Chem8

Question 1

Define 1 Faraday (F) in coulombs.

Answer 1

1 F = 96,500 C (charge to deposit 1 eq. mass).

Question 2

How many Faradays are needed to deposit 1 mole of Al?

Answer 2

3 F (Al³⁺ + 3e⁻ → Al; Z = 3).

Question 3

Calculate mass of Ag deposited by 2 F (Ag = 108 g/mol).

Answer 3

2 F × 108 g/eq = 216 g.

Question 4

Second Law of Faraday formula.

Answer 4

Mass(A)/Mass(B) = Eq.mass(A)/Eq.mass(B).

Question 5

If 0.5 F deposits 5.9 g Cu, find Cu’s eq. mass (Z = 2).

Answer 5

Eq.mass = Mass/F = 5.9g/0.5 = 31.75 g/eq (matches Cu²⁺).

Question 6

Current (I) to deposit 27 g Al in 10 hours (Z = 3).

Answer 6

Moles Al = 1 → 3 F = 289,500 C; I = Q/t = 289,500/(10×3600) = 8.04 A.

Question 7

Prove Faraday’s 2nd Law using Al, Cu, Ag.

Answer 7

Mass ratio = 9:31.75:108 (their eq. masses).

Question 8

Time to deposit 108 g Ag with 5 A current.

Answer 8

1 F = 96,500 C; t = Q/I = 96,500/5 = 19,300 s (~5.36 hrs).

Question 9

Equivalent mass of Fe³⁺ (Fe = 56 g/mol).

Answer 9

Eq.mass = 56/3 = 18.67 g/eq.

Question 10

Why does 1 F deposit 1 eq. mass, not 1 mole?

Answer 10

1 eq. mass = mass per mole of electrons (Z-dependent).

Question 11

Charge on ion in FeCl₃?

Answer 11

Fe³⁺ (from formula or name: iron(III) chloride).

Question 12

Grams of Cu deposited by 48,250 C (Cu²⁺).

Answer 12

48,250 C = 0.5 F → 0.5 × 31.75 g = 15.875 g.