Practicing Success

Target Exam

CUET

Subject

Chemistry

Chapter

Physical: Electro Chemistry

Question:

Electrochemical cell consists of two metallic electrodes (anode and cathode) dipping in electrolytic solutions. At anode, oxidation takes place while at cathode, reduction takes place. These cells are of two types:

1) Galvanic cell in which chemical energy of a spontaneous reaction is converted into electrical energy. Eo(Cells) = EoCathode - Eoanode where Eocell is standard potential of the cells. ΔrGo = -nFEocell where ΔrGo is standard Gibbs energy change.

2) Electrolytic cells in which electrical energy is used to carry out non-spontaneous redox reactions. The amount of substance produced at a particular electrode depends upon quantity of electricity passed, Q = I x t (where I is current in ampere and t-time in seconds) one faraday in the quantity of electricity. It is the charge carried by 1 mole of electrons = 96500 C, If conductivity K of an electrolytic solution depends on concentration of electrolyte, nature of kohlrausch Law of independent migration of ions Λmo(NaCl) = λoNa+ + λoCl

where λom represent limiting molar conductivity. This can be used for calculation of molar conductivity for weak electrolytes.

Λom for NaCl, HCl and CH3COONa are 126.4, 425.9 and 91.0 S cm2 mol-1 respectively. Calculate Λo for CH3COOH.

Options:

516.9 S cm2 mol-1 

653.3 S cm2 mol-1 

390.5 S cm2 mol-1 

460.9 S cm2 mol-1 

Correct Answer:

390.5 S cm2 mol-1 

Explanation:

Λom for NaCl, HCl and CH3COONa are 126.4, 425.9 and 91.0 S cm2 mol-1 respectively.

Λom NaCl = Λom Na+ + Λom Cl- = 126.4 S cm2 mol-1 ...(i)

Λom HCl = Λom H+ + Λom Cl- = 425.9 S cm2 mol-1 ...(ii)

Λom CH3COONa = Λom CH3COO- + Λom Na+ = 91.0 S cm2 mol-1 ...(iii)

Λo for CH3COOH = (iii) + (ii) - (i)

Λo for CH3COOH = Λom CH3COONa + Λom HCl - Λom NaCl = 91.0 + 425.9 - 126.4 = 390.5 S cm2 mol-1