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2 years ago

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5. When a ruthenium chloride solution was electrolyzed for 500 s with a 120mA current, 31 mg of ruthenium was deposited. What is

the oxidation number of ruthenium in the ruthenium chloride?

1 answer:

Leni [432]2 years ago

5 0

Explanation:

Relation between equivalent weight, current, time is as follows.

W = $\frac{ECt}{96500}$

where, W = mass of the metal deposited = 31.0 mg = $31.0 \times 10^{-3} g$

E = Equivalent weight of Ru = ?

C = current = 120 mA = $120 \times 10^{-3} A$

t = time taken = 500 sec

Therefore, calculate the value of equivalent weight as follows.

E = $\frac{96500 \times W}{Ct}$

= $\frac{96500 \times \times 31.0 \times 10^{-3}}{120 \times 10^{-3} \times 500 sec}$

= 49.86 g eqv

Since, we know that the molar mass of ruthenium is 101.1 g/mol.

Therefore, the oxidation number of ruthenium is calculated as follows.

$\frac{101.1}{49.86}$

= 2.027

= 2 (approximately)

Thus, we can conclude that the oxidation number of ruthenium in the ruthenium chloride is 2.

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