Assertion: EAg^+/Ag increases with increase in concentration of Ag^+ ions. Reason: EAg^+/Ag has a positive val

NCERT Exemplar · Q67

E_{Ag^+/Ag}

(i) Both assertion and reason are true and the reason is the correct explanation of assertion.

(ii) Both assertion and reason are true and the reason is not the correct explanation of assertion.

(iii) Assertion is true but the reason is false.

(iv) Both assertion and reason are false.

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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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The Nernst equation shows that $E_{Ag^+/Ag}$ depends on $\log[Ag^+]$ , so increasing $[Ag^+]$ raises the potential. The reason given — that $E^\circ$ is positive — is true but does not explain this change; it only describes the standard value. Hence assertion true, reason true but not the correct explanation.

The heart of this question is the Nernst equation for a metal/metal-ion electrode. For the half-cell reaction

Ag^+ + e^- \rightarrow Ag(s)

the electrode potential is not fixed — it varies with the concentration of $Ag^+$ ions in solution. The standard reduction potential $E^\circ_{Ag^+/Ag} = +0.80\ \text{V}$ is just the reference value when $[Ag^+] = 1\ \text{M}$ at 298 K.

The Nernst equation at 298 K gives:

E_{Ag^+/Ag} = E^\circ_{Ag^+/Ag} - \frac{0.059}{1} \log \frac{1}{[Ag^+]}

which simplifies to

E_{Ag^+/Ag} = E^\circ_{Ag^+/Ag} + 0.059 \log [Ag^+]

Since $\log [Ag^+]$ increases when $[Ag^+]$ increases, the whole expression $E_{Ag^+/Ag}$ increases. That is the direct, quantitative reason the assertion is true. …

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Here are the common mistakes students make on this assertion-reason question and how to avoid each.

Mistake 1: Confusing “increase in $E$ ” with “more positive $E$ ”

What students do wrong:

They think that because the standard reduction potential $E^\circ_{Ag^+/Ag} = +0.80\ \text{V}$ is positive, any increase in concentration will make $E$ even more positive — so they mark option (i) or (ii).

Why it’s wrong:

The Nernst equation for $Ag^+ + e^- \rightarrow Ag$ is:

E = E^\circ - \frac{0.059}{1} \log \frac{1}{[Ag^+]}

Simplifying:

E = E^\circ + 0.059 \log [Ag^+]

If $[Ag^+]$ increases, $\log [Ag^+]$ increases → $E$ increases (assertion is true).
But the reason says “ $E_{Ag^+/Ag}$ has a positive value” — this is a fact about $E^\circ$ , not an explanation of why $E$ changes with concentration.

How to avoid:

Always write the Nernst equation explicitly. The reason must directly cause the assertion. A positive $E^\circ$ does not explain the change in $E$ with concentration — the Nernst equation does.

Mistake 2: Thinking the reason is false

What students do wrong:

They see that $E^\circ$ is positive but think “ $E$ can become negative if $[Ag^+]$ is very low” — so they mark option (iii).

Why it’s wrong:

The reason says “ $E_{Ag^+/Ag}$ has a positive value” — this is true for the standard value ( $E^\circ = +0.80\ \text{V}$ ). The statement does not say “always positive under all conditions.” It is a true statement about the standard potential.

How to avoid:

Read the reason carefully. If it states a known standard value, it is true unless it says “under all conditions.” Here, it simply states a fact about $E^\circ$ .

Mistake 3: Choosing (i) — “reason is correct explanation”

What students do wrong:

They think: “Positive $E^\circ$ means the reaction is spontaneous, so increasing $[Ag^+]$ makes it even more spontaneous → $E$ increases.”

Why it’s wrong:

The reason (positive value) does not explain the change in $E$ with concentration. The change is governed by the log term in the Nernst equation, not by the sign of $E^\circ$ .

How to avoid: …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

GUJCET 2025Set 031 markMCQ
$\Delta G$
›Reveal solutionSolution
[!TLDR]

$\Delta G$ depends on amount (extensive); $E_{cell}$ does not (intensive).

Concept

An extensive property depends on the quantity of matter; an intensive property does not. The Gibbs energy change of a cell reaction is linked to cell potential by $\Delta G = -nFE_{cell}$ .

Solution …
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GSEB Higher Secondary Certificate (HSC) Examination 2025Set ANNUAL1 markMCQ
$Q. Select the correct Nernst Equation for the given cell - Pt | H2(g) | H+(aq) || Br-(aq) | Br2(l) | Pt$
(a) Ecell = E0cell - (0.059/2) log[H+][Br-]
(b) Ecell = E0cell - 0.059 log([H+]/[Br-])
(c) Ecell = E0cell - (0.059/2) log([H+]^2/[Br-]^2)
(d) Ecell = E0cell - 0.059 log[H+][Br-]
›Reveal solutionSolution

Writing the Nernst equation for the cell reaction H2(g) + Br2(l) -> 2H+(aq) + 2Br-(aq) (n = 2 electrons) and simplifying the log term of squared concentrations gives the 0.059 (not 0.059/2) coefficient.

Cell: Pt | H2(g) | H+(aq) || Br-(aq) | Br2(l) | Pt

Anode (oxidation): H2(g) -> 2H+(aq) + 2e-

Cathode (reduction): Br2(l) + 2e- -> 2Br-(aq)

Overall: H2(g) + Br2(l) -> 2H+(aq) + 2Br-(aq), n = 2

Nernst equation: Ecell = E0cell - (0.059/n) log Q, where Q = [H+]^2[Br-]^2 / ([H2][Br2]). Since H2(g) is taken at unit activity/pressure and Br2 is a pure liquid (activity = 1):

Ecell = E0cell - (0.059/2) log([H+]^2[Br-]^2)

…

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GSEB Higher Secondary Certificate (HSC) Examination 2024Set ANNUAL1 markMCQ
$Q. Which Nernst equation is correct for the following cell? Al(s) | Al3+(aq) || Zn2+(aq) | Zn(s)$
(a) Ecell = E-cell-degree - (0.059/6) log([Al3+]^2 / [Zn2+]^3)
(b) Ecell = E-cell-degree - (0.059/6) log([Zn2+]^3 / [Al3+]^2)
(c) Ecell = E-cell-degree - (0.059/3) log([Al3+]^3 / [Zn2+]^2)
(d) Ecell = E-cell-degree - (0.059/2) log([Al3+]^2 / [Zn2+]^3)
›Reveal solutionSolution

Balancing the cell reaction to equalise electrons transferred (n=6) gives the correct Nernst equation form.

Cell: Al(s) | Al3+(aq) || Zn2+(aq) | Zn(s)

Anode (oxidation): Al -> Al3+ + 3e-, multiplied by 2: 2Al -> 2Al3+ + 6e-

Cathode (reduction): Zn2+ + 2e- -> Zn, multiplied by 3: 3Zn2+ + 6e- -> 3Zn

Overall: 2Al + 3Zn2+ -> 2Al3+ + 3Zn, with n = 6 electrons transferred.

…

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GUJCET 2021Set 151 markMCQ
$Mg_{(s)} + Cl_{2(g)} \rightarrow Mg^{2+}_{(aq)} + 2Cl^{-}_{(aq)}$
›Reveal solutionSolution
Anode (oxidation, Mg) on the left, cathode (Cl2, needs inert Pt) on the right.

Concept: Cell notation writes the anode half on the left and cathode on the right, with the phase boundaries and the salt bridge $(\|)$ in between.

Anode: $Mg \rightarrow Mg^{2+} + 2e^-$ → $Mg\,|\,Mg^{2+}(1M)$ .

Cathode: $Cl_2 + 2e^- \rightarrow 2Cl^-$ on an inert Pt electrode → $Cl^-(1M)\,|\,Cl_2(1\text{bar})\,|\,Pt$ . …
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GUJCET 2019Set 131 markMCQ
$\text{Zn}_{(s)} / \text{Zn}_{(aq)}(1\text{M}) // \text{Ni}_{(aq)}(1\text{M}) / \text{Ni}_{(s)}$
›Reveal solutionSolution

A Daniel cell is specifically Zn|Cu; this Zn|Ni cell is NOT a Daniel cell.

Concept: Any spontaneous cell converting chemical energy to electrical energy is a galvanic (= voltaic = electrochemical) cell. The Daniel cell is one particular galvanic cell using the $\text{Zn}/\text{Zn}^{2+}$ and $\text{Cu}/\text{Cu}^{2+}$ electrodes. The given cell uses nickel, not copper.

Steps: …
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←Previous questionQ66 — Assertion: Current stops flowing when ECell = 0. Reason: Equilibrium o…→Next questionQ68 — Assertion: Copper sulphate can be stored in zinc vessel. Reason: Zinc …

Q.Assertion: $E_{Ag^+/Ag}$ increases with increase in concentration of $Ag^+$ ions.
Reason: $E_{Ag^+/Ag}$ has a positive value.

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: Nernst Equation Analysis for Electrode Potential

Mistake 1: Confusing “increase in $E$ ” with “more positive $E$ ”

Mistake 2: Thinking the reason is false

Mistake 3: Choosing (i) — “reason is correct explanation”

Concept

Solution …

Q.Assertion: EAg+/AgE_{Ag^+/Ag}EAg+/Ag​ increases with increase in concentration of Ag+Ag^+Ag+ ions. Reason: EAg+/AgE_{Ag^+/Ag}EAg+/Ag​ has a positive value.

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Q.Assertion: $E_{Ag^+/Ag}$ increases with increase in concentration of $Ag^+$ ions.
Reason: $E_{Ag^+/Ag}$ has a positive value.

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