Ecell for the given redox reaction is 2.71 V Mg(s) + Cu²⁺(0.01M) arrow Mg²⁺(0.001M) + Cu(s) Calculate Ecell fo

Question

E_{cell}

(i) less than 2.71 V and

(ii) greater than 2.71 V

(OR)

(a) A steady current of 2 amperes was passed through two electrolytic cells X and Y connected in series containing electrolytes

FeSO_4

and

ZnSO_4

until 2.8 g of Fe deposited at the cathode of cell X. How long did the current flow? Calculate the mass of Zn deposited at the cathode of cell Y. (Molar mass: Fe = 56

g\,mol^{-1}

, Zn = 65.3

g\,mol^{-1}

, 1F = 96500

C\,mol^{-1}

)

(b) In the plot of molar conductivity (

\Lambda_m

, y-axis) vs square root of concentration (

c^{1/2}

, x-axis), two curves are obtained for electrolytes A and B: curve A stays high and decreases only slightly and almost linearly with

c^{1/2}

, whereas curve B lies lower and rises steeply as

c^{1/2}

approaches zero. Answer the following:

(i) Predict the nature of electrolytes A and B.

(ii) What happens on extrapolation of

\Lambda_m

to concentration approaching zero for electrolytes A and B?

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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 — Faraday's Laws of Electrolysis

Faraday's Laws of Electrolysis – From Intuition to Precision

Imagine you are plating a spoon with silver. You dip it in a silver salt solution, connect it to a battery, and silver metal starts coating the spoon. Two questions naturally arise: How much silver will deposit? And does the amount depend only on the battery's strength, or also on the time?

Faraday answered both with two beautifully simple laws.

The Core Intuition

Electrolysis is about moving electrons. Each silver ion ( $\text{Ag}^+$ ) arriving at the spoon grabs one electron and becomes a neutral silver atom. So the mass of silver deposited is directly proportional to the number of electrons that have flowed — that is, to the total charge passed.

But different ions need different numbers of electrons. A copper ion ( $\text{Cu}^{2+}$ ) needs two electrons to become copper metal. So for the same charge, you get half as many copper atoms as silver atoms. That is why the chemical nature of the substance matters — specifically, its equivalent weight (the mass that reacts with one mole of electrons).

The Two Laws – Precise Statements

m \propto Q \quad \text{or} \quad m = Z Q

\frac{m_1}{E_1} = \frac{m_2}{E_2}

Putting Them Together – The Combined Equation

The two laws merge into one powerful formula:

m = \frac{Q \times E}{F}

where:

$m$ = mass deposited (g)
$Q$ = total charge passed (coulombs) = $I \times t$
$E$ = equivalent weight (g/eq)
$F$ = Faraday's constant = 96485 C/mol (charge of one mole of electrons)

Tip

A quick way to remember: $m = \frac{I t E}{F}$ . The charge $I t$ is just current × time.

Worked Example – Silver Plating

Problem: A current of 2.0 A is passed through a silver nitrate solution for 30 minutes. How much silver deposits? (Atomic mass of Ag = 107.9 g/mol, valency = 1)

Step 1 – Find the charge:

Q = I \times t = 2.0 \times (30 \times 60) = 3600 \text{ C}

Step 2 – Find equivalent weight:

E = \frac{107.9}{1} = 107.9 \text{ g/eq}

Step 3 – Apply the combined law:

$m = \frac{Q \times E}{F} = \frac{3600 \times 107.9}{96485} \approx 4.03 \text{ g}$ …

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Part (a): Nernst gives $E_{cell}=2.71+\tfrac{0.0591}{2}=2.74$ V; current flows spontaneously (Mg→Cu) when the opposing potential is below $E_{cell}$ and reverses above it. Part (b): series electrolysis — charge $=9650$ C so $t=4825$ s and $3.265$ g Zn deposits; A is a strong electrolyte (extrapolate to $\Lambda_m^\circ$ ), B a weak one (use Kohlrausch's law).

Part (a)

Nernst equation

Half-reactions: $Mg\to Mg^{2+}+2e^-$ (anode) and $Cu^{2+}+2e^-\to Cu$ (cathode), so $n=2$ . Solids have unit activity, hence

Q=\frac{[Mg^{2+}]}{[Cu^{2+}]}=\frac{0.001}{0.01}=0.1.

E_{cell}=E^\circ_{cell}-\frac{0.0591}{n}\log Q=2.71-\frac{0.0591}{2}\log(0.1).

Since $\log(0.1)=-1$ ,

E_{cell}=2.71+\frac{0.0591}{2}=2.71+0.0296=2.7396\approx \mathbf{2.74\ V}.

Dilute products (low $[Mg^{2+}]$ ) raise the EMF slightly above $E^\circ$ .

Effect of an opposing external potential

(i) External potential $<$ cell EMF: the cell wins, the spontaneous reaction continues; electrons flow from the Mg anode to the Cu cathode through the wire (conventional current Cu→Mg externally).
(ii) External potential $>$ cell EMF: the external source forces the reverse (electrolytic) reaction — Cu is oxidised, $Mg^{2+}$ reduced — and the current reverses direction. …

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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 30 on this concept.

CBSE 2026Set A1 markMCQ
$Q. Which of the following equations represents the Faraday's first law of electrolysis ?$
(a) mz = c.t
(b) m = c.z.t
(c) mc = z.t
(d) c = m.z.t
›Reveal solutionSolution

Faraday's first law: mass deposited m is proportional to the quantity of charge, m = z x Q = z x c x t (c = current, t = time, z = electrochemical equivalent).

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. Assertion (A) : Reduction of 1 mole of Cu2+ ions requires 2 Faraday of charge. Reason (R) : 1 Faraday is equal to the charge of 1 mole of electrons.$
(a) Both (A) and (R) are true and (R) is the correct explanation of (A)
(b) Both (A) and (R) are true but (R) is not the correct explanation of (A)
(c) (A) is true but (R) is false.
(d) (A) is false but (R) is true.
›Reveal solutionSolution

Cu²⁺ + 2e⁻ → Cu needs 2 moles of electrons, i.e. 2 Faraday of charge, and 1 Faraday is defined as the charge carried by 1 mole of electrons — so the reason directly explains the assertion.

Assertion: Reduction of 1 mole of Cu²⁺ requires 2 Faraday of charge.

The reduction half-reaction is:

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

To reduce 1 mole of Cu²⁺ ions, 2 moles of electrons are needed. Since 1 Faraday (F) is exactly the amount of charge carried by 1 mole of electrons (F = N_A × e ≈ 96500 C/mol), 2 moles of electrons corresponds to 2 Faraday of charge. So the assertion is true.

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. The charge required to reduce 1 mol of MnO₄⁻ to MnO₂ is-$
(a)
(i) 1F
(b)
(ii) 3F
(c)
(iii) 5F
(d)
(iv) 6F
›Reveal solutionSolution
$\text{Mn}$ goes from $+7$ (in $\text{MnO}_4^{-}$ ) to $+4$ (in $\text{MnO}_2$ ), a gain of 3 electrons per Mn; 1 mole requires 3 F. Correct option: (ii).

Concept. By Faraday's laws, the charge needed to reduce 1 mole of a species equals (number of electrons gained per ion) $\times$ 1 F, where $1\ \text{F} = 96500\ \text{C}$ is the charge of 1 mole of electrons.

Steps.

Oxidation state of Mn in $\text{MnO}_4^{-}$ : $x + 4(-2) = -1 \Rightarrow x = +7$ .

Oxidation state of Mn in $\text{MnO}_2$ : $x + 2(-2) = 0 \Rightarrow x = +4$ .

Change $= +7 \rightarrow +4$ , so each Mn gains $7-4 = 3$ electrons. …
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CBSE 2026Set ANNUAL1 markMCQ
$Q. How much charge is required for the 1 mol Al3+ to Al?$
(a) 1F
(b) 2F
(c) 4F
(d) 3F
›Reveal solutionSolution

Al3+ + 3e- -> Al, so 1 mol Al needs 3 mol electrons = 3F.

The reduction half-reaction is: Al3+ + 3e- -> Al.

…

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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
$Q. If 96500 coulomb of electricity is passed through CuSO4 solution, it will liberate-$
(a) 63.5 gm copper
(b) 100 gm copper
(c) 96500 gm copper
(d) None of the these
›Reveal solutionSolution

96500 C = 1 Faraday = 1 mole of electrons, but Cu2+ needs 2 electrons per atom, so only half a mole (31.75 g) of copper is deposited.

At the cathode: $Cu^{2+} + 2e^- \to Cu$ .

$96500\ C = 1\ F = 1$ mole of electrons. Since 2 moles of electrons are needed to deposit 1 mole (63.5 g) of copper, 1 mole of electrons (96500 C) deposits only:

…

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CBSE 2025Set D1 markMCQ
$Q. The quantity of electricity required to liberate 32 g of oxygen is$
(a) 1 faraday
(b) 2 faraday
(c) 3 faraday
(d) 4 faraday
›Reveal solutionSolution

32 g O2 = 1 mol; the electrode reaction transfers 4 electrons per O2, so 4 faraday are needed.

32 g of oxygen (O2, molar mass 32 g/mol) is 1 mole of O2 molecules.

At the anode, oxygen is liberated by:

2H2O → O2 + 4H+ + 4e- (or 4OH- → O2 + 2H2O + 4e-)

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. The number of electrons with one coulomb of charge will be:$
(a) 6.29 x 10^11
(b) 1.6 x 10^19
(c) 6.24 x 10^18
(d) 5.46 x 10^29
›Reveal solutionSolution

Since the charge on one electron is 1.6 × 10^-19 C, the number of electrons carrying 1 C of charge is 1 / (1.6 × 10^-19) ≈ 6.24 × 10^18.

The charge on a single electron is e = 1.6 × 10^-19 coulomb (this value itself is option b, which is a distractor — it is the charge of ONE electron, not the count of electrons).

To find how many electrons (n) are needed to make up a total charge of 1 C:

n = Total charge / charge per electron = 1 / (1.6 × 10^-19) = 6.25 × 10^18, which rounds to the listed 6.24 × 10^18.

…

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CBSE 2025Set ANNUAL1 markMCQ
$AlCl_3$
(a) 0.33 faraday
(b) 1 faraday
(c) 3 faraday
(d) 1 ampere
›Reveal solutionSolution

Aluminium exists as $Al^{3+}$ in $AlCl_3$ , so depositing one mole of Al at the cathode needs 3 moles of electrons, i.e. 3 faradays.

Reasoning

The cathodic reduction half-reaction is:

$Al^{3+} + 3e^- \rightarrow Al$

By Faraday's first law, the quantity of electricity needed to deposit 1 mole of a substance equals (number of electrons transferred per ion) $\times F$ , where $F = 96500\ \text{C mol}^{-1}$ .

Here $n = 3$ , so the charge required is:

$Q = nF = 3F$

…

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CBSE 2024Set D1 markMCQ
$Q. A charge of 96500 coulomb liberates .............. from the solution of CuSO4.$
(a) 63.5 gm copper
(b) 31.76 gm copper
(c) 96500 gm copper
(d) 100 gm copper
›Reveal solutionSolution

96500 C = 1 faraday = 1 mole of electrons. Depositing Cu requires 2 electrons per Cu atom, so 1 F deposits 63.5/2 = 31.76 g Cu.

Electrode reaction: Cu2+ + 2e- -> Cu.

To deposit 1 mole of copper (63.5 g) you need 2 moles of electrons = 2 x 96500 C = 193000 C.

Therefore the charge passed here, 96500 C (1 faraday, i.e. 1 mole of electrons), deposits half a mole of copper: …

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CBSE 2024Set B1 mark
$Q. Fill in the blank: One Faraday electricity equals to ______ coulomb.$
›Reveal solutionSolution

1 Faraday = charge carried by one mole of electrons = 96,500 C (more precisely 96,487 C, usually rounded to 96,500 C).

One Faraday (F) is defined as the quantity of electric charge carried by one mole (Avogadro's number, 6.022x10^23) of electrons:

$1F = N_A \times e = 6.022\times10^{23} \times 1.602\times10^{-19}\text{ C} \approx 96{,}500\text{ C mol}^{-1}$

…

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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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Part (a)

Nernst equation

Effect of an opposing external potential

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