### Related Formula textNumber of equivalents deposited = fracW, E = fracQ, F$\text{Number of equivalents deposited} = \frac{W}, {E} = \frac{Q}, {F}$ textEquivalent Weight (E) = fractextAtomic Mass, ntext-factor$\text{Equivalent Weight } (E) = \frac{\text{Atomic Mass}}, {n\text{-factor}}$ ### Core Logic In the reduction of gold from the tetrachloroaurate(III$III$) complex anion: mathrmAuCl_4^- + 3e^- rightarrow mathrmAu(s) + 4mathrmCl^- implies ntext-factor = 3$\mathrm{AuCl}_4^- + 3e^- \rightarrow \mathrm{Au}(s) + 4\mathrm{Cl}^- \implies n\text{-factor} = 3$ Calculate the equivalent weight (E$E$) of Gold: E = frac197, 3$E = \frac{197}, {3}$ Set up the Faraday equivalence relation to solve for charge (Q$Q$ in Faradays): frac1.314, left(frac197, 3right) = Q$\frac{1.314}, {\left(\frac{197}, {3}\right)} = Q$ ### Step 1: Arithmetic Resolution Q = frac1.314 times 3, 197 = frac3.942, 197 = 0.02text F = 2 times 10^-2text F$Q = \frac{1.314 \times 3}, {197} = \frac{3.942}, {197} = 0.02\text{ F} = 2 \times 10^{-2}\text{ F}$ 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. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Electrochemistry

### Related Formula By Faraday's Second Law of Electrolysis: textEquivalents of Ag = textEquivalents of O_2$\text{Equivalents of Ag} = \text{Equivalents of O}_2$ textEquivalents = fractextMasstextEquivalent Mass = textMoles times ntext-factor$\text{Equivalents} = \frac{\text{Mass}}{\text{Equivalent Mass}} = \text{Moles} \times n\text{-factor}$ ### Step 1: Calculate equivalents using standard metrics Let x$x$ grams of Silver be displaced. Using the older STP molar volume baseline (22.4text L$22.4\text{ L}$ or 22400text mL$22400\text{ mL}$): textMoles of O_2 = frac560022400 = 0.25text moles$\text{Moles of O}_2 = \frac{5600}{22400} = 0.25\text{ moles}$ Since the n$n$-factor of O_2$O_2$ is 4 (2textO^2- rightarrow textO_2 + 4texte^-$2\text{O}^{2-} \rightarrow \text{O}_2 + 4\text{e}^-$): textEquivalents of O_2 = 0.25 times 4 = 1$\text{Equivalents of O}_2 = 0.25 \times 4 = 1$ ### Step 2: Equating equivalents for silver mass textEquivalents of Ag = fracx108 times 1 = 1 implies x = 108text g$\text{Equivalents of Ag} = \frac{x}{108} \times 1 = 1 \implies x = 108\text{ g}$ ### Step 3: Alternative calculation using current STP metric Using modern STP volume metrics (22.7text L$22.7\text{ L}$): fracx times 1108 = frac5.622.7 times 4 implies x approx 106.57text g rightarrow 107text g$\frac{x \times 1}{108} = \frac{5.6}{22.7} \times 4 \implies x \approx 106.57\text{ g} \rightarrow 107\text{ g}$ ### 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

### Core Logic The Standard Reduction Potential (E^circ$E^\circ$) measures a species' tendency to undergo reduction (gain electrons). A higher, more positive E^circ$E^\circ$ value means the species has a stronger tendency to be reduced, which in turn makes it a stronger oxidizing agent. Comparing the given E^circ$E^\circ$ values: mathrmBrO_4^-$\mathrm{BrO}_4^-$: 1.74mathrmV$1.74\mathrm{V}$ mathrmIO_4^-$\mathrm{IO}_4^-$: 1.65mathrmV$1.65\mathrm{V}$ mathrmClO_4^-$\mathrm{ClO}_4^-$: 1.19mathrmV$1.19\mathrm{V}$ The order of oxidizing power follows the magnitude of the reduction potential: mathrmBrO_4^- > mathrmIO_4^- > mathrmClO_4^-$\mathrm{BrO}_4^- > \mathrm{IO}_4^- > \mathrm{ClO}_4^-$ ### Pattern Recognition Higher +ve Standard Reduction Potential (SRP) = Stronger Oxidising Agent. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Electrochemistry Class 12 Chemistry: The p Block Elements

### Core Logic Mn, Ni, and Cd metals are predominantly used in battery industries. - Mn is used in dry cells (Leclanche cell). - Ni and Cd are used in Nickel-Cadmium (Ni-Cd) rechargeable batteries. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Electrochemistry