced to flow from Cu to Zn, so Cu is oxidised and Zn2+ is reduced — exactly the reverse of the spontaneous cell reaction. The cell now behaves as an electrolytic cell, consuming electrical energy to drive a non-spontaneous reaction (this is the working principle of charging a rechargeable/secondary cell such as a lead storage battery).

✓Final answer
$The cell behaves as an electrolytic cell when Eext > Ecell (option c).$
CBSE 2024Set ANNUAL1 mark
$Q. Write True/False: A hydrogen bridge is used to maintain continuity of ion flow in a Daniell cell.$
›Reveal solutionSolution

This statement is FALSE. A Daniell cell uses a salt bridge (e.g. containing KCl or KNO3 in agar-agar gel), not any "hydrogen bridge", to complete the internal circuit.

A Daniell cell consists of a Zn electrode dipped in ZnSO4 solution (anode) and a Cu electrode dipped in CuSO4 solution (cathode), connected externally by a wire and internally by a salt bridge. The salt bridge allows ions to migrate between the two half-cells, maintaining electrical neutrality in each compartment as the cell reaction proceeds, and completes the internal circuit without letting the two solutions mix directly. There is no such thing as a "hydrogen bridge" performing this role.

✓Final answer
$False -- it is a salt bridge, not a hydrogen bridge, that maintains continuity of ion flow in a Daniell cell.$
CBSE 2023Set 56/1/11 markMCQ
$Zn + 2Ag^+ \rightarrow Zn^{2+} + 2Ag$
›Reveal solutionSolution
By convention the anode (oxidation) is written on the left and the cathode (reduction) on the right. Zinc is oxidised and silver ions are reduced, so the cell is $Zn \mid Zn^{2+} \parallel Ag^+ \mid Ag$ — option (D).

A cell diagram is written anode (left) $\parallel$ cathode (right), with each half-cell running from the electrode metal outward and the double bar $\parallel$ marking the salt bridge.

For the reaction
$Zn + 2Ag^+ \rightarrow Zn^{2+} + 2Ag$

Zinc loses electrons: $Zn \rightarrow Zn^{2+} + 2e^-$ (oxidation, anode, left).

Silver ions gain electrons: $Ag^+ + e^- \rightarrow Ag$ (reduction, cathode, right).

Writing the anode as metal $\mid$ ion and the cathode as ion $\mid$ metal gives
$Zn \mid Zn^{2+} \parallel Ag^+ \mid Ag$
Checking the options:

(A) $2Ag \mid Ag^+ \parallel Zn \mid Zn^{2+}$ — electrodes reversed and coefficients are never used.

(B) $Ag^+ \mid Ag \parallel Zn^{2+} \mid Zn$ — anode/cathode reversed.

(C) $Ag \mid Ag^+ \parallel Zn \mid Zn^{2+}$ — anode/cathode reversed.

(D) $Zn \mid Zn^{2+} \parallel Ag^+ \mid Ag$ — Zn (anode) on the left, Ag (cathode) on the right. Correct.

✓Final answer
$Zn \mid Zn^{2+} \parallel Ag^+ \mid Ag$
CBSE 2022Set E1 markMCQ
$Q. The standard electrode potentials for the following reactions are given ( At 25°C ): Ag+(aq) + e- -> Ag(s), E° Ag+/Ag = +0.80 V ; Sn2+(aq) + 2e -> Sn(s), E° Sn2+/Sn = -0.14 V. The electromotive force (EMF) of the given cell Sn | Sn2+ (1M) || Ag+ (1M) | Ag is$
(a) 0.66 V
(b) 0.80 V
(c) 1.08 V
(d) 0.94 V
›Reveal solutionSolution
For Sn | Sn2+ || Ag+ | Ag, EMF = E°(Ag+/Ag) - E°(Sn2+/Sn) = 0.80 - (-0.14) = 0.94 V.

In the cell notation the left electrode is the anode (oxidation) and the right is the cathode (reduction):

Cathode (reduction): Ag+ + e- -> Ag, E° = +0.80 V

Anode (oxidation): Sn -> Sn2+ + 2e-, E°(Sn2+/Sn) = -0.14 V

E°cell = E°cathode - E°anode = (+0.80) - (-0.14) = +0.94 V.

(The positive value confirms the reaction Sn + 2Ag+ -> Sn2+ + 2Ag is spontaneous. Standard potentials are intensive, so they are not multiplied by the number of electrons.)

✓Final answer
$(d) 0.94 V.$
CBSE 2022Set ANNUAL1 markMCQ
$Q. Which one of the following statements is incorrect for a voltaic cell ?$
(a) It converts chemical energy to electrical energy.
(b) It uses electrical energy to carry out chemical changes.
(c) It is based on a redox reaction.
(d) It has $-\Delta G$ .
›Reveal solutionSolution
A voltaic cell produces electricity from a spontaneous redox reaction — it does not consume electrical energy, so statement (b) describes an electrolytic cell instead.

Checking each option against what a voltaic (galvanic) cell actually does:

(a) True — a voltaic cell converts chemical energy into electrical energy.

(b) False — this describes an electrolytic cell, which uses externally supplied electrical energy to force a non-spontaneous chemical change. A voltaic cell does the opposite.

(c) True — it is based on a spontaneous redox (oxidation–reduction) reaction.

(d) True — a spontaneous cell reaction has $\Delta G < 0$ , i.e. $-\Delta G$ .

✓Final answer
$(b) — "It uses electrical energy to carry out chemical changes" is the incorrect statement for a voltaic cell.$
CBSE 2022Set ANNUAL1 markMCQ
$Mg\,|\,Mg^{2+}\,||\,Cu^{2+}\,|\,Cu$
(a) $Mg$ as cathode
(b) $Cu$ as cathode
(c) $Cu$ is oxidizing agent
(d) None of the above
›Reveal solutionSolution
By IUPAC convention the electrode written on the LEFT of a cell is the anode and the one on the RIGHT is the cathode. Here Mg is the anode and Cu is the cathode. Option (B).

The cell is written as $Mg\,|\,Mg^{2+}\,||\,Cu^{2+}\,|\,Cu$ .

Convention: anode (negative, oxidation) is written on the left; cathode (positive, reduction) is written on the right.

The electrode reactions are:

Anode (Mg, oxidation): $Mg \rightarrow Mg^{2+} + 2e^-$

Cathode (Cu, reduction): $Cu^{2+} + 2e^- \rightarrow Cu$

Since $E^\circ_{Mg^{2+}/Mg} = -2.37\ V$ and $E^\circ_{Cu^{2+}/Cu} = +0.34\ V$ , Mg (more negative) is oxidised and acts as the anode; $Cu^{2+}$ is reduced, so Cu is the cathode. ( $Cu^{2+}$ is the oxidising agent here, not Cu metal, so option C is wrong.)

✓Final answer
$Cu$
CBSE 2021Set A1 markMCQ
$Q. Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s) is$
(a) Weston cell
(b) Daniel cell
(c) Calomel cell
(d) None of these
›Reveal solutionSolution
A zinc-copper galvanic cell with this notation is the Daniell cell.

The cell Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s) is the Daniell cell, a galvanic (voltaic) cell.

At the anode (LHS): Zn(s) → Zn2+ + 2e- (oxidation).

At the cathode (RHS): Cu2+ + 2e- → Cu(s) (reduction).

Standard EMF = E°(cathode) - E°(anode) = 0.34 - (-0.76) = +1.10 V.

The Weston cell (standard cell) uses Cd/Hg, and the calomel electrode uses Hg/Hg2Cl2 — neither matches this notation.

✓Final answer
$(B) Daniel cell.$

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