Answer :
Explanation:
Relation between equivalent weight, current, time is as follows.
W = [tex]\frac{ECt}{96500}[/tex]
where, W = mass of the metal deposited = 31.0 mg = [tex]31.0 \times 10^{-3} g[/tex]
E = Equivalent weight of Ru = ?
C = current = 120 mA = [tex]120 \times 10^{-3} A[/tex]
t = time taken = 500 sec
Therefore, calculate the value of equivalent weight as follows.
E = [tex]\frac{96500 \times W}{Ct}[/tex]
= [tex]\frac{96500 \times \times 31.0 \times 10^{-3}}{120 \times 10^{-3} \times 500 sec}[/tex]
= 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.
[tex]\frac{101.1}{49.86}[/tex]
= 2.027
= 2 (approximately)
Thus, we can conclude that the oxidation number of ruthenium in the ruthenium chloride is 2.