Question

A current of 0.44 A is passed through a solution of a ruthenium nitrate salt, causing reduction of the metal ion to the metal. After 25.0 minutes, 0.345 g of Ru(s) has been deposited. What is the oxidation state of ruthenium in the nitrate salt?

Answer 1

Ruthenium has an oxidation state of +2

Step-by-step explanation:

Step 1: Data given

A current = 0.44 A

Time = 25.0 minutes

Mass of Ru(s) = 0.345 grams

Step 2: Calculate charge

Charge = current * time

Charge = 0.44 A * 25 min *60 sec/min

Charge = 660 C

Step 3: Calculate moles electrons

1 mol e- = 660 C * (1 mol e-/ 96500 C)

1 mol e- = 6.84 * 10^-3 mol e-

Step 4: Calculate moles of Ru

Moles Ru = mass Ru / molar mass Ru

Moles Ru = 0.345 grams / 101 g/mol

Moles Ru = 3.416 * 10^-3 moles Ru

Step 5: Calculate oxidation state of ruthenium

(6.84e-3 moles e-) / (3.416e-3 moles Ru) = 2 e- / Ru atom (Ru+2)

Ruthenium has an oxidation state of +2

Since a nitrate ion has a charge of -1, ruthenium nitrate would be

Ru(NO3)2