### Related Formula E_textcell = E^circ_textcell - frac0.06n log Q$E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.06}{n} \log Q$ ### Core Logic First, analyze the electrolysis step to determine final ionic concentrations: 1. **For mathrmCu^2+/mathrmCu$\mathrm{Cu}^{2+}/\mathrm{Cu}$ half-cell**: - Initial moles of mathrmCu^2+ = 1.5 text M times 1 text L = 1.5 text mol$\mathrm{Cu}^{2+} = 1.5 \text{ M} \times 1 \text{ L} = 1.5 \text{ mol}$. - Reductive half-reaction: mathrmCu^2+ + 2mathrme^- rightarrow mathrmCu$\mathrm{Cu}^{2+} + 2\mathrm{e}^- \rightarrow \mathrm{Cu}$. - Passing 1 text Faraday$1 \text{ Faraday}$ converts: frac12 = 0.5 text mol$\frac{1}{2} = 0.5 \text{ mol}$ of mathrmCu^2+$\mathrm{Cu}^{2+}$. - Remaining moles of mathrmCu^2+ = 1.5 - 0.5 = 1.0 text mol$\mathrm{Cu}^{2+} = 1.5 - 0.5 = 1.0 \text{ mol}$. - Final concentration [mathrmCu^2+] = 1.0 text M$[\mathrm{Cu}^{2+}] = 1.0 \text{ M}$. 2. **For mathrmAg^+/mathrmAg$\mathrm{Ag}^{+}/\mathrm{Ag}$ half-cell**: - Initial moles of mathrmAg^+ = 0.2 text M times 1 text L = 0.2 text mol$\mathrm{Ag}^+ = 0.2 \text{ M} \times 1 \text{ L} = 0.2 \text{ mol}$. - Reductive half-reaction: mathrmAg^+ + mathrme^- rightarrow mathrmAg$\mathrm{Ag}^+ + \mathrm{e}^- \rightarrow \mathrm{Ag}$. - Passing 0.1 text Faraday$0.1 \text{ Faraday}$ converts: 0.1 text mol$0.1 \text{ mol}$ of mathrmAg^+$\mathrm{Ag}^+$. - Remaining moles of mathrmAg^+ = 0.2 - 0.1 = 0.1 text mol$\mathrm{Ag}^+ = 0.2 - 0.1 = 0.1 \text{ mol}$. - Final concentration [mathrmAg^+] = 0.1 text M$[\mathrm{Ag}^+] = 0.1 \text{ M}$. Now, connect the two components into a galvanic cell: - Anode reaction: mathrmCu(s) rightarrow mathrmCu^2+mathrm(aq) + 2mathrme^-$\mathrm{Cu(s)} \rightarrow \mathrm{Cu}^{2+}\mathrm{(aq)} + 2\mathrm{e}^-$ - Cathode reaction: 2mathrmAg^+mathrm(aq) + 2mathrme^- rightarrow 2mathrmAg(s)$2\mathrm{Ag}^{+}\mathrm{(aq)} + 2\mathrm{e}^- \rightarrow 2\mathrm{Ag(s)}$ - Net cell reaction: mathrmCu(s) + 2mathrmAg^+mathrm(aq) rightarrow mathrmCu^2+mathrm(aq) + 2mathrmAg(s)$\mathrm{Cu(s)} + 2\mathrm{Ag}^{+}\mathrm{(aq)} \rightarrow \mathrm{Cu}^{2+}\mathrm{(aq)} + 2\mathrm{Ag(s)}$ - n = 2$n = 2$ Calculate standard cell potential: E^circ_textcell = E^circ_mathrmAg^+/mathrmAg - E^circ_mathrmCu^2+/mathrmCu = 0.80 - 0.34 = 0.46 text V$E^{\circ}_{\text{cell}} = E^{\circ}_{\mathrm{Ag}^+/\mathrm{Ag}} - E^{\circ}_{\mathrm{Cu}^{2+}/\mathrm{Cu}} = 0.80 - 0.34 = 0.46 \text{ V}$ Applying Nernst Equation: E_textcell = E^circ_textcell - frac0.062 log left( frac[mathrmCu^2+][mathrmAg^+]^2 right)$E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.06}{2} \log \left( \frac{[\mathrm{Cu}^{2+}]}{[\mathrm{Ag}^+]^2} \right)$ E_textcell = 0.46 - 0.03 log left( frac1(0.1)^2 right) = 0.46 - 0.03 log(100)$E_{\text{cell}} = 0.46 - 0.03 \log \left( \frac{1}{(0.1)^2} \right) = 0.46 - 0.03 \log(100)$ E_textcell = 0.46 - 0.03(2) = 0.46 - 0.06 = 0.40 text V$E_{\text{cell}} = 0.46 - 0.03(2) = 0.46 - 0.06 = 0.40 \text{ V}$ (Note: The potential is 0.4text V$0.4\text{ V}$ or 400text mV$400\text{ mV}$). ### Pattern Recognition Electrolysis modifies the bulk concentrations. First, use Faraday's laws to get the new concentration values ([Cu^2+] = 1.0text M$[Cu^{2+}] = 1.0\text{ M}$, [Ag^+] = 0.1text M$[Ag^+] = 0.1\text{ M}$). Then plug these straight into standard Nernst equations. ### Evaluation Rubric / Model Answer Requires complete calculations showing concentrations updated by electrolysis, followed by a double-transfer Nernst equation calculation. ### Chapter Mix Class 12 Chemistry: Electrochemistry

### Related Formula The Nernst equation for a reduction half-cell is: E = E^circ - frac2.303RTnF log Q$E = E^\circ - \frac{2.303RT}{nF} \log Q$ For this reaction, the reaction quotient Q$Q$ is: Q = frac[textCr^3+]^2[textCr_2textO_7^2-] cdot [textH^+]^14$Q = \frac{[\text{Cr}^{3+}]^2}{[\text{Cr}_2\text{O}_7^{2-}] \cdot [\text{H}^+]^{14}}$ ### Execution Step 1: Identify the number of transferred electrons (n = 6$n = 6$) and substitute the condition E = 0$E = 0$: 0 = 1.33 - frac0.0596 log left( frac10^-6[textH^+]^14 right)$0 = 1.33 - \frac{0.059}{6} \log \left( \frac{10^{-6}}{[\text{H}^+]^{14}} \right)$ Step 2: Isolate the logarithmic term: 1.33 = frac0.0596 left[ log(10^-6) - log([textH^+]^14) right]$1.33 = \frac{0.059}{6} \left[ \log(10^{-6}) - \log([\text{H}^+]^{14}) \right]$ frac1.33 times 60.059 = -6 - 14 log[textH^+]$\frac{1.33 \times 6}{0.059} = -6 - 14 \log[\text{H}^+]$ Step 3: Perform the arithmetic division: 135.254 = -6 - 14 log[textH^+]$135.254 = -6 - 14 \log[\text{H}^+]$ Step 4: Rearrange the terms using the definition of pH (-log[textH^+] = textpH$-\log[\text{H}^+] = \text{pH}$): 135.254 + 6 = 14 cdot textpH$135.254 + 6 = 14 \cdot \text{pH}$ 141.254 = 14 cdot textpH$141.254 = 14 \cdot \text{pH}$ textpH = frac141.25414 = 10.089$\text{pH} = \frac{141.254}{14} = 10.089$ Rounding to the nearest integer value gives **10**. ### Pattern Recognition The exponent of the hydrogen ion concentration ([textH^+]^14$[\text{H}^+]^{14}$) heavily influences the cell potential. A small shift in pH causes a large change in EMF due to this factor of 14, which explains why the potential drops to zero even in a highly basic environment (textpH approx 10$\text{pH} \approx 10$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Electrochemistry Class 11 Chemistry: Ionic Equilibrium

### Core Logic Matching applications to their respective electrochemical cells: * Transistors use standard dry cells: Anode is Zn container, Cathode is carbon rod coated with MnO_2$MnO_2$ ightarrow$ ightarrow$ (III). * Hearing aids require compact voltage outputs over time, matching Mercury cells: Anode Zn/Hg, Cathode HgO + C ightarrow$ ightarrow$ (I). * Invertors utilize rechargeable systems, matching Lead-storage batteries: Anode Pb, Cathode Pb | PbO_2$PbO_2$ ightarrow$ ightarrow$ (IV). * Apollo space ship dynamically powered via Hydrogen-Oxygen Fuel cells ightarrow$ ightarrow$ (II). ### Pattern Recognition Space missions universally trigger fuel cell pairs in standard test patterns due to the secondary requirement of gathering pure drinking water byproduct. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Electrochemistry

### Core Logic Analyzing anodic reactions during electrolysis: * Case (A): With active Ag electrodes, silver oxidation occurs at the anode (Ag ightarrow Ag^+ + e^-$Ag ightarrow Ag^{+} + e^{-}$). No oxygen is evolved. * Case (B): With inert Pt electrodes, oxidation of water occurs preferentially at the anode over NO_3^-$NO_3^-$ ions:2H_2O ightarrow O_2 + 4H^+ + 4e^-$2H_2O ightarrow O_2 + 4H^{+} + 4e^{-}$ * Case (C): In dilute H_2SO_4$H_2SO_4$, water oxidation takes place, releasing O_2$O_2$ gas at the anode. * Case (D): In concentrated H_2SO_4$H_2SO_4$, oxidation of SO_4^2-$SO_4^{2-}$ creates peroxodisulphate ions (S_2O_8^2-$S_2O_8^{2-}$), inhibiting oxygen evolution. ### Pattern Recognition Remember that active electrodes participate directly in redox reactions, whereas inert electrodes (Pt, Graphite) yield oxygen gas when water is oxidized in the presence of oxoanions like NO_3^-$NO_3^-$ or dilute SO_4^2-$SO_4^{2-}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Electrochemistry

### Related Formula Conductivity relationship with cell parameters: kappa = G cdot G^* = fracG^*R$\kappa = G \cdot G^* = \frac{G^*}{R}$ lambda_m = frackappa times 1000C$\lambda_m = \frac{\kappa \times 1000}{C}$ where G^*$G^*$ represents the static cell constant. ### Step 1: Setting Up Ratios Using concentration subscripts c$c$ (concentrated) and d$d$ (dilute): frackappa_ckappa_d = fracR_dR_c$\frac{\kappa_c}{\kappa_d} = \frac{R_d}{R_c}$ Expressing conductivity through molar conductivity values: kappa = fraclambda_m cdot C1000$\kappa = \frac{\lambda_m \cdot C}{1000}$ frac(lambda_m cdot C)_c(lambda_m cdot C)_d = fracR_dR_c$\frac{(\lambda_m \cdot C)_c}{(\lambda_m \cdot C)_d} = \frac{R_d}{R_c}$ Substituting the graphical read coordinates (C_c = 0.15^2, C_d = 0.1^2$C_c = 0.15^2, C_d = 0.1^2$ with scaled lambda_m$\lambda_m$ parameters): frac100 cdot (0.15)^2150 cdot (0.1)^2 = fracR_d100$\frac{100 \cdot (0.15)^2}{150 \cdot (0.1)^2} = \frac{R_d}{100}$ R_d = 150\,Omega$R_d = 150\,\Omega$ ### Pattern Recognition Sees: Resistance correlation across specific graph coordinates. Shortcut: Equate cell parameters through kappa propto frac1R$\kappa \propto \frac{1}{R}$ and solve for the target resistance directly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Electrochemistry