Question

An aqueous solution of an unknown salt of ruthenium is electrolyzed by a current of 2.50 A passing for 50.0 min. If 2.618 g Ru is produced at the cathode, what is the charge on the ruthenium ions in solution?

Short answer

The charge present in ruthenium ion is +3

Step-by-step solution

Given Information

The mass of Ru is 2.618 g. The time value is 50 min. The current can be given as 2.5 A. The reaction for Ru formation can be given as:

$R{u}^{x+}+x{e}^{-}\to Ru$

Mole number of electrons released

The number of moles of electron released with 2.5 A current and 50 mins can be expressed as:

$$\begin{gathered} Moles of electron=(50 \min )\left(\frac{60s}{1 \min }\right)\left(\frac{2.5C}{1s}\right)\left(\frac{1 mol}{96485 C}\right)\backslash \mathrm{hfill} \\ =0.07773 mol \end{gathered}$$

The moles of Ruthenium produced in electrolysis can be expressed as:

$$\begin{gathered} Mole=(2.618 g)\left(\frac{1 mol}{101.07 g}\right) \\ =0.0259 mol \end{gathered}$$

Mole of electrons observed

The mole of electrons observed in reduction process can be given as:

$$\begin{gathered} Moles of electrons Ru=\frac{0.07773 mol e}{0.0259 mol Ru} \\ =3 e^{-}/Ru \end{gathered}$$

Thecharge which is present in ruthenium ion is +3.