For the given cell: Fe²⁺(aq) + Ag⁺(aq) arrow Fe³⁺(aq) + Ag(s)$$Fe^{2+}(aq) + Ag^{+}(aq) \rightarrow Fe^{3+}(aq) + Ag(s)$$ The standard cell potential of the above reaction is given by: Ag⁺ + e⁻ arrow Ag E⁰ = x V$$Ag^{+} + e^{-} \rightarrow Ag \quad E^{0} = x\text{ V}$$ Fe²⁺ + 2e⁻ arrow Fe E⁰ = y V$$Fe^{2+} + 2e^{-} \rightarrow Fe \quad E^{0} = y\text{ V}$$ Fe³⁺ + 3e⁻ arrow Fe E⁰ = z V$$Fe^{3+} + 3e^{-} \rightarrow Fe \quad E^{0} = z\text{ V}$$

Related Formula

Core Logic

• At the cathode, the metal ion with the highest standard reduction potential (E⁰$E^0$) gets reduced and deposited first. Arranging the given potentials:

Thus, deposition follows the order Ag arrow Hg arrow Cu$\text{Ag} \rightarrow \text{Hg} \rightarrow \text{Cu}$ as voltage is steadily increased, confirming Statement I.

• For Mg²⁺$\text{Mg}^{2+}$, its reduction potential is highly negative (-2.37 V$-2.37\text{ V}$), much lower than that of water (-0.83 V$-0.83\text{ V}$). Consequently, water undergoes reduction at the cathode instead of magnesium:

This results in the evolution of Hydrogen gas at the cathode, not oxygen gas. Oxygen gas is evolved at the anode via water oxidation. Thus, Statement II is incorrect. [cite: 1042, 1044]

Step 1: Conclusion Match

Since Statement I is correct and Statement II is incorrect, we select option (2).

Pattern Recognition

Cathode vs Anode Gas Trap: During the aqueous electrolysis of highly reactive metals (Groups 1, 2, and Al$\text{Al}$), H₂$\text{H}_2$ gas is always discharged at the cathode due to water's easier reduction profile. Oxygen gas (O₂$\text{O}_2$) is an anodic product generated by water oxidation.

Chapter Mix

Class 12 Chemistry: Electrochemistry

Related Formula

Faraday's law of electrolysis equation:

Water ion product relation:

Core Logic

During the electrolysis of brine (NaCl(aq)$NaCl(aq)$), hydroxide ions (OH^-$OH^-$) are generated at the cathode:

Given metrics:

• Final pH = 12 pOH = 14 - 12 = 2$\text{pH} = 12 \implies \text{pOH} = 14 - 12 = 2$
• [OH^-] = 10⁻² M$[OH^-] = 10^{-2}\mathrm{\ M}$
• Volume = 600 mL = 0.6 L$\text{Volume} = 600\mathrm{\ mL} = 0.6\mathrm{\ L}$
• Time = 5 = 300 s$\text{Time} = 5\mathrm{\ \min} = 300\mathrm{\ s}$

Step 1: Calculate Moles of Hydroxide Produced

Find the absolute moles of OH^-$OH^-$ ions generated:

Step 2: Relate to Electrical Current

Since 1$1$ mole of electrons produces 1$1$ mole of OH^-$OH^-$, the moles of charge equals 6 × 10⁻³$6 \times 10^{-3}$. Applying Faraday's equation:

Rounding to the nearest integer gives 2$2$.

Pattern Recognition

Always convert a given pH$\text{pH}$ value into [OH^-]$[OH^-]$ concentration when dealing with cathodic water reduction. Tracking the relationship where 1 e^- ≡ 1 OH^-$1\ e^- \equiv 1\ OH^-$ provides a direct shortcut to link pH$\text{pH}$ changes to current flow.

Chapter Mix

Class 12 Chemistry: Electrochemistry

Related Formula

Core Logic

A higher positive value of standard reduction potential (E$E^{\ominus}$) indicates a stronger tendency to undergo reduction, hence behaving as a stronger oxidising agent. Comparing the given values:

• ESn⁴⁺/Sn²⁺ = +1.15V$\mathrm{E}_{\mathrm{Sn}^{4+}/\mathrm{Sn}^{2+}}^{\ominus} = +1.15\mathrm{V}$
• ETl³⁺/Tl = +1.26V$\mathrm{E}_{\mathrm{T}l^{3+}/\mathrm{T}l}^{\ominus} = +1.26\mathrm{V}$
• EAl³⁺/Al = -1.66V$\mathrm{E}_{\mathrm{Al}^{3+}/\mathrm{Al}}^{\ominus} = -1.66\mathrm{V}$
• EPb⁴⁺/Pb²⁺ = +1.67V$\mathrm{E}_{\mathrm{Pb}^{4+}/\mathrm{Pb}^{2+}}^{\ominus} = +1.67\mathrm{V}$

Since +1.67V$+1.67\mathrm{V}$ is the highest value, Pb⁴⁺$\mathrm{Pb}^{4+}$ possesses the strongest oxidising capacity.

Pattern Recognition

Strongest oxidising agent = Most positive reduction potential. Weakest oxidising agent = Most negative reduction potential.

Chapter Mix

Class 12 Chemistry: Electrochemistry Class 12 Chemistry: p-Block Elements

Related Formula

For weak electrolytes, the degree of dissociation α$\alpha$ increases sharply near infinite dilution according to Ostwald's Dilution Law:

Core Logic

When a weak electrolyte is diluted (concentration C arrow 0$C \rightarrow 0$), its molar conductivity increases steeply. Conversely, when plotted against √(C)$\sqrt{C}$, as concentration increases, the degree of dissociation drops rapidly, causing a sharp decrease in molar conductivity. This matches the curve given below:

Pattern Recognition

Weak electrolyte plots feature a steep asymptotic exponential-like rise towards the y-axis as C arrow 0$C \rightarrow 0$, meaning a sharp decrease occurs with increasing concentration.

Chapter Mix

Class 12 Chemistry: Electrochemistry