Question

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?

Answer 1

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.