Question

When $6\times {10}^{22}$ electrons are used in the electrolysis of a metallic salt, 1.9 gm of the metal is deposited at the cathode. The atomic weight of that metal is 57. So oxidation state of the metal in the salt is:

• A. 2
• B. 3
• C. 1
• D. 4

Answer 1

The correct option is B 3

Number of moles of $e^{-}=\frac{6\times {10}^{22}}{N_A}=\frac{6\times {10}^{22}}{6\times {10}^{23}}=0.1$

$M^{n+}+n{e}^{-}\to M$

$\therefore$ n mole $e^{-}$ give 1 mole metal

0.1 mole $e^{-}$ gives $=\frac{1}{n}\times 0.1mole{e}^{-}$

$\therefore$ Moles of metal $=\frac{1}{10n}$

$\therefore \frac{1}{10\times n}=\frac{1.9}{57}$

$\therefore 57=19\times n$

$\therefore n=3$

Option B is correct.