Consider the following half-cell reduction reaction: Cr₂O₇²⁻(aq) + 6e^- + 14H^+(aq) 2Cr³⁺(aq) + 7H₂O(l)$$\text{Cr}_2\text{O}_7^{2-}\text{(aq)} + 6e^- + 14\text{H}^+\text{(aq)} \longrightarrow 2\text{Cr}^{3+}\text{(aq)} + 7\text{H}_2\text{O(l)}$$ The process is conducted with a concentration ratio of [Cr³⁺]²[Cr₂O₇²⁻] = 10⁻⁶$\frac{[\text{Cr}^{3+}]^2}{[\text{Cr}_2\text{O}_7^{2-}]} = 10^{-6}$. The specific pH value at which the EMF (E$E$) of this reduction half-cell becomes exactly zero is _________ (as the nearest integer value). Given parameters: E^°Cr₂O₇²⁻/Cr³⁺ = 1.33 V$E^\circ_{\text{Cr}_2\text{O}_7^{2-}/\text{Cr}^{3+}} = 1.33 \text{ V}$ and (2.303RT)/(F) = 0.059 V$\frac{2.303RT}{F} = 0.059 \text{ V}$.

Related Formula

Core Logic

During the charging cycle, the discharge chemical reactions are driven in reverse:

• At Anode: Lead sulfate (PbSO₄$PbSO_4$, where Lead is +2$+2$) is reduced back to metallic lead (Pb$Pb$, oxidation state 0$0$):

• At Cathode: Lead sulfate (PbSO₄$PbSO_4$, where Lead is +2$+2$) is oxidized back into lead dioxide (PbO₂$PbO_2$, where Lead is +4$+4$):

Thus, the values are +2, 0, +2, +4$+2, 0, +2, +4$.

Pattern Recognition

Be careful with wording! Discharging consumes Pb$Pb$ and PbO₂$PbO_2$ to create PbSO₄$PbSO_4$. Charging does the exact opposite, converting PbSO₄$PbSO_4$ (+2$+2$) back into its parent elements (0$0$ and +4$+4$).

Chapter Mix

Class 12 Chemistry: Electrochemistry

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

Core Logic

Using Gibbs free energy changes for individual steps to find the target reduction potential:

• Ag⁺ + e⁻ arrow Ag Δ G₁⁰ = -1Fx$$Ag^{+} + e^{-} \rightarrow Ag \quad \Delta G_1^0 = -1Fx$$
• Fe²⁺ + 2e⁻ arrow Fe Δ G₂⁰ = -2Fy$$Fe^{2+} + 2e^{-} \rightarrow Fe \quad \Delta G_2^0 = -2Fy$$
• Fe³⁺ + 3e⁻ arrow Fe Δ G₃⁰ = -3Fz$$Fe^{3+} + 3e^{-} \rightarrow Fe \quad \Delta G_3^0 = -3Fz$$

For the conversion of Fe²⁺ arrow Fe³⁺ + e⁻$$Fe^{2+} \rightarrow Fe^{3+} + e^{-}$$, we compute the free energy change as:

Thus, E⁰Fe²⁺/Fe³⁺ = 2y - 3z$$E^{0}_{Fe^{2+}/Fe^{3+}} = 2y - 3z$$

Combining with silver reduction:

Step 1: Final Calculation

The overall potential equals x + 2y - 3z$x + 2y - 3z$.

Pattern Recognition

Direct application of Δ G⁰$\Delta G^0$ summation. Remember that standard cell potentials cannot be added directly unless the number of electrons involved is identical.

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