Given below is the plot of the molar conductivity vs concentration$\sqrt{\text{concentration}}$ for KCl in aqueous solution.

If, for the higher concentration of KCl solution, the resistance of the conductivity cell is 100Ω$100\Omega$ then the resistance of the same cell with the dilute solution is xΩ$\mathrm{x}\Omega$ The value of x$\mathbf{x}$ is (Nearest integer)

Core Logic

To find 'n', the number of moles of electrons transferred, we must balance the overall redox equation and calculate the net electron exchange.

Step 1: Assign Oxidation States

In BH₄^-$BH_4^-$, Hydrogen is -1$-1$ (hydride), so Boron is +3$+3$. In H₂BO₃^-$H_2BO_3^-$, Boron is +3$+3$. The oxidation state of B doesn't change, but H$H$ oxidizes from -1$-1$ to +1$+1$ (in water/acid).\nAlternatively, treat the whole moiety:\nOxidation half-reaction: BH₄^- + 3H₂O arrow H₂BO₃^- + 8H^+ + 8e^-$BH_4^- + 3H_2O \rightarrow H_2BO_3^- + 8H^+ + 8e^-$\nIn ClO₃^-$ClO_3^-$, Chlorine is +5$+5$. In Cl^-$Cl^-$, Chlorine is -1$-1$.\nReduction half-reaction: ClO₃^- + 6H^+ + 6e^- arrow Cl^- + 3H₂O$ClO_3^- + 6H^+ + 6e^- \rightarrow Cl^- + 3H_2O$

Step 2: Balance Electrons

Multiply the oxidation half-reaction by 3 and the reduction half-reaction by 4 to equalize electrons transferred:\n3 × (BH₄^- + 3H₂O arrow H₂BO₃^- + 8H^+ + 8e^-) 24e^-$3 \times (BH_4^- + 3H_2O \rightarrow H_2BO_3^- + 8H^+ + 8e^-) \implies 24e^-$\n4 × (ClO₃^- + 6H^+ + 6e^- arrow Cl^- + 3H₂O) 24e^-$4 \times (ClO_3^- + 6H^+ + 6e^- \rightarrow Cl^- + 3H_2O) \implies 24e^-$\nThe lowest common multiple of electrons exchanged is 24.

Final Conclusion

The balanced equation transfers 24 electrons, so n = 24$n = 24$ in the Nernst equation.

Pattern Recognition

The total electrons 'n$n$' in the Nernst equation is always the lowest common multiple of electrons from the balanced oxidation and reduction half-reactions.

Chapter Mix

Class 12 Chemistry: Electrochemistry Class 11 Chemistry: Redox Reactions

Related Formula

Λm = Λm° - A√(c)$$\Lambda_{m} = \Lambda_{m}^{\circ} - A\sqrt{c}$$ (Debye-Huckel-Onsager equation)

Core Logic

Using the equation for the first set of values (c = 0.04, √(c) = 0.2$c = 0.04, \sqrt{c} = 0.2$): 96.1 = Λm° - A(0.2)$96.1 = \Lambda_{m}^{\circ} - A(0.2)$ --- (1)

Using the equation for the second set of values (c = 0.09, √(c) = 0.3$c = 0.09, \sqrt{c} = 0.3$): 95.7 = Λm° - A(0.3)$95.7 = \Lambda_{m}^{\circ} - A(0.3)$ --- (2)

Step 1: Solve for A

Subtract eq (2) from eq (1): 96.1 - 95.7 = A(0.3) - A(0.2)$96.1 - 95.7 = A(0.3) - A(0.2)$ 0.4 = 0.1A$0.4 = 0.1A$ A = (0.4)/(0.1) = 4$A = \frac{0.4}{0.1} = 4$

Pattern Recognition

Direct linear interpolation. Change in Λm$\Lambda_{m}$ over change in √(c)$\sqrt{c}$ gives the slope (which is A).

Chapter Mix

Class 12 Chemistry: Electrochemistry

Related Formula

Core Logic

Note: In the question paper, EHSnO₂^-/[Sn(OH)₆]²⁻° = -0.9 V$E_{HSnO_2^-/[Sn(OH)_6]^{2-}}^{\circ} = -0.9\text{ V}$ is given, but standard NTA solution requires assuming E[Sn(OH)₆]²⁻/HSnO₂^-° = -0.9 V$E_{[Sn(OH)_6]^{2-}/HSnO_2^-}^{\circ} = -0.9\text{ V}$ for standard operation. (Our Ans. is Bonus due to this discrepancy, NTA Answer is 78). We will solve using the assumed valid logic.

Ecell° = Ecathode° - Eanode° = -0.44 - (-0.90) = 0.46 V$E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ} = -0.44 - (-0.90) = 0.46\text{ V}$

Oxidation Half: HSnO₂^- + H₂O + 3OH^- arrow [Sn(OH)₆]²⁻ + 2e^-$HSnO_2^- + H_2O + 3OH^- \rightarrow [Sn(OH)_6]^{2-} + 2e^-$ Reduction Half: Bi₂O₃ + 3H₂O + 6e^- arrow 2Bi + 6OH^-$Bi_2O_3 + 3H_2O + 6e^- \rightarrow 2Bi + 6OH^-$ Overall: 3HSnO₂^- + Bi₂O₃ + 6H₂O + 3OH^- arrow 3[Sn(OH)₆]²⁻ + 2Bi$3HSnO_2^- + Bi_2O_3 + 6H_2O + 3OH^- \rightarrow 3[Sn(OH)_6]^{2-} + 2Bi$ Here n = 6$n = 6$.

Nernst Eq: Ecell = Ecell° - (0.059)/(6) ([Sn(OH)₆]²⁻)³([HSnO₂^-]³ [OH^-]³)$E_{cell} = E_{cell}^{\circ} - \frac{0.059}{6} \log \frac{([Sn(OH)_6]^{2-})^3}{([HSnO_2^-]^3 [OH^-]^3)}$ 0.2353 = 0.46 - (0.059)/(6) ((0.5)³)/((0.05)³ [OH^-]³)$0.2353 = 0.46 - \frac{0.059}{6} \log \frac{(0.5)^3}{(0.05)^3 [OH^-]^3}$ 0.2353 = 0.46 - (0.059)/(2) (10)/([OH^-])$0.2353 = 0.46 - \frac{0.059}{2} \log \frac{10}{[OH^-]}$ ( (10)/([OH^-]) ) = ((0.46 - 0.2353) × 2)/(0.059) = (0.2247 × 2)/(0.059) = 7.6$\log \left( \frac{10}{[OH^-]} \right) = \frac{(0.46 - 0.2353) \times 2}{0.059} = \frac{0.2247 \times 2}{0.059} = 7.6$

Step 1: Calculate pH

(10) - [OH^-] = 7.6$\log(10) - \log[OH^-] = 7.6$ 1 + pOH = 7.6 pOH = 6.6$1 + pOH = 7.6 \implies pOH = 6.6$ pH = 14 - 6.6 = 7.4$pH = 14 - 6.6 = 7.4$

Step 2: Buffer Equation

Using Henderson-Hasselbalch equation for buffer of NaHCO₃$NaHCO_3$ and H₂CO₃$H_2CO_3$: pH = pKₐ + nsaltnacid$pH = pK_a + \log \frac{n_{\text{salt}}}{n_{\text{acid}}}$ 7.4 = 6.11 + (5x)/(10 × 2)$7.4 = 6.11 + \log \frac{5x}{10 \times 2}$ 1.29 = (5x)/(20) = (x)/(4)$1.29 = \log \frac{5x}{20} = \log \frac{x}{4}$ Taking antilog: (x)/(4) = Antilog(1.29) = 19.5$\frac{x}{4} = \text{Antilog}(1.29) = 19.5$ x = 19.5 × 4 = 78 mL$x = 19.5 \times 4 = 78\text{ mL}$

Pattern Recognition

Merge Nernst equation finding unknown concentration with Buffer equations. Determine overall cell reaction to find exact stoichiometry and 'n' electrons for Nernst.

Chapter Mix

Class 12 Chemistry: Electrochemistry Class 11 Chemistry: Equilibrium

Related Formula

Core Logic

To compute the molar conductivity, we first calculate the molarity of the solution and the conductivity of the electrolyte from the given resistivity.

Step 1: Calculate Molarity (M)

0.2%$0.2\%$ (w/v) NaOH$\mathrm{NaOH}$ means 0.2~g$0.2~\mathrm{g}$ of NaOH$\mathrm{NaOH}$ is present in 100~mL$100~\mathrm{mL}$ of solution.

Step 2: Calculate Conductivity (kappa) in dm Units

Given resistivity ρ = 870.0~mΩ~m = 870 × 10⁻³~Ω~m = 0.87~Ω~m$\rho = 870.0~\mathrm{m\Omega~m} = 870 \times 10^{-3}~\Omega~m = 0.87~\Omega~m$.

Since 1~m = 10~dm$1~\mathrm{m} = 10~\mathrm{dm}$:

Now, conductivity κ$\kappa$ is:

Step 3: Calculate Molar Conductivity (Lambda_m)

Converting S$\mathrm{S}$ to mS$\mathrm{mS}$ (1~S = 10³~mS$1~\mathrm{S} = 10^3~\mathrm{mS}$):

Rounding off to the nearest integer gives 23.

Pattern Recognition

Ensure careful handling of volumetric conversions. Since concentration is expressed in moles per liter (equivalent to dm⁻³$\mathrm{dm^{-3}}$), expressing conductivity in terms of dm⁻¹$\mathrm{dm^{-1}}$ directly eliminates the need for arbitrary 1000$1000$ multiplication factors.

Chapter Mix

Class 12 Chemistry: Electrochemistry

Related Formula

Nernst equation for the net combined redox cell expression:

Hydrolysis equation for a salt composed of a weak base and strong acid:

Core Logic

Let's compute the operational values line-by-row:

• Combined redox process: QH₂ + 2Ag^+ arrow Q + 2Ag + 2H^+$\mathrm{QH_2 + 2Ag^+ \rightarrow Q + 2Ag + 2H^+}$.
• Standard cell potential difference: E^°cell = E^°Ag^+/Ag - E^°Q/QH₂ = 0.8 - 0.7 = +0.1~V$E^\circ_{\text{cell}} = E^\circ_{\mathrm{Ag^+/Ag}} - E^\circ_{\mathrm{Q/QH_2}} = 0.8 - 0.7 = +0.1\mathrm{~V}$.
• Apply Nernst adjustments using known concentrations ([Ag^+] = 1~M$[\mathrm{Ag}^+] = 1\mathrm{~M}$):

Step 1: Salt Hydrolysis Substitution

Substitute the determined pH$\mathrm{pH}$ along with salt molarity (C = 0.01~M = 10⁻²~M$C = 0.01\mathrm{~M} = 10^{-2}\mathrm{~M}$) into the hydrolysis equation:

Pattern Recognition

Quinhydrone electrodes act as excellent pH indicators in electrochemical cells. Note that each change of 1 pH unit shifts the cell output potential by exactly 0.06~V$0.06\mathrm{~V}$ at standard ambient conditions.

Chapter Mix

Class 12 Chemistry: Electrochemistry Class 11 Chemistry: Equilibrium