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

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

Related Formula

Core Logic

Let us explicitly formulate the complete chemical oxidation-reduction equations : Anode oxidation: Mg(s) arrow Mg²⁺(aq) + 2e^-$\mathrm{Mg}_{(s)} \rightarrow \mathrm{Mg}^{2+}_{(aq)} + 2e^-$ Cathode reduction: 2Ag⁺(aq) + 2e^- arrow 2Ag(s)$2\mathrm{Ag}^{+}_{(aq)} + 2e^- \rightarrow 2\mathrm{Ag}_{(s)}$

Net total equation :

Total transferred moles of electrons n = 2$n = 2$ . Reaction quotient :

Substituting into Nernst form :

Inverting the inside quotient changes the sign of the logarithm term from negative to positive:

Pattern Recognition

A standard negative logarithmic quotient can always toggle into an addition configuration by inverting the products/reactants variables concentration ratio.

Related Formula

For the Standard Hydrogen Electrode half-reaction: 2H^+ + 2e^- arrow H₂$2H^+ + 2e^- \rightarrow H_2$

Step 1: Substitute the given values

E^°H^+/H₂ = 0.00 ~V$E^\circ_{H^+/H_2} = 0.00 \mathrm{~V}$ (by definition) [H^+] = 1 ~M$[H^+] = 1 \mathrm{~M}$ PH₂ = 2 ~atm$P_{H_2} = 2 \mathrm{~atm}$ n = 2$n = 2$ electrons

Step 2: Solve the calculation

E = -0.03 2$E = -0.03 \log 2$ Given 2 = 0.3$\log 2 = 0.3$ E = -0.03 × 0.3$E = -0.03 \times 0.3$ E = -0.009 ~V$E = -0.009 \mathrm{~V}$ E = -0.9 × 10⁻² ~V$E = -0.9 \times 10^{-2} \mathrm{~V}$

Step 3: Match the requested format

The question asks for (-) × 10⁻² ~V$(-) \dots \times 10^{-2} \mathrm{~V}$. This gives exactly 0.9$0.9$. For NAT type with integer expected, 0.9$0.9$ can be rounded to 1$1$. However, exact calculation yields 0.9$0.9$. According to official JEE rounding, 0.9 ≈ 1$0.9 \approx 1$.

Pattern Recognition

Hydrogen electrode non-standard potential depends strictly on pressure of H₂$H_2$ and concentration of H^+$H^+$. If [H^+]=1$[H^+]=1$, increasing H₂$H_2$ pressure lowers the potential below zero (makes it negative).

Chapter Mix

Class 12 Chemistry: Electrochemistry

Related Formula

Core Logic

In the reduction of gold from the tetrachloroaurate(III$III$) complex anion:

Calculate the equivalent weight (E$E$) of Gold:

Set up the Faraday equivalence relation to solve for charge (Q$Q$ in Faradays):

Step 1: Arithmetic Resolution

Thus, the required integer value is 2.

Pattern Recognition

Always determine the correct change in oxidation state (+3$+3$ to 0$0$) to establish the proper n$n$-factor value for calculations using Faraday's laws.

Chapter Mix

Class 12 Chemistry: Electrochemistry

Related Formula

By Faraday's Second Law of Electrolysis:

Step 1: Calculate equivalents using standard metrics

Let x$x$ grams of Silver be displaced. Using the older STP molar volume baseline (22.4 L$22.4\text{ L}$ or 22400 mL$22400\text{ mL}$):

Since the n$n$-factor of O₂$O_2$ is 4 (2O²⁻ arrow O₂ + 4e^-$2\text{O}^{2-} \rightarrow \text{O}_2 + 4\text{e}^-$):

Step 2: Equating equivalents for silver mass

Step 3: Alternative calculation using current STP metric

Using modern STP volume metrics (22.7 L$22.7\text{ L}$):

Pattern Recognition

Equivalents equations bypass complex current/time measurements. Always link volume fractions directly to n$n$-factor equivalents.

Chapter Mix

Class 12 Chemistry: Electrochemistry Class 11 Chemistry: Some Basic Concepts of Chemistry