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Question

Chemistry Question on Electrolysis

A current of 2.0A2.0\, A passed for 55 hours through a molten metal salt deposits 22.2g22.2\, g of metal (At wt. = 177177). The oxidation state of the metal in the metal salt is

A

1

B

2

C

3

D

4

Answer

3

Explanation

Solution

Weight =E×i×tF= M it n factor ×FE=Mn factor =\frac{E \times i \times t}{F}=\frac{\text { M it }}{n-\text { factor } \times F} \mid E=\frac{M}{n-\text { factor }}
22.2=177×5×2×60×60n factor ×96500\Rightarrow 22.2=\frac{177 \times 5 \times 2 \times 60 \times 60}{ n -\text { factor } \times 96500}
n\therefore n - factor =2.973=2.97 \simeq 3
Hence oxidation state of metal =+3=+3