A current of 2.0 A passed for 5 hours through a molten metal salt deposits 22.2 g of metal (At wt. = 177). The oxidation state of the metal in the metal salt is

Question

A current of 2.0 A passed for 5 hours through a molten metal salt deposits 22.2 g of metal (At wt. = 177). The oxidation state of the metal in the metal salt is (A) + 1 (B) +2 (C) +3 (D) +4

Tardigrade Admin · Accepted Answer

Weight =(E × i × t/F)=( text M it /n- text factor × F) mid E=(M/n- text factor ) ⇒ 22.2=(177 × 5 × 2 × 60 × 60/ n - text factor × 96500) ∴ n - factor =2.97 ≃ 3 Hence oxidation state of metal =+3

Q. A current of $2.0 A$ passed for $5$ hours through a molten metal salt deposits $22.2 g$ of metal (At wt. = $177$ ). The oxidation state of the metal in the metal salt is

+ 1

+2

+3

+4

Weight $= \frac{E \times i \times t}{F} = \frac{M it}{n - factor \times F} ∣ E = \frac{M}{n - factor}$
$\Rightarrow 22.2 = \frac{177 \times 5 \times 2 \times 60 \times 60}{n - factor \times 96500}$
$∴ n -$ factor $= 2.97 ≃ 3$
Hence oxidation state of metal $= + 3$

Q. A current of 2.0A passed for 5 hours through a molten metal salt deposits 22.2g of metal (At wt. = 177). The oxidation state of the metal in the metal salt is

+ 1

+2

+3

+4

Weight =FE×i×t​=n− factor ×F M it ​∣E=n− factor M​ ⇒22.2=n− factor ×96500177×5×2×60×60​ ∴n− factor =2.97≃3 Hence oxidation state of metal =+3

Q. A current of $2.0 A$ passed for $5$ hours through a molten metal salt deposits $22.2 g$ of metal (At wt. = $177$ ). The oxidation state of the metal in the metal salt is

Weight $= \frac{E \times i \times t}{F} = \frac{M it}{n - factor \times F} ∣ E = \frac{M}{n - factor}$
$\Rightarrow 22.2 = \frac{177 \times 5 \times 2 \times 60 \times 60}{n - factor \times 96500}$
$∴ n -$ factor $= 2.97 ≃ 3$
Hence oxidation state of metal $= + 3$