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

Step 1: Set spontaneous cell reaction threshold (Ecell = 0$E_{\text{cell}} = 0$):

Step 2: Anodic reaction:

Step 3: Equate oxidation potential:

Pattern Recognition

Sees: Minimum pH for gas evolution at anode. Shortcut: Apply Ecell = 0$E_{\text{cell}} = 0$ threshold equation -0.994 = -1.23 + 0.059 pH pH = 4$-0.994 = -1.23 + 0.059 \text{pH} \implies \text{pH} = 4$.

Chapter Mix

Class 12 Chemistry: Electrochemistry

Related Formula

Core Logic

First, write the complete balanced cell reaction. At Anode (Oxidation): Ag(s) + Cl^-(aq) arrow AgCl(s) + e^-$Ag_{(s)} + Cl^-_{(aq)} \rightarrow AgCl_{(s)} + e^-$ At Cathode (Reduction): Fe³⁺(aq) + e^- arrow Fe²⁺(aq)$Fe^{3+}_{(aq)} + e^- \rightarrow Fe^{2+}_{(aq)}$

Overall cell reaction:

Step 1: Applying the Nernst Equation

Applying the Nernst equation at 298 K$298 \text{ K}$ with n = 1$n = 1$:

Step 2: Analysis of Variables

To increase Ecell$E_{\text{cell}}$, the value of the logarithmic term [Fe²⁺][Cl^-][Fe³⁺]$\log \frac{[Fe^{2+}]}{[Cl^-][Fe^{3+}]}$ must decrease. This happens if the numerator decreases or the denominator increases.

• Decreasing [Fe²⁺]$[Fe^{2+}]$ (Statement C)
• Increasing [Fe³⁺]$[Fe^{3+}]$ (Statement D)
• Increasing [Cl^-]$[Cl^-]$ (Statement E)

Thus, statements C, D, and E will increase the cell potential.

Pattern Recognition

Nernst equation trick: Ecell$E_{\text{cell}}$ increases when product concentrations decrease or reactant concentrations increase (Le Chatelier's perspective of pushing the forward reaction).

Chapter Mix

Class 12 Chemistry: Electrochemistry

Related Formula

For a concentration cell, E°cell = 0$E^{\circ}_{\text{cell}} = 0$, so:

Core Logic

For a concentration cell to have a positive cell potential (Ecell > 0$E_{\text{cell}} > 0$), the ratio [Anode][Cathode]$\frac{[\text{Anode}]}{[\text{Cathode}]}$ must be less than 1$1$. This implies that [Anode] < [Cathode]$[\text{Anode}] < [\text{Cathode}]$.

Let's evaluate the given conditions:

Case 1: If c₁$c_1$ is at the anode. Then c₂$c_2$ is at the cathode. Cell reaction: M(s) + M⁺(c₂) arrow M(s) + M⁺(c₁)$M(s) + M^{+}(c_2) \rightarrow M(s) + M^{+}(c_1)$

For Ecell > 0$E_{\text{cell}} > 0$, we need c₁ < c₂$c_1 < c_2$. Option 1 says c₁ = c₂$c_1 = c_2$ (Incorrect). Option 4 says c₁ > c₂$c_1 > c_2$ (Incorrect).

Step 1: Check Cathode Conditions

Case 2: If c₁$c_1$ is at the cathode. Then c₂$c_2$ is at the anode. Cell reaction: M(s) + M⁺(c₁) arrow M(s) + M⁺(c₂)$M(s) + M^{+}(c_1) \rightarrow M(s) + M^{+}(c_2)$

For Ecell > 0$E_{\text{cell}} > 0$, we need (c₂)/(c₁) < 1 c₂ < c₁ c₁ > c₂$\frac{c_2}{c_1} < 1 \implies c_2 < c_1 \implies c_1 > c_2$. Option 2 says c₁ < c₂$c_1 < c_2$ (Incorrect). Option 3 says c₁ > c₂$c_1 > c_2$ (Correct).

Pattern Recognition

In any spontaneous concentration cell, ions flow from the higher concentration compartment to the lower concentration compartment. Thus, for a positive voltage, the cathode must always have the higher concentration.

Chapter Mix

Class 12 Chemistry: Electrochemistry

Related Formula

Core Logic

Phase 1: Deposition of 300 mg$300 \text{ mg}$ Cu. By Faraday's Laws, the equivalents of copper deposited at the cathode must equal the equivalents of oxygen evolved at the anode during this period. Equivalents of Cu = (W)/(E) = 300 × 10⁻³(63.54)/(2)$Cu = \frac{W}{E} = \frac{300 \times 10^{-3}}{\frac{63.54}{2}}$ Equivalents of O₂ = nO₂ × 4$O_2 = n_{O_2} \times 4$ 300 × 10⁻³ × 263.54 = nO₂ × 4$\frac{300 \times 10^{-3} \times 2}{63.54} = n_{O_2} \times 4$ 2.36 × 10⁻³ = nO₂$2.36 \times 10^{-3} = n_{O_2}$ (moles of O₂$O_2$ in Phase 1)

Phase 2: Current continued. I = 600 mA = 0.6 A$I = 600 \text{ mA} = 0.6 \text{ A}$, t = 28 mins = 28 × 60 seconds$t = 28 \text{ mins} = 28 \times 60 \text{ seconds}$. Moles of electrons passed = (0.6 × 28 × 60)/(96500) = 0.010445 moles of e⁻$= \frac{0.6 \times 28 \times 60}{96500} = 0.010445 \text{ moles of } e^{-}$ Equivalents of O₂$O_2$ evolved in Phase 2 = Moles of e⁻$e^{-}$ passed = 0.010445$0.010445$ nO₂ (Phase 2) = (0.010445)/(4) = 2.611 × 10⁻³ mol$n_{O_2} \text{ (Phase 2)} = \frac{0.010445}{4} = 2.611 \times 10^{-3} \text{ mol}$

Step 1: Calculate Total Volume of Oxygen

Total moles of O₂$O_2$ evolved = (2.36 × 10⁻³) + (2.611 × 10⁻³) = 4.971 × 10⁻³ mol$= (2.36 \times 10^{-3}) + (2.611 \times 10^{-3}) = 4.971 \times 10^{-3} \text{ mol}$

Total volume at STP: VO₂ = ntotal × 22400 mL$V_{O_2} = n_{total} \times 22400 \text{ mL}$ VO₂ = 4.971 × 10⁻³ × 22400 mL = 111.35 mL$V_{O_2} = 4.971 \times 10^{-3} \times 22400 \text{ mL} = 111.35 \text{ mL}$ Rounding to nearest integer 111 mL$\implies 111 \text{ mL}$.

Pattern Recognition

Equivalents of products at cathode and anode are always equal in any given time span. Valency factor for O₂$O_2$ evolution from water is 4.

Chapter Mix

Class 12 Chemistry: Electrochemistry

Related Formula

Core Logic

For weak acid HQ: α₁ = λm(HQ)λm∞(HQ)$\alpha_1 = \frac{\lambda_m(\mathrm{HQ})}{\lambda_m^{\infty}(\mathrm{HQ})}$ Kₐ(HQ) = C₁ α₁² = 0.18 ( λm(HQ)λm∞(HQ))²$K_a(\mathrm{HQ}) = C_1 \alpha_1^2 = 0.18 \left(\frac{\lambda_m(\mathrm{HQ})}{\lambda_m^{\infty}(\mathrm{HQ})}\right)^2$

For weak acid HZ: α₂ = λm(HZ)λm∞(HZ)$\alpha_2 = \frac{\lambda_m(\mathrm{HZ})}{\lambda_m^{\infty}(\mathrm{HZ})}$ Kₐ(HZ) = C₂ α₂² = 0.02 ( λm(HZ)λm∞(HZ))²$K_a(\mathrm{HZ}) = C_2 \alpha_2^2 = 0.02 \left(\frac{\lambda_m(\mathrm{HZ})}{\lambda_m^{\infty}(\mathrm{HZ})}\right)^2$

Step 1: Evaluate Ratios

Since limiting molar conductivities of anions λQ^-⁰$\lambda_{\mathrm{Q}^-}^0$ and λZ^-⁰$\lambda_{\mathrm{Z}^-}^0$ are equal, and both have H^+$H^+$ as the cation: λm∞(HQ) = λm∞(HZ)$\lambda_m^{\infty}(\mathrm{HQ}) = \lambda_m^{\infty}(\mathrm{HZ})$

Take the ratio of their ionization constants:

We are given λm(HQ) = (1)/(30) λm(HZ)$\lambda_m(\mathrm{HQ}) = \frac{1}{30} \lambda_m(\mathrm{HZ})$, so the ratio inside the bracket is (1)/(30)$\frac{1}{30}$.

Step 2: Logarithmic Difference

Taking the negative logarithm on both sides:

Pattern Recognition

When λₐₙᵢₒₙ⁰$\lambda_{\text{anion}}^0$ is identical for both acids, λm^∞$\lambda_m^\infty$ cancels out entirely in comparative ratios. Use Kₐ = C · (λm / λm^∞)²$K_a = C \cdot (\lambda_m / \lambda_m^\infty)^2$ directly.

Chapter Mix

Class 12 Chemistry: Electrochemistry Class 11 Chemistry: Equilibrium