Question

An acidic solution containing ruthenium ions is electrolyzed, producing gaseous oxygen (from water) at the anode and ruthenium at the cathode. For every 15.3 g of oxygen generated, 96.6 g of ruthenium plates out. What is the oxidation state of the ruthenium in the solution?

Answer 1

Let n+ be the oxidation state of ruthenium. Reduction half reaction at cathode is shown below.

$Ru^{n+}+ne^- \to Ru$

At anode, water is oxidised to oxygen.

$2H_2O \to O_2+4H^++4e^-$

Number of moles of electrons involved for reduction of $Ru^{n+}$ ions are equal to number of moles of electrons released during oxidation of water to oxygen.

Mass of oxygen is15.3 g.

Molar mass of oxygen is 32 g/mol

Number of moles of oxygen $=\frac{15.3 \ g}{32 \ g/mol}=0.478 \ mol$

Number of moles of electrons released during oxidation of water to oxygen $=4 \times$ number of moles of oxygen obtained.

Number of moles of electrons released during oxidation of water to oxygen $=4 \times 0.478=1.91 \ mol$

.

Number of moles of electrons involved for reduction of $Ru^{n+}$ ions are equal to number of moles of electrons released during oxidation of water to oxygen.

Number of moles of electrons involved for reduction of $Ru^{n+}$ ions

Mass of Ru obtained is 96.6 g

Atomic mass of Ru is 101 g/mol

Number of moles of Ru $=\frac{96.6 \ g}{101 \ g/mol}=0.956 \ mol$

When we divide number of moles of electrons involved for reduction of $Ru^{n+}$ ions with number of moles of Ru we get value of oxidation state n.

Value of oxidation state n $=\frac{ 1.91 \ mol}{0.956 \ mol}=2$

Hence, the oxidation state of the ruthenium in the solution is .