Electrochemistry Class XII Solved Numericals

**Que 1. 0.05 M NaOH solutions offered resistance of 31.16 Ω in a conductivity cell at 298 K. If cell constant is 0.367 cm ^{–1}. Find out the molar conductivity of NaOH solution.**

**Ans 1.**Given Molarity = 0.05 M

Cell Constant = 0.367 cm

^{-1}

**Que 2. Consider the following cell reaction:**

**2Fe(s) + O _{2} (g) + 4H^{+} (aq) → 2Fe^{2+} (aq) + 2H_{2}O(l); E^{0} = 1.67V. At [Fe^{2+}] = 10^{–3} M, p(O_{2}) = 0.1 atm and pH = 3. What will be the cell potential at 25^{0}C?**

**Ans 2.**Here n = 4, and [H

^{+}] = 10

^{– 3}(as pH = 3)

Applying Nernst equation

**Que 3. A salt solution of 0.30 N placed in a cell whose electrodes are 1.9 cm apart and 3.6 cm ^{2} in area offers a resistance of 20 Ω. Calculate equivalent conductivity of solution. **

**Ans 3.**Resistance (R) = 20 Ω

Length (l) = 1.9 cm a = 3.6 cm2 , Normality (N) = 0.30 N

Cell constant = l/a = 1.9/3.6 = 0.528 cm

^{-1}

Specific conductance (K) = Conductance × Cell constant = (1/R) × cell constant

= (1/20) × 0.528 = 0.0264 S cm

^{–1}

Electrochemistry Class XII Solved Numericals

**Que 4. The electrode potential E _{(Zn2+/Zn)} of a zinc electrode at 25°C with an aqueous solution of 0.1 M ZnSO_{4} is.**

**Ans 4.**

**Que 5. The molar conductivities of CH _{3}COOH at 25°C at the concentration of 0.1 M and 0.001 M are 5.20 and 49.2 cm^{2} mol^{–1} respectively. Calculate the degree of dissociation of CH_{3}COOH at these concentrations. (Λ^{∞}_{m} (CH_{3}COOH) = 390.7 S cm^{2} mol^{–1}).**

**Ans 5.**

**Que 6. ****If 0.01 M solution of an electrolyte has a resistance of 40 ohms in a cell having a cell constant of 0.4 cm ^{–1}, then find its molar conductance in ohm^{–1} cm^{2} mol^{–1}.**

**Ans 6.**Molarity = 0.01 M

Resistance = 40 ohm

Cell constant l/A = 0.4cm

^{-1}.

Electrochemistry Class XII Solved Numericals

**Que 7. If Λ ^{∞}_{m} of HCl, NaCl, and CH_{3}COONa is 425, 128, and 96 Ω^{–1} cm^{2} mol^{–1} respectively, calculate the value of Λ^{∞}_{m} for acetic acid.**

**Ans 7.**Λ

^{∞}

_{m}CH

_{3}COOH = Λ

^{∞}

_{m}CH

_{3}COONa + Λ

^{∞}

_{m}HCl – Λ

^{∞}

_{m}NaCl

= 96.0 + 425.0 – 128.0 = 393.0 Ω

^{–1}cm

^{2}mol

^{–1}.

**Que 8. **

Electrolyte |
KCl |
KNO_{3} |
HCl |
NaOAc |
NaCl |

Λ^{∞} (Scm^{2} mol^{–1}) |
149.9 |
145 |
426.2 |
91 |
126.5 |

**Calculate ****Λ ^{∞}**

_{HOAc}**using appropriate molar conductance of the electrolytes listed above at infinite dilution in H**

_{2}O at 25°C.**Ans 8.**Λ

^{∞}

_{Hcl}= 426.2 ………………(1)

Λ

^{∞}

_{AcONa }= 91.0 …………………………….(2)

Λ

^{∞}

_{NaCl}= 126.5 ………………………….….(3)

Λ

^{∞}

_{HOAc}= (1) + (2) + (3)

= [426.2 + 91.0 + 126.5] = 390.7 Scm

^{2}mol

^{–1}

**Que 9. ****How many hours does it take to reduce 3 moles of Fe ^{3+} to Fe^{2+} with the 2.0-ampere current? (F= 96500 C mol^{–1}).**

**Ans 9.**Fe

^{3+}+ e

^{– }→ Fe

^{2+}

Electrochemistry Class XII Solved Numericals

**Que 10. A****t 25°C, the molar conductance at infinite dilution for the strong electrolytes NaOH, NaCl, and BaCl _{2} is 248 × 10^{–4}, 126 × 10^{–4} and 280 × 10^{–4} Sm^{2}mol^{–1} respectively. Find **

**Λ**

^{0}

_{m}**Ba(OH)**

_{2}in S m^{2}mol^{–1}.**Ans 10.**

**Que 11****. How many grams of silver can be plated out on a serving tray by electrolysis of a solution containing silver in +1 oxidation state for a period of 8.0 hours at a current of 8.46 amperes? What is the area of the tray if the thickness of the silver plating is 0.00254 cm? The density of silver is 10.5 g/cm ^{3}.**

**Ans 11.**

**Que 12. Find ****the number of coulombs required to reduce 12.3 g of nitrobenzene to aniline.**

**Ans 12. **As per the equation

Electrochemistry Class XII Solved Numericals

**Que 13. ****Calculate the quantity of electricity required to reduce 6.15 g of nitrobenzene to aniline if the current efficiency is 68%. If potential drops across the cell is 7.0 volts. Calculate the energy consumed in the process.**

**Ans 13. **C_{6}H_{5}NO_{2} + 6H^{+} + 6e^{–} → C_{6}H_{5}NH_{2} + 2H_{2}O

** **

**Que 14. ****Calculate the cell potential of the given cell at 25°C. (R = 8.31 J K ^{–1} mol^{–1}, F = 96500 C mol^{–1}). **

**Ni (s)|Ni**

^{2+}(0.01 M) || Cu^{2+}(0.1 M)|Cu(s) Given: E^{0}Cu^{2+}/Cu = + 0.34 V E^{0}(Ni^{2+}/Ni) = – 0.25 V.**Ans 14.**

**Que 15. ****Calculate the cell potential for the cell containing 0.10 M Ag ^{+} and 4.0 M Cu^{2+} at 298 K. Given E^{0}Ag^{+} /Ag = 0.80 V, E^{0}Cu^{2+}/Cu = 0.34 V**

**Ans 15.**Cu (s)|Cu

^{2+}(4.0 M) || Ag

^{+}(0.1 M)|Ag (s) Here n = 2

Electrochemistry Class XII Solved Numericals