Question

Match List-I with List-II:

List-I (Conversion)	List-II (Number of Faraday required)
A. 1 mol of H2O to O2	I. 3F
B. 1 mol of \[\ce{MnO^-4}\] to Mn2+	II. 2F
C. 1.5 mol of Ca from molten CaCl2	III. 1F
D. 1 mol of FeO to Fe2O3	IV. 5F

Choose the correct answer from the options given below:

विकल्प

• A - II, B - III, C - I, D - IV

• А - III, B - IV, C - II, D - I

• A - II, B - IV, C - I, D - III

• A - III, B - IV, C - I, D - II

Answer 1

A - II, B - IV, C - I, D - III

Explanation:

List-I (Conversion)	List-II (Number of Faraday required)
A. 1 mol of H2O to O2	II. 2F
B. 1 mol of \[\ce{MnO^-4}\] to Mn2+	IV. 5F
C. 1.5 mol of Ca from molten CaCl2	I. 3F
D. 1 mol of FeO to Fe2O3	III. 1F

A. \[\ce{H2O -> 2H+ + 1/2O2 + 2e-}\]

Q = nF (n = number of electrons given by Q)

= 2F

B. \[\ce{MnO^-4 + 5e- -> Mn^{2+}}\]

Q = nF

Q = 5F

C. \[\ce{CaCl2 -> Ca^2+ + 2Cl-}\]

\[\ce{Ca^{2+} + 2e- -> Ca}\]

1 mol = 2e⁻

1.5 mol = 3e⁻

Q = nF

= 3F

D. Oxidation number of Fe in FeO is +2.

Oxidation number of Fe in Fe₂O₃ is +3.

1 electron is required to conversion:

Q = nF

= 1F