(a) Answer the following : (i) Why is the Equilibrium Constant (Kc) related to E°cell and not to Ecell ? (ii)

Question

Q. (a) Answer the following :

(i) Why is the Equilibrium Constant (

K_c

) related to

E^{\circ}_{cell}

and not to

E_{cell}

(ii) Two metals ‘A’ and ‘B’ have standard electrode potential values of

-0.24\ V

and

+0.80\ V

respectively. Which of these will liberate hydrogen gas from dil.

H_2SO_4

(iii) Write the cell reaction which occurs in lead storage battery when it is in charging.

(OR)

(b) What type of battery is Mercury cell ? Why it is more advantageous than dry cell ? Write overall reaction taking place in Mercury cell.

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Part (a)Concept understanding — Nernst Equation

The Nernst Equation: Why Batteries Don't Always Give Their Rated Voltage

Imagine you have a fresh AA battery. It says 1.5 V on the side. But if you measure it with a voltmeter, you might get 1.58 V when it's new, and 1.2 V when it's almost dead. Why does the voltage change? The Nernst equation is the tool that tells you exactly why.

The Core Idea: Concentration Drives Voltage

Every electrochemical cell works because of a chemical reaction that wants to happen. But here's the key: how badly the reaction wants to happen depends on how much of each chemical is present.

Think of it like a slope. A steep hill gives you more energy when you roll down. A shallow hill gives you less. In a battery, the "hill" is the difference in concentration (or more precisely, activity) of ions between the two electrodes. When the battery is fresh, the hill is steep — lots of reactants, few products. As the battery runs, reactants get used up, products build up, the hill flattens, and the voltage drops.

The Nernst equation is the mathematical formula that calculates the exact voltage for any given set of concentrations.

The Precise Statement

For a general electrochemical reaction:

aA + bB \rightarrow cC + dD

The cell potential $E$ under non-standard conditions is:

E = E^\circ - \frac{RT}{nF} \ln Q

Where:

$E$ = cell potential under the given conditions (what you actually measure)
$E^\circ$ = standard cell potential (the voltage when all reactants and products are at 1 M concentration, 1 atm pressure, 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 reaction
$F$ = Faraday constant (96,485 C/mol)
$Q$ = reaction quotient = $\frac{[C]^c [D]^d}{[A]^a [B]^b}$ (using concentrations for now)

At 25°C (298 K), the constants combine into a simpler form:

E = E^\circ - \frac{0.0592}{n} \log_{10} Q

Note

The 0.0592 comes from $\frac{2.303RT}{F}$ at 298 K. The 2.303 converts natural log to base-10 log, which is more convenient for calculations.

What It Actually Means

The equation has three parts:

$E^\circ$ — the "ideal" voltage when everything is at standard conditions. This is what you'd get in a textbook table.
$\frac{RT}{nF}$ — a scaling factor. It tells you how sensitive the voltage is to concentration changes. More electrons transferred ( $n$ ) means less sensitivity.
$\ln Q$ — the "concentration penalty". When $Q$ is small (lots of reactants, few products), $\ln Q$ is negative, so $E$ is higher than $E^\circ$ . When $Q$ is large (products building up), $\ln Q$ is positive, so $E$ drops below $E^\circ$ .

A Concrete Example

Consider the Daniell cell: $\text{Zn} | \text{Zn}^{2+} || \text{Cu}^{2+} | \text{Cu}$

The reaction is: $\text{Zn} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cu}$

$E^\circ = 1.10 \text{ V}$ , $n = 2$

If $[\text{Cu}^{2+}] = 0.1 \text{ M}$ and $[\text{Zn}^{2+}] = 1.0 \text{ M}$ :

Q = \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} = \frac{1.0}{0.1} = 10

$E = 1.10 - \frac{0.0592}{2} \log_{10}(10) = 1.10 - 0.0296 \times 1 = 1.07 \text{ V}$ …

Part (b)Concept understanding — Electrochemical Series

The Intuition: A "Tug-of-War" for Electrons

Imagine a chemical reaction where two substances are fighting over electrons. One substance wants to give away electrons (get oxidised), the other wants to take electrons (get reduced). Who wins? That depends on how strongly each substance holds onto its electrons.

Some metals, like sodium, are extremely generous — they practically throw their electrons at anyone. Others, like gold, are miserly — they cling to their electrons and rarely let go.

The Electrochemical Series is simply a ranking of substances based on how badly they want to keep their electrons (or how eagerly they want to give them up). It's a leaderboard of electron greed.

The Precise Statement

The Electrochemical Series (also called the Activity Series or Reactivity Series) is a list of elements — mostly metals — arranged in order of their standard electrode potentials ( $E^\ominus$ ). These potentials are measured in volts, relative to the Standard Hydrogen Electrode (SHE), which is arbitrarily assigned a value of $0.00\ \text{V}$ .

E^\ominus_{\text{SHE}} = 0.00\ \text{V}

A more negative $E^\ominus$ means the element is a stronger reducing agent — it readily loses electrons and gets oxidised. A more positive $E^\ominus$ means the element is a stronger oxidising agent — it readily gains electrons and gets reduced.

Here is the series for some common metals and hydrogen:

Element	Half-Reaction (Reduction)	$E^\ominus$ (V)
Lithium	$\text{Li}^+ + e^- \rightarrow \text{Li}$	$-3.04$
Potassium	$\text{K}^+ + e^- \rightarrow \text{K}$	$-2.93$
Calcium	$\text{Ca}^{2+} + 2e^- \rightarrow \text{Ca}$	$-2.87$
Sodium	$\text{Na}^+ + e^- \rightarrow \text{Na}$	$-2.71$
Magnesium	$\text{Mg}^{2+} + 2e^- \rightarrow \text{Mg}$	$-2.37$
Aluminium	$\text{Al}^{3+} + 3e^- \rightarrow \text{Al}$	$-1.66$
Zinc	$\text{Zn}^{2+} + 2e^- \rightarrow \text{Zn}$	$-0.76$
Iron	$\text{Fe}^{2+} + 2e^- \rightarrow \text{Fe}$	$-0.44$
Tin	$\text{Sn}^{2+} + 2e^- \rightarrow \text{Sn}$	$-0.14$
Lead	$\text{Pb}^{2+} + 2e^- \rightarrow \text{Pb}$	$-0.13$
Hydrogen	$\text{2H}^+ + 2e^- \rightarrow \text{H}_2$	$0.00$
Copper	$\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$	$+0.34$
Silver	$\text{Ag}^+ + e^- \rightarrow \text{Ag}$	$+0.80$
Gold	$\text{Au}^{3+} + 3e^- \rightarrow \text{Au}$	$+1.50$

Watch out

The half-reactions are written as reductions (gaining electrons). A more negative $E^\ominus$ means the reverse reaction — oxidation (losing electrons) — is more favourable. So lithium, with $-3.04\ \text{V}$ , is the best at giving away electrons, not keeping them.

What the Series Tells You

1. Predicting Reactivity

The lower (more negative) a metal is in the series, the more reactive it is. Lithium, potassium, and sodium react violently with water. Gold and platinum sit at the bottom and do almost nothing.

2. Displacement Reactions

A metal higher in the series (more negative $E^\ominus$ ) can displace a metal lower in the series from its salt solution. For example, zinc can displace copper from copper sulphate:

\text{Zn}(s) + \text{CuSO}_4(aq) \rightarrow \text{ZnSO}_4(aq) + \text{Cu}(s)

Why? Zinc loses electrons more readily than copper. The zinc atoms donate electrons to the $\text{Cu}^{2+}$ ions, turning them into copper metal.

Tip

To check if a displacement happens: the metal doing the displacing must have a more negative $E^\ominus$ than the metal being displaced. If both are on the same side of hydrogen, the one with the more negative value wins.

3. Direction of Redox Reactions …

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Part (a): $K_c$ is set by the constant $E^\circ_{cell}$ (not the variable $E_{cell}$ ), only metal A ( $-0.24$ V) liberates $H_2$ , and charging the lead battery gives $2PbSO_4+2H_2O\rightarrow Pb+PbO_2+2H_2SO_4$ . Part (b): the mercury cell is a primary cell giving a constant ~1.35 V; $Zn(Hg)+HgO\rightarrow ZnO+Hg$ .

Part (a)

(i) Why $K_c$ relates to $E^\circ_{cell}$ and not $E_{cell}$

$K_c$ is a thermodynamic equilibrium constant fixed at a given temperature, exactly like the standard EMF $E^\circ_{cell}$ (measured at unit activities). The Nernst equation is

E_{cell}=E^\circ_{cell}-\frac{0.0591}{n}\log Q.

At equilibrium the reaction quotient $Q$ becomes $K_c$ and no net current flows, so $E_{cell}=0$ . Substituting,

\log K_c=\frac{nE^\circ_{cell}}{0.0591}.

Since $E_{cell}$ changes with concentration and is zero at equilibrium, it cannot determine $K_c$ ; only the fixed standard value $E^\circ_{cell}$ can.

(ii) Which metal liberates $H_2$ from dil. $H_2SO_4$

For $2H^+ + 2e^- \rightarrow H_2$ , $E^\circ = 0$ V. A metal can displace hydrogen only if its standard reduction potential is negative (it is oxidised more easily than $H_2$ ).

Metal A: $E^\circ=-0.24$ V (negative) → liberates $H_2$ .
Metal B: $E^\circ=+0.80$ V (positive, a noble metal) → does not.

(iii) Cell reaction during charging of a lead storage battery

Discharge produces lead sulphate at both plates: $Pb + PbO_2 + 2H_2SO_4 \rightarrow 2PbSO_4 + 2H_2O$ . Charging supplies external energy to reverse this: …

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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 16 on this concept.

CBSE 2026Set A1 markMCQ
$Q. The standard reduction potentials of metals A, B, C and D are -3.05, -1.66, -0.40 and +0.80 volt respectively. Which metal of the following would have the highest reducing power ?$
(a) A
(b) B
(c) C
(d) D
›Reveal solutionSolution

The more negative the standard reduction potential, the stronger the tendency to be oxidised, i.e. the greater the reducing power. A has -3.05 V, the most negative.

A species with a very negative reduction potential is easily oxidised (gives up electrons readily), so it is a strong reducing agent. …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. Copper sulphate cannot be stored in zinc vessel. This is because$
(a) EMF of the reaction is negative
(b) EMF of the reaction is positive
(c) reduction potential value of Zn is more than Cu
(d) reduction potential value of Cu is negative
›Reveal solutionSolution

Zinc's much lower (more negative) standard reduction potential than copper's makes the displacement reaction's EMF positive, i.e. spontaneous — which is exactly why $Zn$ metal corrodes when it touches $CuSO_4$ solution.

Standard reduction potentials: $E^{\circ}(Cu^{2+}/Cu) = +0.34\ V$ ; $E^{\circ}(Zn^{2+}/Zn) = -0.76\ V$ . Since $Zn^{2+}/Zn$ has the far more negative (lower) reduction potential, $Zn$ metal is the stronger reducing agent and is oxidised preferentially:

$Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)$

$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = 0.34-(-0.76) = +1.10\ V$

Since $\Delta G^{\circ} = -nFE^{\circ}_{cell}$ , a positive $E^{\circ}_{cell}$ gives a negative $\Delta G^{\circ}$ — the reaction is thermodynamically spontaneous. So $CuSO_4$ solution spontaneously reacts with a zinc container, dissolving the zinc (corroding it) and depositing copper metal on it.

Why the other options are wrong: …

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CBSE 2025Set 56/4/11 markMCQ
$2Cu^{+} (aq) + Zn (s) \rightarrow 2Cu (s) + Zn^{2+} (aq)$
›Reveal solutionSolution
The cell voltage depends on the reaction quotient via the Nernst equation. As the reaction proceeds, $[Zn^{2+}]$ increases and $[Cu^+]$ decreases, so the voltage decreases. The correct option is (B).

The Nernst equation tells us that the actual voltage of an electrochemical cell under non-standard conditions is:
$E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{n} \log Q$
where $Q$ is the reaction quotient. For the given reaction:
$2Cu^+ (aq) + Zn (s) \rightarrow 2Cu (s) + Zn^{2+} (aq)$
the reaction quotient is:
$Q = \frac{[Zn^{2+}]}{[Cu^+]^2}$
(Remember: pure solids like Zn and Cu have activity = 1, so they don’t appear in $Q$ .)

The number of electrons transferred, $n$ , is 2 (each Cu⁺ gains one electron, and two Cu⁺ ions are reduced; Zn loses two electrons).

So the Nernst equation becomes:
$E_{\text{cell}} = 1.28 - \frac{0.059}{2} \log \frac{[Zn^{2+}]}{[Cu^+]^2}$
Now, as the reaction progresses:

$[Zn^{2+}]$ increases — Zn metal is oxidised to Zn²⁺, so its concentration in solution rises.

$[Cu^+]$ decreases — Cu⁺ ions are reduced to Cu metal, so their concentration falls.

Both changes make the fraction $\frac{[Zn^{2+}]}{[Cu^+]^2}$ larger.

A larger $Q$ means $\log Q$ is larger (more positive).

Since we subtract this term, $E_{\text{cell}}$ decreases.

Watch out

A common mistake is to think that because $[Zn^{2+}]$ appears in the numerator, the voltage might increase. But the Nernst equation has a minus sign in front of the log term — so anything that increases $Q$ actually lowers the voltage. …
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CBSE 2024Set ANNUAL1 mark
$Q. In an electrochemical cell the free energy change is related to EMF of the cell as ______.$
›Reveal solutionSolution

The free energy change of a cell reaction is related to its EMF by Delta G = -nFE, which is the thermodynamic basis for the Nernst equation.

The electrical work done by a galvanic cell is equal to the product of the total charge passed and the EMF of the cell. The total charge passed when n moles of electrons flow is nF (F = Faraday constant = 96500 C/mol).

Maximum electrical work obtainable = nFE (E = EMF of the cell)

This maximum work done by the system equals the decrease in Gibbs free energy of the system, so:

Delta G = -nFE

…

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CBSE 2023Set F1 markMCQ
$Q. The standard electrode potentials of A, B, C and D metals are -3.05 V, -1.66 V, -0.40 V and +0.8 V respectively. Which of the following would have the highest reducing power?$
(a) A
(b) B
(c) C
(d) D
›Reveal solutionSolution

The more negative the standard reduction potential, the stronger the reducing agent — so A (-3.05 V) has the highest reducing power.

Reducing power measures the tendency of a metal to lose electrons (get oxidised). A more negative standard electrode (reduction) potential means the metal is more easily oxidised and is therefore a stronger reducing agent.

…

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CBSE 2021Set A1 markMCQ
$Q. The standard reduction potential values of elements A, B and C are + 0.68 V, -2.50 V and -0.50 V respectively. The order of their reducing power is$
(a) A > B > C
(b) A > C > B
(c) C > B > A
(d) B > C > A
›Reveal solutionSolution
Lower (more negative) standard reduction potential means a stronger tendency to be oxidised, i.e. a stronger reducing agent.

Reducing power measures how readily a species gives up electrons (undergoes oxidation). The more negative the standard reduction potential (E°), the greater the tendency to lose electrons, so the stronger the reducing agent.

Given:

A: E° = +0.68 V (least negative → weakest reducing agent) …
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CBSE 2021Set A1 markMCQ
$Q. The Electromotive force (EMF) of the cell for the cell reaction at equilibrium state is$
(a) positive
(b) zero
(c) negative
(d) none of these
›Reveal solutionSolution

At equilibrium a galvanic cell is fully discharged, so its EMF = 0.

As a galvanic cell operates, the concentrations change until the reaction reaches equilibrium (the cell is 'dead').

From the Nernst equation: Ecell = E°cell − (0.059/n) log Q.

At equilibrium Q = K and Ecell = 0, giving the relation E°cell = (0.059/n) log K.

…

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CBSE 2020Set WA1 markMCQ
$+1.5\,V$
(a) $A > C > D > B$
(b) $B > D > C > A$
(c) $A > B > D > C$
(d) $D > A > B > C$
›Reveal solutionSolution

Lower (more negative) $E^\circ$ ⇒ stronger reducing agent ⇒ more reactive metal; so $B>D>C>A$ .

Concept: A metal's chemical activity (its tendency to lose electrons and get oxidised) increases as its standard reduction potential becomes more negative. Metals high in the activity series (like the alkali/alkaline-earth metals) have strongly negative $E^\circ$ .

…

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CBSE 2020Set ANNUAL1 mark
$Q. What is the relation between standard Gibbs' free energy and standard emf of the cell?$
›Reveal solutionSolution

The standard Gibbs free energy change of a cell reaction is related to the standard cell emf by $\Delta G^\circ = -nFE^\circ_{cell}$ .

The maximum electrical work obtainable from a galvanic cell equals the decrease in Gibbs free energy of the cell reaction. The electrical work done is the product of the total charge passed (nF, where n is the number of moles of electrons transferred in the balanced cell reaction and F is Faraday's constant, 96500 C/mol) and the cell's emf:

Electrical work = nFE_cell

Since this work is done at the expense of the free energy of the system, $\Delta G = -nFE_{cell}$ , and under standard conditions:

…

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CBSE 2020Set ANNUAL1 mark
$Q. How is equilibrium constant related to standard Gibb's energy?$
›Reveal solutionSolution

The standard Gibbs energy change of a reaction and its equilibrium constant are linked by $\Delta G^\circ = -RT\ln K$ .

For any reaction at equilibrium, thermodynamics gives the relation

$\Delta G^\circ = -RT\ln K = -2.303\,RT\log K$

where $R$ is the gas constant, $T$ the absolute temperature, and $K$ the equilibrium constant.

This connects directly to electrochemistry through the Nernst equation. Since $\Delta G^\circ = -nFE^\circ_{cell}$ (where $n$ = number of electrons transferred, $F$ = Faraday constant, $E^\circ_{cell}$ = standard cell potential), equating the two expressions for $\Delta G^\circ$ gives

$-nFE^\circ_{cell} = -RT\ln K \quad\Rightarrow\quad E^\circ_{cell} = \frac{RT}{nF}\ln K = \frac{0.0591}{n}\log K \text{ (at 298 K)}$

…

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CBSE 2019Set ANNUAL1 mark
$Q. Write the Nernst equation for following cell: Sn(s) | Sn²⁺ || H⁺ | H₂(g)(1bar) | Pt(s).$
›Reveal solutionSolution

For the cell Sn(s) | Sn²⁺ || H⁺ | H₂(g)(1 bar) | Pt(s), n = 2 and the Nernst equation is $E_{cell}=E^\circ_{cell}-\frac{0.0591}{2}\log\frac{[Sn^{2+}]p_{H_2}}{[H^+]^2}$ .

From the cell notation, the left electrode (Sn) is the anode (oxidation) and the right electrode (Pt, with H₂/H⁺) is the cathode (reduction):

Anode (oxidation): $Sn(s) \rightarrow Sn^{2+}(aq) + 2e^-$

Cathode (reduction): $2H^+(aq) + 2e^- \rightarrow H_2(g)$

Overall cell reaction: $Sn(s) + 2H^+(aq) \rightarrow Sn^{2+}(aq) + H_2(g)$

Here the number of electrons transferred, $n = 2$ .

The general Nernst equation is $E_{cell}=E^\circ_{cell}-\frac{0.0591}{n}\log Q$ , where $Q$ is the reaction quotient (products over reactants, each raised to its stoichiometric coefficient, gases as partial pressure, solids/pure liquids omitted):

$Q = \dfrac{[Sn^{2+}]\, p_{H_2}}{[H^+]^2}$

So:

…

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CBSE 2019Set ANNUAL1 markMCQ
$A^+/A$
(a) $A^+ + B^+ \longrightarrow A + B$
(b) $A^+ + B \longrightarrow A + B^+$
(c) $A + B^+ \longrightarrow A^+ + B$
(d) $A + B \longrightarrow A^+ + B^+$
›Reveal solutionSolution

$A$ has the more negative (lower) reduction potential, so it is the better reducing agent and is oxidised, while $B^+$ (higher reduction potential) is reduced.

…

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Q.(a) Answer the following :

Part (a)

(i) Why KcK_cKc​ relates to Ecell∘E^\circ_{cell}Ecell∘​ and not EcellE_{cell}Ecell​

(ii) Which metal liberates H2H_2H2​ from dil. H2SO4H_2SO_4H2​SO4​

(iii) Cell reaction during charging of a lead storage battery

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(i) Why $K_c$ relates to $E^\circ_{cell}$ and not $E_{cell}$

(ii) Which metal liberates $H_2$ from dil. $H_2SO_4$

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