Q.Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10.
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→cC+dD
The cell potential E under non-standard conditions is given by:
E=E∘−nFRTlnQ
Where:
- E = cell potential under the given conditions (in volts)
- E∘ = 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 = [A]a[B]b[C]c[D]d (using concentrations for dilute solutions)
At 25°C (298 K), the equation simplifies to a very practical form:
E=E∘−n0.0591log10Q
The 0.0591 comes from F2.303RT at 298 K. Notice it uses log10 (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:
Zn(s)∣Zn2+(aq)∥Cu2+(aq)∣Cu(s)
The single vertical line ∣ represents a phase boundary (solid electrode | solution). The double line ∥ 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): Zn(s)→Zn2+(aq)+2e−
- Cathode (reduction): Cu2+(aq)+2e−→Cu(s)
Overall: Zn(s)+Cu2+(aq)→Zn2+(aq)+Cu(s)
Here n=2 (two electrons transferred). The reaction quotient is:
Q=[Cu2+][Zn2+]
So the Nernst equation becomes:
E=E∘−20.0591log10[Cu2+][Zn2+]
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 [Zn2+]=0.1 M and [Cu2+]=1.0 M at 25°C. E∘ for the cell is 1.10 V.
E=1.10−20.0591log101.00.1
E=1.10−20.0591log10(0.1)
log10(0.1)=−1
E=1.10−20.0591(−1)=1.10+0.02955=1.1296 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).
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 [Cu2+]) decreases E
- Diluting the product side (lowering [Zn2+]) increases E
- At equilibrium, E=0 and Q=K (the equilibrium constant), giving lnK=RTnFE∘
This last point connects electrochemistry directly to thermodynamics — the Nernst equation is just the Gibbs free energy equation (ΔG=−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 (ΔG):
ΔG=−nFEcell
Where:
- n = moles of electrons transferred
- F = Faraday constant (96,485C mol−1)
- Ecell = cell potential (volts)
Why negative? A spontaneous reaction has ΔG<0 and Ecell>0 — the negative sign makes this consistent.
2. The Chemical Side: ΔG Depends on Concentration
For a general redox reaction:
aA+bB→cC+dD
The Gibbs free energy under non-standard conditions is:
ΔG=ΔG∘+RTlnQ
Where Q is the reaction quotient:
Q=[A]a[B]b[C]c[D]d
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:
−nFEcell=−nFEcell∘+RTlnQ
Divide both sides by −nF:
Ecell=Ecell∘−nFRTlnQ
This is the Nernst equation.
4. The "Why" in Plain Terms
| Concept | Physical Meaning |
|---|---|
| Ecell∘ | Voltage when all species are at 1 M (standard state) |
| −nFRTlnQ | Correction factor — adjusts voltage for real concentrations |
| Q | Tells you how far the reaction is from equilibrium |
Key insight: When Q=K (equilibrium), Ecell=0 — the battery is dead because no net reaction occurs.
5. The Common Form (log base 10)
At 25∘C (298K):
FRTln10≈0.0592V
So:
Ecell=Ecell∘−n0.0592log10Q
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:
Zn(s)∣Zn2+(aq)∥Cu2+(aq)∣Cu(s)
The reaction is:
Zn(s)+Cu2+(aq)→Zn2+(aq)+Cu(s)
Solids are omitted from Q (activity = 1):
Q=[Cu2+][Zn2+]
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∘ for real conditions |
| 4 | Use log10 at 25∘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 key idea is the Nernst equation applied to the hydrogen electrode, where the potential depends on the concentration (activity) of H+ ions, which is directly related to pH.
For the hydrogen electrode reaction: 2H++2e−→H2(g), the Nernst equation at 298 K is:
E=E∘−20.059log[H+]2PH2
Given E∘=0V (standard hydrogen electrode) and PH2=1atm (standard condition), this simplifies to:
E=0−20.059log[H+]21=−0.059log[H+]1=−0.059×pH
Substituting pH = 10:
E=−0.059×10=−0.59V
The potential of the hydrogen electrode is −0.59V.
The Nernst equation for the hydrogen electrode directly links its potential to the pH of the solution. For pH = 10, the potential is –0.591 V (vs. SHE).
The hydrogen electrode is the reference point for all electrochemistry — its potential under standard conditions (1 M H⁺, 1 atm H₂, 298 K) is defined as exactly 0 V. But when the solution isn't acidic, the H⁺ concentration changes, and so does the electrode's potential. The Nernst equation tells us exactly how.
For the half‑cell reaction
2H++2e−→H2(g)
the Nernst equation at 298 K is:
E=E∘−n0.0591log[H+]2PH2
Here E∘=0 V, n=2, and we take PH2=1 atm (standard pressure). That simplifies things beautifully.
- Substitute the known values
E=0−20.0591log[H+]21
- Simplify the log term
log[H+]21=log[H+]−2=−2log[H+]
So
E=−20.0591×(−2log[H+])=0.0591log[H+]
- Connect to pH By definition, pH=−log[H+], so log[H+]=−pH. Therefore
E=0.0591×(−pH)=−0.0591×pH
This is a clean, linear relationship: every increase of 1 pH unit makes the hydrogen electrode potential more negative by 59.1 mV.
- Plug in pH = 10
E=−0.0591×10=−0.591 V
A common mistake is forgetting the sign. The Nernst equation gives E=0.0591log[H+], and since log[H+] is negative for pH > 0, the potential is negative. A pH 10 solution is basic — the hydrogen electrode should be less able to reduce H⁺, so its potential drops below zero.
The result E=−0.0591×pH is a handy shortcut for any hydrogen electrode problem at 298 K and 1 atm H₂. Just multiply pH by –0.0591 and you're done.
The potential of the hydrogen electrode at pH 10 is −0.591 V (vs. SHE).
Method: Nernst Equation for a Hydrogen Electrode
This is a direct application of the Nernst equation to a single electrode (half-cell), not a full cell.
Step 1: Write the half-cell reaction
For a hydrogen electrode:
2H(aq)++2e−→H2(g)
Step 2: Write the Nernst equation for this half-cell
The general Nernst equation at 298 K (25°C) is:
E=E∘−n0.0591logQ
Where:
- E∘ = standard electrode potential = 0 V (by definition for SHE)
- n = number of electrons transferred = 2
- Q = reaction quotient = [H+]2PH2
Step 3: Substitute conditions
Given pH = 10, so:
[H+]=10−10 M
For a standard hydrogen electrode, PH2=1 atm.
Therefore:
Q=(10−10)21=1020
Step 4: Calculate the potential
E=0−20.0591log(1020)
E=−0.02955×20
E=−0.591 V
Final Answer
The potential of the hydrogen electrode at pH 10 is −0.591 V.
Key Concept Check
- The negative potential makes sense: at pH 10 (basic), H+ concentration is very low, so the reduction of H+ to H2 is less favourable than under standard conditions — hence the potential is lower (more negative).
- If you ever forget the formula, remember: E=−0.0591×pH for a hydrogen electrode at 25°C (since n=1 per H+ in the simplified form). Here: E=−0.0591×10=−0.591 V ✓
Common Mistakes: Potential of Hydrogen Electrode at pH 10
The Correct Approach First
For a hydrogen electrode, the half-cell reaction is:
2H++2e−→H2(g)
The Nernst equation for this electrode (at 298 K) is:
E=E∘−20.0591log[H+]2PH2
Since E∘=0V and PH2=1atm (standard conditions), this simplifies to:
E=0−20.0591log[H+]21=−0.0591log[H+]1
E=−0.0591×pH
At pH = 10: E=−0.591V
✗ Mistake 1: Forgetting the Negative Sign
What students do: They calculate E=+0.0591×pH and write +0.591V.
Why it's wrong: The Nernst equation gives E=−0.0591×pH. At high pH (low [H+]), the reduction potential becomes more negative — the electrode is less likely to gain electrons.
How to avoid: Always write the Nernst equation in full before simplifying. The negative sign comes from log(1/[H+])=−log[H+].
✗ Mistake 2: Using the Wrong Form of the Nernst Equation
What students do: They use E=E∘−n0.0591logQ but write Q=PH2[H+]2 (inverted).
Why it's wrong: For the reduction half-reaction 2H++2e−→H2, the reaction quotient is:
Q=[H+]2PH2
Products over reactants (excluding solids and pure liquids), with gases in atm.
How to avoid: Memorise the pattern: products (gases, aqueous) / reactants (aqueous, gases). For reduction, products are on the right side of the half-reaction.
✗ Mistake 3: Confusing pH with [H+]
What students do: They substitute pH = 10 directly into the equation without converting.
Why it's wrong: The Nernst equation uses [H+] in mol/L, not pH. You must use:
[H+]=10−pH=10−10M
How to avoid: Write the conversion step explicitly: pH=10⇒[H+]=10−10M.
✗ Mistake 4: Using n=1 Instead of n=2
What students do: They write E=−0.0591×pH but think it's because n=1.
Why it's wrong: The half-reaction involves 2 electrons (2e−), so n=2. The simplification works because:
20.0591×2=0.0591
The factor of 2 from log(1/[H+]2)=2log(1/[H+]) cancels with n=2.
How to avoid: Always state n from the balanced half-reaction before simplifying.
✗ Mistake 5: Forgetting Standard Conditions for PH2
What students do: They assume PH2 is not 1 atm and try to include it.
Why it's wrong: The problem states "hydrogen electrode" — by convention, this means PH2=1atm unless specified otherwise.
How to avoid: Remember: standard hydrogen electrode (SHE) always uses PH2=1atm and [H+]=1M for E∘=0V.
✓ Quick Checklist to Avoid All Mistakes
| Step | Action |
|---|---|
| 1 | Write balanced half-reaction: 2H++2e−→H2 |
| 2 | Identify n=2 |
| 3 | Write Nernst equation: E=0−20.0591log[H+]2PH2 |
| 4 | Set PH2=1 atm |
| 5 | Convert pH to [H+]=10−10 M |
| 6 | Simplify: E=−0.0591×pH |
| 7 | Calculate: E=−0.591V |
Final answer: −0.591V
Showing the 12 most recent of 13 on this concept.
- CBSE 2026Set ANNUAL1 markQ.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, Ecell.
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:
Ecell=Ecathode−Eanode
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 answerIt is called the electromotive force (EMF) of the cell — equivalently, the cell potential Ecell.
- CBSE 2025Set ANNUAL1 markQ.For the electrochemical cell Zn(s)+Cu2+(aq)→Zn2+(aq)+Cu(s) the cell produces an electrical potential of 1.1 volt, when [Zn2+] and [Cu2+] are unity. State the direction of flow of current on applying external potential of 1.1 volt.
›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)∣Zn2+(aq)∥Cu2+(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 answerSince 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 markMCQQ.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 markMCQQ.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 markMCQQ.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 answerThe cell behaves as an electrolytic cell when Eext > Ecell (option c).
-
- CBSE 2024Set ANNUAL1 markQ.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 answerFalse -- it is a salt bridge, not a hydrogen bridge, that maintains continuity of ion flow in a Daniell cell.
- CBSE 2023Set 56/1/11 markMCQQ.The correct cell to represent the following reaction is : Zn+2Ag+→Zn2++2Ag (A) 2Ag∣Ag+∣∣Zn∣Zn2+ (B) Ag+∣Ag∣∣Zn2+∣Zn (C) Ag∣Ag+∣∣Zn∣Zn2+ (D) Zn∣Zn2+∣∣Ag+∣Ag
›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∣Zn2+∥Ag+∣Ag — option (D).
A cell diagram is written anode (left) ∥ cathode (right), with each half-cell running from the electrode metal outward and the double bar ∥ marking the salt bridge.
For the reaction
Zn+2Ag+→Zn2++2Ag
- Zinc loses electrons: Zn→Zn2++2e− (oxidation, anode, left).
- Silver ions gain electrons: Ag++e−→Ag (reduction, cathode, right).
Writing the anode as metal ∣ ion and the cathode as ion ∣ metal gives
Zn∣Zn2+∥Ag+∣Ag
Checking the options:
- (A) 2Ag∣Ag+∥Zn∣Zn2+ — electrodes reversed and coefficients are never used.
- (B) Ag+∣Ag∥Zn2+∣Zn — anode/cathode reversed.
- (C) Ag∣Ag+∥Zn∣Zn2+ — anode/cathode reversed.
- (D) Zn∣Zn2+∥Ag+∣Ag — Zn (anode) on the left, Ag (cathode) on the right. Correct.
✓Final answerThe correct representation is (D) Zn∣Zn2+∥Ag+∣Ag.
- CBSE 2022Set E1 markMCQQ.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 markMCQQ.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 −ΔG.
›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 ΔG<0, i.e. −ΔG.
✓Final answer(b) — "It uses electrical energy to carry out chemical changes" is the incorrect statement for a voltaic cell.
-
- CBSE 2022Set ANNUAL1 markMCQQ.For the given cell reaction Mg∣Mg2+∣∣Cu2+∣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∣Mg2+∣∣Cu2+∣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→Mg2++2e−
- Cathode (Cu, reduction): Cu2++2e−→Cu
Since EMg2+/Mg∘=−2.37 V and ECu2+/Cu∘=+0.34 V, Mg (more negative) is oxidised and acts as the anode; Cu2+ is reduced, so Cu is the cathode. (Cu2+ is the oxidising agent here, not Cu metal, so option C is wrong.)
✓Final answer(B) Cu acts as the cathode.
- CBSE 2021Set A1 markMCQQ.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.
- CBSE 2018Set ANNUAL1 markQ.What is salt bridge?
›Reveal solutionSolution
The salt bridge completes the electrical circuit of a galvanic cell by letting ions migrate between the two half-cells (maintaining charge neutrality) while physically keeping the two electrolyte solutions from mixing.
A salt bridge is typically a U-shaped glass tube filled with a gel (agar-agar or gelatin) containing a concentrated solution of an inert electrolyte (one whose ions do not react with the cell's other ions and are not involved in the electrode reactions), commonly KCl, KNO3 or NH4NO3. Its two functions:
- It completes the internal electrical circuit of the cell, allowing ions to flow between the two half-cells, so that current continues to flow through the external circuit.
- As oxidation at the anode generates excess positive ions and reduction at the cathode depletes positive ions (or vice-versa), the salt bridge's ions migrate to maintain electrical neutrality in both half-cell solutions (anions flow toward the anode compartment, cations toward the cathode compartment) — without this, charge would build up and stop the current flow almost immediately.
- It prevents direct mixing of the two electrode solutions, which would otherwise allow direct (non-useful) chemical reaction between the species instead of a controlled flow of electrons through the external wire.
✓Final answerA salt bridge is a U-tube containing an inert electrolyte in gel form (e.g. KCl-agar) that connects the two half-cells of a galvanic cell — it completes the circuit and maintains electrical neutrality of both half-cell solutions by ion migration, without letting the solutions mix.
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- ✓PYQ mapping + timed mock tests
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