(a) Calculate Δ G for the reaction Zn(s) + Cu²⁺(aq) arrow Zn²⁺(aq) + Cu(s). Given : E^° for Zn²⁺/Zn = -0·76 V

Question

\Delta G

(b) Give two advantages of fuel cells.

(OR)

(a) Out of the following pairs, predict with reason which pair will allow greater conduction of electricity:

(i) Silver wire at

30\,°C

or silver wire at

60\,°C

(ii) 0·1 M

CH_3COOH

solution or 1 M

CH_3COOH

solution.

(iii) KCl solution at

20\,°C

or KCl solution at

50\,°C

(b) Give two points of differences between electrochemical and electrolytic cells.

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Part (a)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.

--- …

Part (b)Concept understanding — Conductivity of Electrolytes

Conductivity of Electrolytes – From Intuition to Precision

Think of a copper wire. You know it conducts electricity because electrons flow through it. Now imagine dipping two metal plates into a beaker of salt water and connecting them to a battery. The bulb glows. The salt water is conducting — but not with electrons. Something else is carrying the charge.

That something is ions. When an electrolyte (like NaCl, HCl, or NaOH) dissolves in water, it splits into positive and negative ions. These ions are free to move. When you apply a voltage, positive ions drift toward the negative electrode, negative ions toward the positive electrode. That directed motion of charged particles is an electric current. That is the core idea: electrolytic conductivity is the ability of a solution to carry current via the movement of its ions.

Note

The more ions present, and the faster they can move, the higher the conductivity. But it is not that simple — concentration changes both the number of ions and how they interact with each other.

The Precise Definition

Conductivity (symbol $\kappa$ , units S m $^{-1}$ or S cm $^{-1}$ ) is the reciprocal of resistivity. For a solution placed between two parallel electrodes of area $A$ and separation $l$ , the resistance $R$ is:

R = \frac{1}{\kappa} \cdot \frac{l}{A}

So $\kappa$ is the conductance of a 1 m × 1 m × 1 m cube of the solution. It depends on:

Number of ions per unit volume (concentration)
Charge on each ion
How fast the ions move (their mobility)

But here is the catch: as you dilute a solution, $\kappa$ does not simply drop proportionally. Why? Because dilution changes both the number of ions and the degree of dissociation (for weak electrolytes). To compare the conducting power of different electrolytes fairly, we need a quantity that normalises for concentration.

Molar Conductivity – The Fair Comparison

Molar conductivity ( $\Lambda_m$ , units S m $^2$ mol $^{-1}$ ) is defined as:

\Lambda_m = \frac{\kappa}{c}

where $c$ is the concentration in mol m $^{-3}$ . It tells you: if I had exactly one mole of electrolyte dissolved in a solution, what would the conductivity of that entire solution be?

\Lambda_m = \frac{\kappa}{c}

For a strong electrolyte like KCl, $\Lambda_m$ decreases slowly as concentration increases. Why? Because at higher concentrations, ions are closer together — they feel each other's electric fields, slow each other down (ion-ion interactions). At infinite dilution ( $c \to 0$ ), ions are so far apart they move independently, and $\Lambda_m$ reaches a maximum value called $\Lambda_m^\infty$ (limiting molar conductivity).

For a weak electrolyte like acetic acid, $\Lambda_m$ rises sharply on dilution. That is because dilution increases the degree of dissociation — more ions are formed from the same number of molecules. At infinite dilution, all molecules are dissociated, and $\Lambda_m^\infty$ is the sum of the individual ion contributions (Kohlrausch's law). …

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Part (a): $E^\circ_{\text{cell}} = 1.10$ V, $n=2$ , so $\Delta G^\circ = -212.3\ \text{kJ mol}^{-1}$ ; fuel cells are efficient and clean. Part (b): better conductors are Ag at 30 °C, 0.1 M $CH_3COOH$ , KCl at 50 °C; a galvanic cell converts chemical→electrical energy spontaneously, an electrolytic cell does the reverse using external power.

Part (a)

(a) $\Delta G^\circ$ from the cell potential

Relation: $\Delta G^\circ = -nFE^\circ_{\text{cell}}$ .

Step 1 — half-reactions: anode (oxidation) $Zn \to Zn^{2+} + 2e^-$ , $E^\circ = -0.76$ V; cathode (reduction) $Cu^{2+} + 2e^- \to Cu$ , $E^\circ = +0.34$ V.

Step 2 — cell potential:

E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = (+0.34) - (-0.76) = +1.10\ \text{V}

Step 3 — electrons transferred: $n = 2$ .

Step 4 — Gibbs energy (using $1\ \text{V} = 1\ \text{J C}^{-1}$ ):

\Delta G^\circ = -(2)(96500\ \text{C mol}^{-1})(1.10\ \text{V}) = -212300\ \text{J mol}^{-1} = -212.3\ \text{kJ mol}^{-1}

The negative value confirms the reaction is spontaneous.

(b) Two advantages of fuel cells

High efficiency — chemical energy is converted directly to electrical energy without a heat-engine (Carnot) limit, reaching ~40–70%. …

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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 29 on this concept.

CBSE 2026Set ANNUAL1 mark
$Q. Conductivity of electrolytic solutions ______ with increase of temperature (fill in the blank).$
›Reveal solutionSolution

Raising the temperature increases ionic mobility (lower viscosity, faster ion movement) and, for weak electrolytes, increases the degree of dissociation, so conductivity goes up.

…

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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 mark
$Q. Write True or False: Conductivity of a solution increases with dilution.$
›Reveal solutionSolution

False - conductivity decreases on dilution (molar conductivity increases).

Conductivity (specific conductance, kappa) is the conductance of ions present in 1 cm3 of solution. On dilution the total number of ions per unit volume decreases, so kappa decreases.

…

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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
$Q. Why does the conductivity of a solution decrease with dilution?$
›Reveal solutionSolution

Conductivity (κ) is conductance per unit volume; diluting spreads the same ions over more volume, so ion density falls.

Conductivity ( $\kappa$ ) is a measure of the conducting power per unit volume of the solution, i.e. it depends on the number of ions present per unit volume.

On dilution, the total volume of the solution increases while the total number of ions (for a fixed amount of electrolyte) essentially stays the same (or increases only slightly for weak electrolytes due to greater dissociation) — so the concentration of ions per unit volume decreases. Since $\kappa$ is directly proportional to ion concentration, $\kappa$ decreases with d …

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CBSE 2025Set ANNUAL1 markMCQ
$Q. A weak electrolyte:$
(a) Does not dissociate into ions.
(b) Dissociate completely into ions.
(c) Dissociate into ions but not completely.
(d) None of these.
›Reveal solutionSolution
A weak electrolyte partially ionizes in solution (degree of dissociation $\alpha \ll 1$ ), unlike a strong electrolyte which ionizes almost completely.

Electrolytes are substances whose aqueous solutions conduct electricity because they furnish ions. They are classified by how completely they ionize:

Strong electrolytes (e.g. NaCl, HCl, KOH) dissociate almost completely (~100%) into ions in solution. …
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$Zn(s)+Cu^{2+}(aq) \rightarrow Zn^{2+}(aq)+Cu(s)$
›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 2025Set ANNUAL1 markMCQ
$Q. Which of the following statements is true regarding the conductivity of electrolytic solutions ?$
(a) It is independent of the size of the ions
(b) It is independent of the viscosity of the solution
(c) It decreases with temperature
(d) It depends on the solvation of ions present in the solution
›Reveal solutionSolution

Ionic conductivity depends on ion size (and its solvated/hydrated size), charge, and how easily ions move through the solvent — all linked to solvation.

Conductivity of an electrolytic solution is governed by the mobility of the ions, which in turn depends on: the size of the bare ion, the extent of solvation (a heavily solvated ion behaves as a larger, slower-moving species), the viscosity of the solvent (higher viscosity hinders ion movement), and temperature (conductivity of electrolytic solutions INCREASES with temperature, because increased thermal motion decreases the extent of ion solvation and sol …

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CBSE 2025Set ANNUAL1 markMCQ
$Q. Which of the following will have the highest conductivity at 298 K ?$
(i) 0.01 M HCl solution
(ii) 0.1 M HCl solution
(iii) 0.01 M CH₃COOH solution
(iv) 0.1 M CH₃COOH solution
›Reveal solutionSolution
Conductivity (unlike molar conductivity) increases with both the concentration of ions AND their degree of dissociation; a more concentrated strong electrolyte therefore conducts more than a dilute one or a weak electrolyte.

Conductivity ( $\kappa$ ) depends on the number of current-carrying ions per unit volume of solution — so, unlike molar conductivity, it generally increases with concentration for a given electrolyte, provided the electrolyte is fully (or mostly) dissociated.

HCl is a strong acid — fully dissociated into H⁺ and Cl⁻ at both 0.01 M and 0.1 M.

CH₃COOH is a weak acid — only partially dissociated, so it has far fewer ions per unit volume than HCl at the same concentration. …
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CBSE 2024Set ANNUAL1 markMCQ
$Q. Conductivity of an Electrolytic solution depends on -$
(a) Nature of electrolyte
(b) Power of AC Source
(c) Distance between the electrodes
(d) Surface area of electrodes
›Reveal solutionSolution

Conductivity is an intrinsic property of the electrolyte itself, not of the cell's geometry.

Conductivity (specific conductance, $\kappa$ ) of an electrolytic solution depends on the nature of the electrolyte — i.e. its degree of dissociation/ionisation, the number, charge and mobility of the ions it produces, and the concentration of the solution. Options (ii)-(iv) (power of AC source, distance between electrodes, surface area of electrodes) instead af …

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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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Part (a)

(a) ΔG∘\Delta G^\circΔG∘ from the cell potential

(b) Two advantages of fuel cells

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(a) $\Delta G^\circ$ from the cell potential

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