(a) (i) Calculate emf of the following cell at 25ºC : Z — CBSE Class 12 Chemistry 2024 PYQ

Question

Q. (a)

(i) Calculate emf of the following cell at 25ºC :

Zn(s) | Zn^{2+}(0·001\,M) || Cd^{2+}(0·1\,M) | Cd(s)

Given :

E^o_{Zn^{2+}/Zn}

= –0·76 V,

E^o_{Cd^{2+}/Cd}

= –0·40 V [log 10 = 1]

(ii) State Faraday's second law of electrolysis. How will the pH of aqueous NaCl solution be affected when it is electrolysed ?

(OR)

(b)

(i) Calculate the

\Delta_rG^o

and log

K_c

for the following cell reaction :

Fe(s) + Ag^+(aq) \rightarrow Fe^{2+}(aq) + Ag(s)

Given :

E^o_{Fe^{2+}/Fe}

= –0·44 V,

E^o_{Ag^+/Ag}

= +0·80 V, 1 F = 96500 C

mol^{-1}

(ii) Write any two advantages of the fuel cells over primary and secondary batteries.

(iii) How many Faradays are required for the oxidation of 1 mole of

H_2O

O_2

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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 — Standard Electrode Potential

The Intuition: Why Does a Metal "Want" to Dissolve?

Imagine a strip of zinc metal dipped in water. Some zinc atoms on the surface have a strong urge to leave the solid and go into solution as $Zn^{2+}$ ions, leaving their two electrons behind on the metal strip. That strip now has a surplus of negative charge. The water near the strip, meanwhile, gets a slight positive charge from the dissolved $Zn^{2+}$ ions.

This separation of charge creates an electric potential difference between the metal and the solution. That difference is the electrode potential of the zinc electrode. Different metals have different "urges" to lose electrons. Copper, for example, has almost no urge — in fact, $Cu^{2+}$ ions in solution prefer to grab electrons from the metal and plate out as copper atoms.

So the electrode potential is a measure of how strongly a metal (or any electrode) tends to lose or gain electrons relative to its own ions in solution.

The Problem: We Can Only Measure Differences

You cannot measure the absolute potential of a single electrode. If you connect a voltmeter to a zinc strip in a beaker, you get nothing — the circuit is incomplete. You need a second electrode to complete the circuit, and the voltmeter reads the difference between the two electrode potentials.

This is like measuring altitude. You cannot say "this hill is 500 meters tall" without a reference point — sea level. For electrode potentials, we need a universal "sea level."

The Reference: Standard Hydrogen Electrode (SHE)

The agreed-upon zero is the Standard Hydrogen Electrode. It consists of a platinum wire (coated with finely divided platinum) dipped in a solution of $H^+$ ions at 1 M concentration, with hydrogen gas at 1 bar pressure bubbling over the platinum surface. The temperature is fixed at 298 K (25 °C).

The half-reaction at this electrode is:

2H^+(aq) + 2e^- \rightarrow H_2(g)

By international convention, the potential of this electrode is defined as exactly 0.00 V under standard conditions.

Important

The SHE is the universal reference. Every standard electrode potential you see in tables is measured against this zero point.

The Precise Definition

Standard Electrode Potential ( $E^\ominus$ ) is the potential difference developed between an electrode and its surrounding electrolyte solution, measured against the Standard Hydrogen Electrode, when all species involved in the half-reaction are at unit activity (effectively 1 M concentration for dissolved ions, 1 bar pressure for gases) and the temperature is 298 K.

The notation $E^\ominus$ uses the plimsoll symbol (a superscript zero with a horizontal bar) to indicate standard conditions.

How It Works in Practice

To measure the standard electrode potential of zinc, you construct an electrochemical cell:

Left electrode: Zinc strip dipped in 1 M $ZnSO_4$ solution
Right electrode: Standard Hydrogen Electrode
Salt bridge: Connects the two solutions

The voltmeter reads the cell potential. For zinc, the reading is -0.76 V. The negative sign tells you that the zinc electrode has a stronger tendency to lose electrons than the SHE — electrons flow from the zinc electrode to the SHE through the external circuit.

For copper, the reading is +0.34 V. The positive sign means copper has a weaker tendency to lose electrons than the SHE — electrons flow from the SHE to the copper electrode. …

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(a)(i) $E_{cell} = 0.419$ V; (a)(ii) Faraday's 2nd law: masses ∝ equivalent weights, and electrolysing aqueous NaCl raises the pH. (b)(i) $\Delta_r G^\circ = -239.32$ kJ mol⁻¹, $\log K_c \approx 42.0$ ; (b)(ii) fuel cells are highly efficient and run continuously/pollution-free; (b)(iii) 2 Faradays oxidise 1 mol $H_2O$ to $O_2$ .

Part (a)

EMF by the Nernst equation. The cell reaction (Zn oxidised, Cd reduced) is $Zn(s) + Cd^{2+} \rightarrow Zn^{2+} + Cd(s)$ , with $n = 2$ . $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = (-0.40) - (-0.76) = +0.36\ \text{V}$ With $Q = \dfrac{[Zn^{2+}]}{[Cd^{2+}]} = \dfrac{0.001}{0.1} = 10^{-2}$ : $E_{cell} = 0.36 - \frac{0.059}{2}\log(10^{-2}) = 0.36 - 0.0295 \times (-2) = 0.36 + 0.059 = \mathbf{0.419\ V}$
Tip

Because $[Zn^{2+}]$ (product) is much lower than $[Cd^{2+}]$ (reactant), $Q < 1$ , the log term is negative, and the emf comes out higher than $E^\circ_{cell}$ — the concentration difference pushes the reaction forward.
Faraday's second law and NaCl electrolysis. Faraday's second law of electrolysis: when the same quantity of electricity is passed through different electrolytes, the masses of substances liberated at the electrodes are directly proportional to their chemical equivalent weights ( $m \propto E$ ). For aqueous NaCl, at the cathode water is reduced in preference to $Na^+$ : $2H_2O + 2e^- \rightarrow H_2 + 2OH^-$ …

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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 20 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$Q. Electrode potential of any Electrode depends on:$
(a) Nature of metal
(b) Temp. of solution
(c) Concentration of solution
(d) All of the above
›Reveal solutionSolution
Electrode potential is governed jointly by the identity of the metal, the temperature, and the concentration of the ions in solution around the electrode (Nernst equation).

The electrode potential of a metal electrode M dipped in a solution of its own ions $M^{n+}$ is given by the Nernst equation:

$E_{M^{n+}/M} = E^\ominus_{M^{n+}/M} - \dfrac{0.059}{n}\log\dfrac{1}{[M^{n+}]} \quad \text{(at 298 K)}$

Nature of the metal: $E^\ominus$ (the standard electrode potential) is an intrinsic property of the metal — different metals have very different tendencies to lose or gain electrons, so $E^\ominus$ differs from metal to metal.

Concentration of the solution: the log term shows the potential varies with the ion concentration $[M^{n+}]$ around the electrode. …
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CBSE 2026Set ANNUAL1 mark
$E^0_{Zn^{2+}/Zn} = -0.76$
›Reveal solutionSolution

A negative standard reduction potential means the species is easier to oxidise (a weaker oxidising agent) than the reference $H^+/H_2$ couple.

Meaning of the sign

Standard electrode potentials are measured relative to the Standard Hydrogen Electrode (SHE), $E^0_{H^+/H_2}=0.00\ V$ , for the reduction half-reaction $2H^++2e^-\rightarrow H_2$ .

For zinc, $E^0_{Zn^{2+}/Zn}=-0.76\ V$ refers to $Zn^{2+}+2e^-\rightarrow Zn$ . A negative value means this reduction is less favourable (occurs with a lower tendency) than the reduction of $H^+$ to $H_2$ . Equivalently, the reverse (oxidation) reaction,

$Zn\rightarrow Zn^{2+}+2e^-$ …

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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 2025Set ANNUAL1 markMCQ
$Q. The electrode potential of SHE (standard hydrogen electrode) is arbitrarily fixed as:$
(a) zero
(b) 0.34 V
(c) – 0.34 V
(d) 0.76 V
›Reveal solutionSolution

By international convention, the standard hydrogen electrode is assigned an electrode potential of exactly 0 V so it can serve as the reference against which all other electrode potentials are measured.

Since the absolute potential of a single electrode cannot be measured directly (only the potential difference between two electrodes, i.e. a full cell, can be measured), chemists needed a universal reference point. The Standard Hydrogen Electrode (SHE) — Pt(s) | H₂(g, 1 bar) | H⁺(aq, 1 M) — was chosen as this reference and its standard reduct …

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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 A1 mark
$Q. Fill in the blank: The value of potential of standard hydrogen electrode is ______.$
›Reveal solutionSolution

The standard hydrogen electrode is the universal reference electrode, and its electrode potential is arbitrarily fixed at exactly zero volts at all temperatures.

The standard hydrogen electrode (SHE) consists of a platinum electrode coated with platinum black, dipped in a 1 M H⁺ solution, with H2 gas bubbled at 1 bar pressure and 298 K.

$2H^{+}(aq, 1M) + 2e^{-} \rightleftharpoons H_2(g, 1\ bar)$

…

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CBSE 2023Set ANNUAL1 markMCQ
$Q. Standard Electrode Potential for Standard Hydrogen Electrode (SHE) is:$
(a) – 0.5 V
(b) + 1.0 V
(c) 0.0 V
(d) + 2.0 V
›Reveal solutionSolution

By convention, the Standard Hydrogen Electrode (SHE) is assigned a standard electrode potential of exactly $0.0\ \text{V}$ .

Since absolute electrode potentials cannot be measured directly, the SHE — a platinum electrode in contact with $\text{H}_2$ gas at 1 bar, dipped in a 1 M $\text{H}^+$ solution at 298 K — is chosen as the reference electrode against which all other …

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CBSE 2023Set ANNUAL1 mark
$Mg^{2+}/Mg$
›Reveal solutionSolution

A single electrode's absolute potential cannot be measured directly; it is obtained by constructing a cell of the $Mg$ electrode against the standard hydrogen electrode (SHE), whose potential is arbitrarily fixed at exactly $0$ volts.

It is impossible to measure the potential of a single (isolated) electrode because any measurement necessarily needs a complete circuit, i.e. a second electrode. Therefore, standard electrode potentials are always measured relative to a reference electrode — the Standard Hydrogen Electrode (SHE), $Pt, H_2(1\ bar)|H^+(1\ M)$ , whose standard reduction potential is, by convention, assigned the value $0.00\ V$ at all temperatures.

Method: A galvanic cell is set up by combining the $Mg^{2+}(aq, 1\ M)|Mg(s)$ electrode with the SHE:

$Mg(s)|Mg^{2+}(1\ M)\ ||\ H^+(1\ M)|H_2(1\ bar), Pt(s)$

…

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CBSE 2022Set M1 mark
$Q. Give an example for inert electrode.$
›Reveal solutionSolution

Platinum is a common inert electrode.

An inert electrode does not take part in the chemical reaction; it only provides a surface for electron transfer (and conducts current). Platinum and graphite are chemically unreactive and are widely used, e.g. the Pt electrode in the standard hydrogen electrode a …
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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 2021Set TERM11 mark
$Q. Potential of an electrode means _______.$
›Reveal solutionSolution

Electrode potential measures how strongly a metal electrode tends to get oxidised or reduced relative to a reference (the standard hydrogen electrode).

When a metal electrode is dipped into a solution of its own ions, an equilibrium is set up at the metal–solution interface, and a potential difference develops between the metal and the solution due to the tendency of the metal to lose electrons (oxidation, giving reduction/oxidation potential) or of the ions to gain electrons (reduction). This tendency, measured relative to the standard hydrogen electrode (taken as 0 V), is called the ** …

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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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Q.(a)

Part (a)

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