A galvanic cell has electrical potential of $1.1\ V$. If an opposing potential of $1.1\ V$ is applied to this cell, what will happen to the cell reaction and current flowing through the cell?

A galvanic cell has electrical potential of 1.1V. If an opposing potential of 1.1V is applied to this cell, wh

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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.

$E = 1.10 - \frac{0.0591}{2} \log_{10} \frac{0.1}{1.0}$ …

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.

--- …

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When the applied opposing potential exactly equals the cell’s EMF, the cell reaction stops (reaches equilibrium) and no current flows — the cell is “balanced” and behaves like an open circuit.

The key here is understanding what happens when you “fight” a galvanic cell with an external voltage. A galvanic cell spontaneously produces electricity because its two half-cells have different reduction potentials — electrons flow from the anode to the cathode through the external circuit. The cell’s electromotive force (EMF), here $1.1\ \text{V}$ , is the driving force for that spontaneous reaction.

Now, if you connect an external power source in opposition — positive to positive, negative to negative — you are essentially trying to push electrons back the other way. This is exactly how electrolysis works: when the applied voltage exceeds the cell’s EMF, you force the reverse reaction. But here the applied voltage is exactly equal to the cell’s EMF, not greater.

Let’s walk through what that means.

The Nernst equation tells us the cell potential depends on concentrations. For a Daniell cell (the classic $1.1\ \text{V}$ example: $\text{Zn}|\text{Zn}^{2+}||\text{Cu}^{2+}|\text{Cu}$ ), the cell reaction is

\text{Zn} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cu}

The Nernst equation at $298\ \text{K}$ is

E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{n} \log Q

where $Q = \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]}$ and $n=2$ . Under standard conditions ( $1\ \text{M}$ each), $E_{\text{cell}} = 1.1\ \text{V}$ .

When you apply an opposing voltage exactly equal to $E_{\text{cell}}$ , the net driving force becomes zero. Think of it like two people pulling a rope with exactly equal force in opposite directions — nothing moves. The external power supply exactly cancels the cell’s tendency to push electrons. Mathematically, the net potential difference across the external circuit is

E_{\text{net}} = E_{\text{cell}} - E_{\text{applied}} = 1.1\ \text{V} - 1.1\ \text{V} = 0\ \text{V}

With zero net voltage, no electrons are driven through the circuit.

No current flows — the cell is at equilibrium.

Current requires a potential difference. Since $E_{\text{net}} = 0$ , the current $I = 0$ . The cell reaction stops because there is no longer any driving force for electron transfer. The system is now at dynamic equilibrium: the forward and reverse rates are equal, but macroscopically, no net reaction occurs.
What about the concentrations? …

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Here are the common mistakes students make when solving this type of question, along with how to avoid each.

Mistake 1: Thinking the cell reaction stops completely

Why it happens

Students assume that when applied voltage equals cell potential ( $E_{\text{cell}} = E_{\text{applied}}$ ), the reaction simply "stops." They forget that at this point, the cell is at equilibrium — the reaction is still occurring in both directions at equal rates.

How to avoid

Remember:

If $E_{\text{applied}} < E_{\text{cell}}$ → cell works normally (spontaneous reaction).
If $E_{\text{applied}} = E_{\text{cell}}$ → no net reaction, but dynamic equilibrium exists.
If $E_{\text{applied}} > E_{\text{cell}}$ → electrolysis (reverse reaction).

So here, with $E_{\text{applied}} = 1.1\ \text{V} = E_{\text{cell}}$ , the net cell reaction stops, but the system is at equilibrium.

Mistake 2: Saying current becomes zero immediately

Why it happens

Students think "no net reaction" means zero current instantly. In reality, at the exact balancing potential, current approaches zero only after equilibrium is reached.

How to avoid

Use the Nernst equation logic:

E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{n} \log Q

At equilibrium, $Q = K$ and $E_{\text{cell}} = 0$ (for a cell at equilibrium). But here, the applied voltage exactly cancels the cell's driving force — so no net electron flow occurs. Current becomes zero (or negligible) because there is no net driving force.

Mistake 3: Confusing this with electrolysis

Why it happens

Students see "opposing potential" and immediately think of electrolytic cells. They forget that equal opposing potential does not reverse the reaction — it only stops it.

How to avoid

Opposing potential < cell potential → cell still discharges (but slower).
Opposing potential = cell potential → no net reaction (equilibrium).
Opposing potential > cell potential → electrolysis (reverse reaction starts).

Here, $1.1\ \text{V}$ exactly equals the cell potential, so no electrolysis occurs.

Mistake 4: Forgetting to mention "no net reaction" vs "reaction stops"

Why it happens …

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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:

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

…

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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 …
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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). …
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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 …

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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 …
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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). …
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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 …

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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: …
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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.

…
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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. …
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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): $Cu^{2+} + 2e^- \rightarrow Cu$ …
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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). …
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Q.A galvanic cell has electrical potential of $1.1\ V$ . If an opposing potential of $1.1\ V$ is applied to this cell, what will happen to the cell reaction and current flowing through the cell?

Concept understanding — Cell Representation Nernst Equation

Cell Representation and the Nernst Equation: From Intuition to Precision

The Intuition First

The Precise Statement

Cell Representation: How We Write It

Applying the Nernst Equation to a Cell Representation

A Worked Example

Why this formula?

Cell Representation & the Nernst Equation: Why It Works

The Core Question

1. The Fundamental Link: Gibbs Free Energy & Cell Potential

2. The Chemical Side: $\Delta G$ Depends on Concentration

3. Combining Both Sides: The Derivation

4. The "Why" in Plain Terms

5. The Common Form (log base 10)

Method: Electrochemical Cell Equilibrium Analysis (using Nernst Equation & Overpotential Concept)

Step 1: Understand the given condition

Step 2: Compare applied potential with cell potential

Step 3: Apply the Nernst equation logic

Step 4: Determine current flow …

Mistake 1: Thinking the cell reaction stops completely

Mistake 2: Saying current becomes zero immediately

Mistake 3: Confusing this with electrolysis

Mistake 4: Forgetting to mention "no net reaction" vs "reaction stops"

Q.A galvanic cell has electrical potential of 1.1 V1.1\ V1.1 V. If an opposing potential of 1.1 V1.1\ V1.1 V is applied to this cell, what will happen to the cell reaction and current flowing through the cell?

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Q.A galvanic cell has electrical potential of $1.1\ V$ . If an opposing potential of $1.1\ V$ is applied to this cell, what will happen to the cell reaction and current flowing through the cell?

Unlock everything free for 14 days