Q.Represent the cell in which the following reaction takes place:
Mg(s)+2Ag+(0.0001 M)→Mg2+(0.130 M)+2Ag(s)
Calculate its Ecell if Ecell∘=3.17 V.
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🔒 Start your 14-day free trial to unlock the full solution →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 …
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.
--- …
The key idea is the Nernst equation, which relates cell potential to concentration for a spontaneous redox reaction.
Step 1 – Write the cell representation
Anode (oxidation): Mg(s) | Mg²⁺(aq)
Cathode (reduction): Ag⁺(aq) | Ag(s)
The cell is:
Mg(s) ∣ Mg2+(0.130 M) ∣∣ Ag+(0.0001 M) ∣ Ag(s)
Step 2 – Determine n and the reaction quotient Q
Two electrons are transferred: n=2.
Q=[Ag+]2[Mg2+]=(0.0001)20.130=1×10−80.130=1.3×107
Step 3 – Apply the Nernst equation
Ecell=Ecell∘−n0.0591logQ …
Cell: Mg(s)∣Mg2+(0.130 M)∣∣Ag+(0.0001 M)∣Ag(s). Applying the Nernst equation with n=2 and Q=[Ag+]2[Mg2+] gives Ecell=2.96 V.
Cell representation
Oxidation (anode) is written on the left, reduction (cathode) on the right:
Mg(s) ∣ Mg2+(0.130 M) ∣∣ Ag+(0.0001 M) ∣ Ag(s)
Here Mg is oxidised to Mg2+ and Ag+ is reduced to Ag; two electrons are transferred, so n=2.
Nernst equation
Ecell=Ecell∘−n0.0591logQ,Q=[Ag+]2[Mg2+]
Step 1 — Reaction quotient. …
Method: Nernst Equation for Electrochemical Cell Representation and EMF Calculation
Step 1: Identify the Cell Representation (Cell Notation)
The reaction is:
Mg(s)+2Ag+(0.0001 M)→Mg2+(0.130 M)+2Ag(s)
Rules for cell notation:
- Anode (oxidation) is written on the left
- Cathode (reduction) is written on the right
- Single vertical line
|represents phase boundary - Double vertical line
||represents salt bridge
Oxidation half-reaction (Anode):
Mg(s)→Mg2+(aq)+2e−
Reduction half-reaction (Cathode):
2Ag+(aq)+2e−→2Ag(s)
Cell representation:
Mg(s) ∣ Mg2+(0.130 M) ∣∣ Ag+(0.0001 M) ∣ Ag(s)
Step 2: Write the Nernst Equation
For the general cell reaction:
aA+bB→cC+dD
The Nernst equation at 298 K is:
Ecell=Ecell∘−n0.0591log[A]a[B]b[C]c[D]d
Where:
- n = number of electrons transferred
- Ecell∘ = standard cell potential
Step 3: Identify Values
- n=2 (2 electrons are transferred)
- Ecell∘=3.17 V
- Reaction quotient Q:
Q=[Ag+]2[Mg2+]=(0.0001)20.130
Step 4: Calculate Q
Q=1×10−80.130=1.3×107
Step 5: Apply Nernst Equation …
🧠 The Core Concept First
A cell representation is a shorthand notation of an electrochemical cell. The convention is:
- Anode (oxidation) on the left
- Cathode (reduction) on the right
- Single vertical line
|for phase boundary - Double vertical line
||for salt bridge - Concentrations written in parentheses
The Nernst equation for the cell reaction at 298 K is:
Ecell=Ecell∘−n0.0591logQ
where Q is the reaction quotient and n is the number of electrons transferred.
✗ Common Mistake #1: Wrong order of electrodes in cell representation
What students do:
Write Ag∣Ag+∣∣Mg2+∣Mg (reversing anode and cathode).
Why it’s wrong:
Oxidation happens at the anode. Here, Mg is oxidised to Mg²⁺, so Mg is the anode (left side). Ag⁺ is reduced to Ag, so Ag is the cathode (right side).
✓ How to avoid:
Always identify which species is oxidised (loses electrons) and which is reduced (gains electrons). Then write:
Anode (oxidation) ∣∣ Cathode (reduction)
Correct representation:
Mg(s) ∣ Mg2+(0.130 M) ∣∣ Ag+(0.0001 M) ∣ Ag(s)
✗ Common Mistake #2: Using wrong value of n in Nernst equation
What students do:
Take n=1 or n=3 (guessing from coefficients).
Why it’s wrong:
n is the number of electrons transferred in the balanced redox reaction. Here:
- Mg → Mg²⁺ loses 2 electrons
- 2Ag⁺ + 2e⁻ → 2Ag gains 2 electrons
So n=2.
✓ How to avoid:
Write the half-reactions and balance electrons. Count electrons lost = electrons gained.
✗ Common Mistake #3: Writing Q incorrectly
What students do:
Write Q=[Ag+][Mg2+] or Q=[Ag+]2[Mg2+] but forget to raise to stoichiometric coefficients.
Why it’s wrong:
For the reaction:
Mg(s)+2Ag+(aq)→Mg2+(aq)+2Ag(s)
Q=[Ag+]2[Mg2+] (solids are omitted).
✓ How to avoid:
Write the reaction quotient using products over reactants, each raised to its stoichiometric coefficient. Ignore solids and pure liquids.
✗ Common Mistake #4: Forgetting the minus sign or log in Nernst equation
What students do:
Use Ecell=Ecell∘+n0.0591logQ or forget the log entirely.
Why it’s wrong:
The correct form for a cell at 298 K is:
Ecell=Ecell∘−n0.0591logQ
✓ How to avoid:
Memorise the formula with the minus sign. Remember: as the reaction proceeds, Ecell decreases from Ecell∘.
✓ Step-by-Step Solution (for reference) …
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
…
- 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). …
- 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 …
- 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 …
- 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). …
- 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 …
- 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: …
- 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.
…
- 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. …
- 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 …
- 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). …
- 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. …
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