A current of 2 amp when passed for 5 hour through a molten salt deposits 22.2 g of metal of atomic mass 177. The oxidation state of the metal in the metal salt is [Report your answer by rounding it upto nearset whole number]

Question

A current of 2 amp when passed for 5 hour through a molten salt deposits 22.2 g of metal of atomic mass 177. The oxidation state of the metal in the metal salt is [Report your answer by rounding it upto nearset whole number] (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

W=(I t E/96500) E=(W × 96500/I t)=(22.2 × 96500/2 × 5 × 3600)=59.5 Equivalent mass =(A t o m i c textmass/O x i d a t i o n s t a t e) 59.5=(177/n) So n≈ 3 .

Q. A current of 2 amp when passed for 5 hour through a molten salt deposits 22.2 g of metal of atomic mass 177. The oxidation state of the metal in the metal salt is

[Report your answer by rounding it upto nearset whole number]

$W = \frac{I tE}{96500}$

$E = \frac{W \times 96500}{I t} = \frac{22.2 \times 96500}{2 \times 5 \times 3600} = 59.5$

Equivalent mass $= \frac{A t o mi c mass}{O x i d a t i o n s t a t e}$

$59.5 = \frac{177}{n}$

So $n \approx 3$ .

Q. A current of 2 amp when passed for 5 hour through a molten salt deposits 22.2 g of metal of atomic mass 177. The oxidation state of the metal in the metal salt is [Report your answer by rounding it upto nearset whole number]

W=96500ItE​ E=ItW×96500​=2×5×360022.2×96500​=59.5 Equivalent mass =OxidationstateAtomicmass​ 59.5=n177​ So n≈3 .

Q. A current of 2 amp when passed for 5 hour through a molten salt deposits 22.2 g of metal of atomic mass 177. The oxidation state of the metal in the metal salt is

[Report your answer by rounding it upto nearset whole number]

$W = \frac{I tE}{96500}$

$E = \frac{W \times 96500}{I t} = \frac{22.2 \times 96500}{2 \times 5 \times 3600} = 59.5$

Equivalent mass $= \frac{A t o mi c mass}{O x i d a t i o n s t a t e}$

$59.5 = \frac{177}{n}$

So $n \approx 3$ .