Match List-I with List-II:

List-I (Applications)	List-II (Batteries/Cell)
(A) Transistors	(I) Anode - Zn/Hg; Cathode - HgO + C
(B) Hearing aids	(II) Hydrogen fuel cell
(C) Invertors	(III) Anode - Zn; Cathode - Carbon
(D) Apollo space ship	(IV) Anode - Pb; Cathode - Pb | PbO_2$PbO_2$

Choose the correct answer from the options given below:

### Related Formula E_textcell = E_textRed(textcathode) + E_textOxi(textanode) > 0$$E_{\text{cell}} = E_{\text{Red}}(\text{cathode}) + E_{\text{Oxi}}(\text{anode}) > 0$$ E_textOxi(textanode) = -E^0_textO_2/textH_2textO - frac0.0592 log left([textH^+]^2 P_textO_2^1/2right)$$E_{\text{Oxi}}(\text{anode}) = -E^0_{\text{O}_2/\text{H}_2\text{O}} - \frac{0.059}{2} \log \left([\text{H}^+]^2 P_{\text{O}_2}^{1/2}\right)$$ ### Core Logic Step 1: Set spontaneous cell reaction threshold (E_textcell = 0$E_{\text{cell}} = 0$): E_textOxi(textanode) = -E_textRed(textcathode) = -0.994text V$$E_{\text{Oxi}}(\text{anode}) = -E_{\text{Red}}(\text{cathode}) = -0.994\text{ V}$$ Step 2: Anodic reaction: textH_2textO rightarrow 2textH^+ + frac12textO_2 + 2e^-$$\text{H}_2\text{O} \rightarrow 2\text{H}^+ + \frac{1}{2}\text{O}_2 + 2e^-$$ E_textOxi = -1.23 - frac0.0592 log left([textH^+]^2 times 1^1/2right) = -1.23 + 0.059 times textpH$$E_{\text{Oxi}} = -1.23 - \frac{0.059}{2} \log \left([\text{H}^+]^2 \times 1^{1/2}\right) = -1.23 + 0.059 \times \text{pH}$$ Step 3: Equate oxidation potential: -0.994 = -1.23 + 0.059 times textpH$$-0.994 = -1.23 + 0.059 \times \text{pH}$$ 0.059 times textpH = 0.236$$0.059 \times \text{pH} = 0.236$$ textpH = frac0.2360.059 = 4$$\text{pH} = \frac{0.236}{0.059} = 4$$ ### Pattern Recognition Sees: Minimum pH for gas evolution at anode. Shortcut: Apply E_textcell = 0$E_{\text{cell}} = 0$ threshold equation -0.994 = -1.23 + 0.059 textpH implies textpH = 4$-0.994 = -1.23 + 0.059 \text{pH} \implies \text{pH} = 4$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Electrochemistry

### Related Formula E_textcell = E^circ_textcell - frac0.059n log Q$$E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.059}{n} \log Q$$ ### Core Logic First, write the complete balanced cell reaction. At Anode (Oxidation): Ag_(s) + Cl^-_(aq) rightarrow AgCl_(s) + e^-$Ag_{(s)} + Cl^-_{(aq)} \rightarrow AgCl_{(s)} + e^-$ At Cathode (Reduction): Fe^3+_(aq) + e^- rightarrow Fe^2+_(aq)$Fe^{3+}_{(aq)} + e^- \rightarrow Fe^{2+}_{(aq)}$ Overall cell reaction: Ag_(s) + Cl^-_(aq) + Fe^3+_(aq) rightarrow AgCl_(s) + Fe^2+_(aq)$$Ag_{(s)} + Cl^-_{(aq)} + Fe^{3+}_{(aq)} \rightarrow AgCl_{(s)} + Fe^{2+}_{(aq)}$$ ### Step 1: Applying the Nernst Equation Applying the Nernst equation at 298 text K$298 \text{ K}$ with n = 1$n = 1$: E_textcell = E^circ_textcell - frac0.0591 log frac[Fe^2+][Cl^-][Fe^3+]$$E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.059}{1} \log \frac{[Fe^{2+}]}{[Cl^-][Fe^{3+}]}$$ ### Step 2: Analysis of Variables To increase E_textcell$E_{\text{cell}}$, the value of the logarithmic term log frac[Fe^2+][Cl^-][Fe^3+]$\log \frac{[Fe^{2+}]}{[Cl^-][Fe^{3+}]}$ must decrease. This happens if the numerator decreases or the denominator increases. - Decreasing [Fe^2+]$[Fe^{2+}]$ (Statement C) - Increasing [Fe^3+]$[Fe^{3+}]$ (Statement D) - Increasing [Cl^-]$[Cl^-]$ (Statement E) Thus, statements C, D, and E will increase the cell potential. ### Pattern Recognition Nernst equation trick: E_textcell$E_{\text{cell}}$ increases when product concentrations decrease or reactant concentrations increase (Le Chatelier's perspective of pushing the forward reaction). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Electrochemistry

### Related Formula E_textcell = E^circ_textcell - frac0.0591n log Q$$E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.0591}{n} \log Q$$ For a concentration cell, E^circ_textcell = 0$E^{\circ}_{\text{cell}} = 0$, so: E_textcell = - frac0.0591n log left( frac[textAnode][textCathode] right)$$E_{\text{cell}} = - \frac{0.0591}{n} \log \left( \frac{[\text{Anode}]}{[\text{Cathode}]} \right)$$ ### Core Logic For a concentration cell to have a positive cell potential (E_textcell > 0$E_{\text{cell}} > 0$), the ratio frac[textAnode][textCathode]$\frac{[\text{Anode}]}{[\text{Cathode}]}$ must be less than 1$1$. This implies that [textAnode] < [textCathode]$[\text{Anode}] < [\text{Cathode}]$. Let's evaluate the given conditions: **Case 1: If c_1$c_1$ is at the anode.** Then c_2$c_2$ is at the cathode. Cell reaction: M(s) + M^+(c_2) rightarrow M(s) + M^+(c_1)$M(s) + M^{+}(c_2) \rightarrow M(s) + M^{+}(c_1)$ E_textcell = -0.059 log fracc_1c_2$$E_{\text{cell}} = -0.059 \log \frac{c_1}{c_2}$$ For E_textcell > 0$E_{\text{cell}} > 0$, we need c_1 < c_2$c_1 < c_2$. Option 1 says c_1 = c_2$c_1 = c_2$ (Incorrect). Option 4 says c_1 > c_2$c_1 > c_2$ (Incorrect). ### Step 1: Check Cathode Conditions **Case 2: If c_1$c_1$ is at the cathode.** Then c_2$c_2$ is at the anode. Cell reaction: M(s) + M^+(c_1) rightarrow M(s) + M^+(c_2)$M(s) + M^{+}(c_1) \rightarrow M(s) + M^{+}(c_2)$ E_textcell = -0.059 log fracc_2c_1$$E_{\text{cell}} = -0.059 \log \frac{c_2}{c_1}$$ For E_textcell > 0$E_{\text{cell}} > 0$, we need fracc_2c_1 < 1 implies c_2 < c_1 implies c_1 > c_2$\frac{c_2}{c_1} < 1 \implies c_2 < c_1 \implies c_1 > c_2$. Option 2 says c_1 < c_2$c_1 < c_2$ (Incorrect). Option 3 says c_1 > c_2$c_1 > c_2$ (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. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Electrochemistry

### Related Formula textEquivalents of metal deposited = textEquivalents of gas evolved (in Phase 1)$$\text{Equivalents of metal deposited} = \text{Equivalents of gas evolved (in Phase 1)}$$ n_e^- = fracQF = fracI times t96500$$n_{e^-} = \frac{Q}{F} = \frac{I \times t}{96500}$$ ### Core Logic Phase 1: Deposition of 300 text 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 = fracWE = frac300 times 10^-3frac63.542$Cu = \frac{W}{E} = \frac{300 \times 10^{-3}}{\frac{63.54}{2}}$ Equivalents of O_2 = n_O_2 times 4$O_2 = n_{O_2} \times 4$ frac300 times 10^-3 times 263.54 = n_O_2 times 4$\frac{300 \times 10^{-3} \times 2}{63.54} = n_{O_2} \times 4$ 2.36 times 10^-3 = n_O_2$2.36 \times 10^{-3} = n_{O_2}$ (moles of O_2$O_2$ in Phase 1) Phase 2: Current continued. I = 600 text mA = 0.6 text A$I = 600 \text{ mA} = 0.6 \text{ A}$, t = 28 text mins = 28 times 60 text seconds$t = 28 \text{ mins} = 28 \times 60 \text{ seconds}$. Moles of electrons passed = frac0.6 times 28 times 6096500 = 0.010445 text moles of e^-$= \frac{0.6 \times 28 \times 60}{96500} = 0.010445 \text{ moles of } e^{-}$ Equivalents of O_2$O_2$ evolved in Phase 2 = Moles of e^-$e^{-}$ passed = 0.010445$0.010445$ n_O_2 text (Phase 2) = frac0.0104454 = 2.611 times 10^-3 text 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 O_2$O_2$ evolved = (2.36 times 10^-3) + (2.611 times 10^-3) = 4.971 times 10^-3 text mol$= (2.36 \times 10^{-3}) + (2.611 \times 10^{-3}) = 4.971 \times 10^{-3} \text{ mol}$ Total volume at STP: V_O_2 = n_total times 22400 text mL$V_{O_2} = n_{total} \times 22400 \text{ mL}$ V_O_2 = 4.971 times 10^-3 times 22400 text mL = 111.35 text mL$V_{O_2} = 4.971 \times 10^{-3} \times 22400 \text{ mL} = 111.35 \text{ mL}$ Rounding to nearest integer implies 111 text 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 O_2$O_2$ evolution from water is 4. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Electrochemistry

### Related Formula alpha = fraclambda_mlambda_m^infty$$\alpha = \frac{\lambda_m}{\lambda_m^{\infty}}$$ K_a simeq C alpha^2 quad (textfor alpha ll 1)$$K_a \simeq C \alpha^2 \quad (\text{for } \alpha \ll 1)$$ ### Core Logic For weak acid HQ: alpha_1 = fraclambda_m(mathrmHQ)lambda_m^infty(mathrmHQ)$\alpha_1 = \frac{\lambda_m(\mathrm{HQ})}{\lambda_m^{\infty}(\mathrm{HQ})}$ K_a(mathrmHQ) = C_1 alpha_1^2 = 0.18 left(fraclambda_m(mathrmHQ)lambda_m^infty(mathrmHQ)right)^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: alpha_2 = fraclambda_m(mathrmHZ)lambda_m^infty(mathrmHZ)$\alpha_2 = \frac{\lambda_m(\mathrm{HZ})}{\lambda_m^{\infty}(\mathrm{HZ})}$ K_a(mathrmHZ) = C_2 alpha_2^2 = 0.02 left(fraclambda_m(mathrmHZ)lambda_m^infty(mathrmHZ)right)^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 lambda_mathrmQ^-^0$\lambda_{\mathrm{Q}^-}^0$ and lambda_mathrmZ^-^0$\lambda_{\mathrm{Z}^-}^0$ are equal, and both have H^+$H^+$ as the cation: lambda_m^infty(mathrmHQ) = lambda_m^infty(mathrmHZ)$\lambda_m^{\infty}(\mathrm{HQ}) = \lambda_m^{\infty}(\mathrm{HZ})$ Take the ratio of their ionization constants: fracK_a(mathrmHQ)K_a(mathrmHZ) = fracC_1C_2 cdot left[ fraclambda_m(mathrmHQ)lambda_m(mathrmHZ) right]^2$$\frac{K_a(\mathrm{HQ})}{K_a(\mathrm{HZ})} = \frac{C_1}{C_2} \cdot \left[ \frac{\lambda_m(\mathrm{HQ})}{\lambda_m(\mathrm{HZ})} \right]^2$$ We are given lambda_m(mathrmHQ) = frac130 lambda_m(mathrmHZ)$\lambda_m(\mathrm{HQ}) = \frac{1}{30} \lambda_m(\mathrm{HZ})$, so the ratio inside the bracket is frac130$\frac{1}{30}$. fracK_a(mathrmHQ)K_a(mathrmHZ) = frac0.180.02 times left(frac130right)^2$$\frac{K_a(\mathrm{HQ})}{K_a(\mathrm{HZ})} = \frac{0.18}{0.02} \times \left(\frac{1}{30}\right)^2$$ fracK_a(mathrmHQ)K_a(mathrmHZ) = 9 times frac1900 = frac1100$$\frac{K_a(\mathrm{HQ})}{K_a(\mathrm{HZ})} = 9 \times \frac{1}{900} = \frac{1}{100}$$ ### Step 2: Logarithmic Difference Taking the negative logarithm on both sides: -logleft(fracK_a(mathrmHQ)K_a(mathrmHZ)right) = -log(10^-2)$$-\log\left(\frac{K_a(\mathrm{HQ})}{K_a(\mathrm{HZ})}\right) = -\log(10^{-2})$$ pK_a(mathrmHQ) - pK_a(mathrmHZ) = 2$$pK_a(\mathrm{HQ}) - pK_a(\mathrm{HZ}) = 2$$ ### Pattern Recognition When lambda_textanion^0$\lambda_{\text{anion}}^0$ is identical for both acids, lambda_m^infty$\lambda_m^\infty$ cancels out entirely in comparative ratios. Use K_a = C cdot (lambda_m / lambda_m^infty)^2$K_a = C \cdot (\lambda_m / \lambda_m^\infty)^2$ directly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Electrochemistry Class 11 Chemistry: Equilibrium