Problem 69 - Chemistry and Chemical Reactivity

Question

A current of \(0.44 \mathrm{A}\) is passed through a solution of ruthenium nitrate causing reduction of the metal ion to the metal. After \(25.0 \mathrm{min}, 0.345 \mathrm{g}\) of Ru has been deposited. What is the charge on the ruthenium ion, \(\mathrm{Ru}^{n+} ?\) What is the formula for ruthenium nitrate?

Step-by-Step Solution

Answer

The ruthenium ion charge is \( 2+ \) and the formula for ruthenium nitrate is \( \mathrm{Ru(NO_3)_2} \).

Step 1: Calculate Total Charge Passed

First, calculate the total charge passed through the solution using the formula \( Q = I \times t \), where \( I \) is the current in amperes and \( t \) is the time in seconds. Given: \( I = 0.44 \text{ A} \) and \( t = 25.0 \times 60 \text{ s} = 1500 \text{ s} \).\[ Q = 0.44 \times 1500 = 660 \text{ C} \]

Step 2: Calculate Moles of Electrons Transferred

Use the formula \( n_e = \frac{Q}{F} \) to find the moles of electrons transferred, where \( F = 96500 \text{ C/mol} \) is Faraday's constant.\[ n_e = \frac{660}{96500} \approx 0.00684 \text{ mol} \]

Step 3: Calculate Moles of Ruthenium Deposited

Next, calculate the moles of ruthenium deposited. Use the molar mass of ruthenium \( 101.07 \text{ g/mol} \).\[ \text{moles of Ru} = \frac{0.345}{101.07} \approx 0.00341 \text{ mol} \]

Step 4: Determine Charge on Ru Ion

Determine the charge on the ruthenium ion based on the moles of electrons transferred and the moles of ruthenium deposited. The reaction is \( \mathrm{Ru}^{n+} + ne^- \rightarrow \mathrm{Ru} \). The moles of electrons per ruthenium atom is given by the ratio:\[ n = \frac{0.00684}{0.00341} \approx 2 \]Thus, the charge on the ion is \( 2+ \).

Step 5: Determine Formula for Ruthenium Nitrate

Since the charge on the ruthenium ion is \( 2+ \), the formula for ruthenium nitrate is \( \mathrm{Ru(NO_3)_2} \). This is because the nitrate ion has a charge of \( 1- \), so we need two nitrates to balance the charge for each ruthenium with a \( 2+ \) charge.

Key Concepts

Ruthenium Ion

Ruthenium ions are positively charged atoms of the metal ruthenium. When ruthenium nitrate is reduced to ruthenium metal, we are essentially transferring electrons to the ruthenium ions. The problem involves finding the charge of the ruthenium ion, denoted as \( \text{Ru}^{n+} \).

• In our reduction process, we discovered that the charge on the ruthenium ion is \( 2+ \). This means each ruthenium ion gains two electrons during reduction.
• The reaction involved is \( \text{Ru}^{2+} + 2e^- \rightarrow \text{Ru} \), where "e" represents an electron.

Understanding the charge of ruthenium ions is important in electrochemistry because it tells us about the valency and reactivity of the metal.

Faraday's Constant

Faraday's constant is a crucial value in electrochemistry, representing the charge of one mole of electrons. With a value of \( 96,500 \text{ C/mol} \), it connects the quantity of electricity to the amount of substance changed in a reaction.

• In our exercise, we used Faraday's constant to determine the moles of electrons transferred during the reduction process.
• This was calculated using the formula \( n_e = \frac{Q}{F} \) where \( Q \) is the total charge.

Faraday's constant allows us to link macroscopic current flow with microscopic chemical changes, making it indispensable for calculations in electrochemical reactions.

Ruthenium Nitrate

Ruthenium nitrate is a compound composed of ruthenium ions and nitrate ions. Its chemical formula, based on our calculations, is \( \text{Ru(NO}_3\text{)_2} \).

• This formula arises because the ruthenium ion has a charge of \( 2+ \) and the nitrate ion has a charge of \( 1- \).
• To balance the charges, each ruthenium requires two nitrate ions.

Ruthenium nitrate is used in various chemical syntheses and applications, and understanding its composition is essential for predicting its behavior in reactions.

Moles of Electrons

The concept of moles of electrons refers to the amount of charge transferred in a chemical reaction. In electrochemical processes, electrons are transferred, and their quantity can be expressed in moles using Faraday's constant.

• In our example, we calculated that 0.00684 moles of electrons were transferred during the reduction of ruthenium ions.
• This was done by using the formula \( n_e = \frac{Q}{F} \), relating the total charge to moles of electrons.

Understanding moles of electrons is important because it quantifies the electrochemical change, letting us relate charge flow to chemical transformations.