Consider the following electrochemical cell: textPt mid textO_2(textg) (1text bar) mid textHCl(textaq) midmid textM^2+(textaq, 1.0text M) mid textM(texts)$$\text{Pt} \mid \text{O}_2(\text{g}) (1\text{ bar}) \mid \text{HCl}(\text{aq}) \mid\mid \text{M}^{2+}(\text{aq}, 1.0\text{ M}) \mid \text{M}(\text{s})$$ The pH above which, oxygen gas would start to evolve at anode is ____ (nearest integer). Given: E^0_textM^2+/textM = 0.994text V$E^0_{\text{M}^{2+}/\text{M}} = 0.994\text{ V}$, E^0_textO_2/textH_2textO = 1.23text V$E^0_{\text{O}_2/\text{H}_2\text{O}} = 1.23\text{ V}$ standard reduction potential and fracRTF(2.303) = 0.059text V$\frac{RT}{F}(2.303) = 0.059\text{ V}$ at the given condition

### Related Formula kappa = G times fraclA$$\kappa = G \times \frac{l}{A}$$ Lambda_m = frackappa times 1000C$$\Lambda_m = \frac{\kappa \times 1000}{C}$$ alpha = fracLambda_mLambda_m^infty$$\alpha = \frac{\Lambda_m}{\Lambda_m^{\infty}}$$ [H^+] = Calpha$[H^+] = C\alpha$ ### Core Logic Given: pH = 5$pH = 5$, so [H^+] = 10^-5text M$[H^+] = 10^{-5}\text{ M}$. Since [H^+] = C cdot alpha = C cdot fracLambda_mLambda_m^infty$[H^+] = C \cdot \alpha = C \cdot \frac{\Lambda_m}{\Lambda_m^{\infty}}$, we have 10^-5 = C cdot fracLambda_mLambda_m^infty$10^{-5} = C \cdot \frac{\Lambda_m}{\Lambda_m^{\infty}}$. First, calculate conductivity (kappa$\kappa$): Conductance G = 4 times 10^-5text S$G = 4 \times 10^{-5}\text{ S}$. Cell constant G^* = fraclA = frac15text cm1text cm^2 = 15text cm^-1$G^* = \frac{l}{A} = \frac{15\text{ cm}}{1\text{ cm}^2} = 15\text{ cm}^{-1}$. kappa = G cdot G^* = (4 times 10^-5) times 15 = 6 times 10^-4text S cm^-1$$\kappa = G \cdot G^* = (4 \times 10^{-5}) \times 15 = 6 \times 10^{-4}\text{ S cm}^{-1}$$ Molar conductivity (Lambda_m$\Lambda_m$): Lambda_m = frackappa times 1000C = frac6 times 10^-4 times 1000C = frac0.6C$$\Lambda_m = \frac{\kappa \times 1000}{C} = \frac{6 \times 10^{-4} \times 1000}{C} = \frac{0.6}{C}$$ Substitute Lambda_m$\Lambda_m$ into the proton concentration formula: [H^+] = 10^-5 = C cdot frac0.6 / CLambda_m^infty$$[H^+] = 10^{-5} = C \cdot \frac{0.6 / C}{\Lambda_m^{\infty}}$$ 10^-5 = frac0.6Lambda_m^infty$$10^{-5} = \frac{0.6}{\Lambda_m^{\infty}}$$ Lambda_m^infty = frac0.610^-5 = 60000text S cm^2text mol^-1$$\Lambda_m^{\infty} = \frac{0.6}{10^{-5}} = 60000\text{ S cm}^2\text{ mol}^{-1}$$ Convert units to textS m^2text mol^-1$\text{S m}^2\text{ mol}^{-1}$: Since 1text m^2 = 10^4text cm^2$1\text{ m}^2 = 10^4\text{ cm}^2$, Lambda_m^infty = 60000 times 10^-4text S m^2text mol^-1 = 6text S m^2text mol^-1$$\Lambda_m^{\infty} = 60000 \times 10^{-4}\text{ S m}^2\text{ mol}^{-1} = 6\text{ S m}^2\text{ mol}^{-1}$$ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Electrochemistry Class 11 Chemistry: Equilibrium

### Core Logic For a closed circuit Daniell cell operating under given conditions, standard cell potential E_textcell^circ$E_{\text{cell}}^{\circ}$ remains constant with time during discharge until equilibrium or noticeable concentration shifts occur in a specific manner, or cell potential versus time plots reflect constancy of standard electromotive force terms before polarization effects take over. ### Step 1: Final Conclusion Plot 2 accurately depicts the expected trend where E_textcell^circ$E_{\text{cell}}^{\circ}$ remains constant with time. ### Pattern Recognition Sees: graphical representation of electrochemical cell potentials over time. Trap: Confusing standard cell potential E^circ$E^{\circ}$ with operating cell potential E_textcell$E_{\text{cell}}$. ### Chapter Mix Class 12 Chemistry: Electrochemistry

### Related Formula E_textX^-/textMX/textM^circ = E_textM^+/textM^circ + frac0.059n log K_textsp$$E_{\text{X}^-/\text{MX}/\text{M}}^{\circ} = E_{\text{M}^+/\text{M}}^{\circ} + \frac{0.059}{n} \log K_{\text{sp}}$$r ### Core Logic Substituting the given values: E_textX^-/textMX/textM^circ = 0.79 + frac0.0591 log(10^-10)$$E_{\text{X}^-/\text{MX}/\text{M}}^{\circ} = 0.79 + \frac{0.059}{1} \log(10^{-10})$$ E_textX^-/textMX/textM^circ = 0.79 + 0.059(-10) = 0.79 - 0.59 = 0.20 text V$$E_{\text{X}^-/\text{MX}/\text{M}}^{\circ} = 0.79 + 0.059(-10) = 0.79 - 0.59 = 0.20 \text{ V}$$ ### Step 1: Converting to Millivolts 0.20 text V = 200 text mV$$0.20 \text{ V} = 200 \text{ mV}$$ ### Pattern Recognition Sees: relationship between standard reduction potential, solubility product, and metal-insoluble salt electrodes. Trap: Sign or logarithmic base calculation errors. ### Chapter Mix Class 12 Chemistry: Electrochemistry

### Related Formula E_cell = E_cell^circ - frac0.0591n log Q$$E_{cell} = E_{cell}^{\circ} - \frac{0.0591}{n} \log Q$$ (Using 0.06 for approximation if needed, as per standard practice unless specified). ### Core Logic First, calculate E^circ_cell$E^{\circ}_{cell}$: E^circ_cell = E^circ_cathode - E^circ_anode$$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$$ E^circ_cell = (-0.44text V) - (-0.90text V) = +0.46text V$$E^{\circ}_{cell} = (-0.44\text{ V}) - (-0.90\text{ V}) = +0.46\text{ V}$$ Identify the number of transferred electrons (n$n$) for the overall reaction. Anode reaction: Sn^2+ rightarrow Sn^4+ + 2e^-$Sn^{2+} \rightarrow Sn^{4+} + 2e^{-}$ (Tin goes from +2 in HSnO_2^-$HSnO_2^-$ to +4 in Sn(OH)_6^2-$Sn(OH)_6^{2-}$). Cathode reaction: Bi^3+ + 3e^- rightarrow Bi$Bi^{3+} + 3e^{-} \rightarrow Bi$ (Bismuth goes from +3 in Bi_2O_3$Bi_2O_3$ to 0). To balance the electrons, we multiply anode by 3 and cathode by 2, yielding n = 6$n = 6$ total electrons transferred. ### Step 1: Apply Nernst Equation Applying the Nernst equation (approximating 0.0591 to 0.06 for simplicity): E_cell = E^circ_cell - frac0.06n log_10 Q$$E_{cell} = E^{\circ}_{cell} - \frac{0.06}{n} \log_{10} Q$$ E_cell = 0.46 - frac0.066 log_10(10^6)$$E_{cell} = 0.46 - \frac{0.06}{6} \log_{10}(10^6)$$ E_cell = 0.46 - 0.01 times 6$$E_{cell} = 0.46 - 0.01 \times 6$$ E_cell = 0.46 - 0.06 = 0.40text V$$E_{cell} = 0.46 - 0.06 = 0.40\text{ V}$$ ### Step 2: Format Output The problem asks for the answer in the form x times 10^-1text V$x \times 10^{-1}\text{ V}$. 0.40 = 4 times 10^-1$$0.40 = 4 \times 10^{-1}$$ So, x = 4$x = 4$. ### Pattern Recognition Be extremely careful finding 'n'. You must balance the half-reactions. A +2 to +4 change (n=2) and a +3 to 0 change (n=3) resolves to a common multiple of 6. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Electrochemistry

### Related Formula textReducing Power propto frac1textStandard Reduction Potential (E^0_red)$$\text{Reducing Power} \propto \frac{1}{\text{Standard Reduction Potential } (E^0_{red})}$$ ### Core Logic Step 1: List standard reduction potential values: - E^0(textAl^3+/textAl) = -1.66text V$E^0(\text{Al}^{3+}/\text{Al}) = -1.66\text{ V}$ - E^0(textCr^3+/textCr) = -0.74text V$E^0(\text{Cr}^{3+}/\text{Cr}) = -0.74\text{ V}$ - E^0(textFe^3+/textFe^2+) = +0.77text V$E^0(\text{Fe}^{3+}/\text{Fe}^{2+}) = +0.77\text{ V}$ - E^0(textCo^3+/textCo^2+) = +1.81text V$E^0(\text{Co}^{3+}/\text{Co}^{2+}) = +1.81\text{ V}$ Step 2: Arrange species in increasing order of E^0_red$E^0_{red}$ (decreasing reducing power): textAl > textCr > textFe^2+ > textCo^2+$$\text{Al} > \text{Cr} > \text{Fe}^{2+} > \text{Co}^{2+}$$ ### Pattern Recognition Sees: Reduction potential values for reducing strength ordering. Shortcut: More negative standard reduction potential implies$\implies$ stronger reducing agent. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Electrochemistry