3 Faradays of electricity was passed through an aqueous solution of iron (II) bromide. The weight of iron metal (at. wt. = 56) deposited at the cathode (in g) is

Question

3 Faradays of electricity was passed through an aqueous solution of iron (II) bromide. The weight of iron metal (at. wt. = 56) deposited at the cathode (in g) is (A) 65 (B) 84 (C) 112 (D) 168

Tardigrade Admin · Accepted Answer

F e2 ++2e- arrow Fe textEF e=(56/2)=28 textwF e=EF e× number of Faraday =28× 3=84 g

Q. 3 Faradays of electricity was passed through an aqueous solution of iron (II) bromide. The weight of iron metal (at. wt. = 56) deposited at the cathode (in g) is

65

84

112

168

$F e^{2 +} + 2 e^{-} \to F e$

$E_{F e} = \frac{56}{2} = 28$

$w_{F e} = E_{F e} \times$ number of Faraday

$= 28 \times 3 = 84 g$

Q. 3 Faradays of electricity was passed through an aqueous solution of iron (II) bromide. The weight of iron metal (at. wt. = 56) deposited at the cathode (in g) is

65

84

112

168

Fe2++2e−→Fe EFe​=256​=28 wFe​=EFe​× number of Faraday =28×3=84g

$F e^{2 +} + 2 e^{-} \to F e$

$E_{F e} = \frac{56}{2} = 28$

$w_{F e} = E_{F e} \times$ number of Faraday

$= 28 \times 3 = 84 g$