Question

In an electrolysis cell similar to the one employed in this experiment, a student observed that his unknown metal anode lost \(0.208 \mathrm{g}\) while a total volume of \(96.30 \mathrm{mL}\) of \(\mathrm{H}_{2}\) was being produced. The temperature in the laboratory was \(25^{\circ} \mathrm{C}\) and the barometric pressure was \(748 \mathrm{mm}\) Hg. At \(25^{\circ} \mathrm{C}\) the vapor pressure of water is \(23.8 \mathrm{mm}\) Hg. To find the equivalent mass of his metal, he filled in the blanks below. Fill in the blanks as he did. $$P_{\mathrm{II}_{2}}=P_{\mathrm{har}}-V P_{\mathrm{H}_{2} \mathrm{O}}=$$ $$\operatorname{mm} \mathrm{Hg}=$$ atm $$V_{\mathrm{H}_{2}}=$$ $$\mathrm{mL}=$$ $$\mathbf{L}$$ $$T=$$ K$$n_{\mathrm{H}_{2}}=$$ moles $$\left.n_{\mathrm{H}_{2}}=\frac{P V}{R T} \quad \text { (where } P=P_{\mathrm{H}_{2}}\right)$$ 1 mole \(\mathrm{H}_{2}\) requires passage of _faradays No. of faradays passed \(=\) Loss of mass of metal anode \(=\) g No. grams of metal lost per faraday passed \(=\frac{\text { no. grams lost }}{\text { no. faradays passed }}=\quad \mathrm{g}=\mathrm{EM}\) The student was told that his metal anodc was madc of iron. MM Fe = \(g\) The charge \(n\) on the Fe ion is therefore $$P_{\mathrm{II}_{2}}=P_{\mathrm{har}}-V P_{\mathrm{H}_{2} \mathrm{O}}=$$ $$\operatorname{mm} \mathrm{Hg}=$$ atm $$V_{\mathrm{H}_{2}}=$$ $$\mathrm{mL}=$$ $$\mathbf{L}$$ $$T=$$ K$$n_{\mathrm{H}_{2}}=$$ moles $$\left.n_{\mathrm{H}_{2}}=\frac{P V}{R T} \quad \text { (where } P=P_{\mathrm{H}_{2}}\right)$$

Short answer

In the given electrolysis scenario, first, we found the partial pressure of hydrogen and converted it to atm. Then, we converted the volume of hydrogen produced to L and the temperature to K. Using the Ideal Gas Law, we calculated the moles of hydrogen produced and then the number of faradays passed. Next, we found the grams of metal lost per faraday passed and finally determined the charge (n) of the Fe ion using the molar mass of iron and the obtained equivalent mass. \(n=\frac{55.845~g/mol}{EM}\)

Step-by-step solution

Find the partial pressure of hydrogen

First, we'll calculate the partial pressure of hydrogen, which can be found by subtracting the vapor pressure of water from the barometric pressure at the given temperature (25°C). \(P_{H_{2}}=P_{har}-V P_{H_{2} O}\) Plug the given values: \(P_{H_{2}}=(748-23.8) ~mm~Hg\)

Convert pressure to atm

Next, we'll convert the partial pressure of hydrogen to atm by dividing the pressure in mm Hg by 760. \(P_{H_{2}}= \frac{(748-23.8)}{760} ~atm\)

Convert volume to L and temperature to K

Now, convert the volume of hydrogen produced from mL to L by dividing by 1000 and the temperature from Celsius to Kelvin by adding 273.15. \( V_{H_{2}} = 96.30~mL = \frac{96.30}{1000}~L \) \(T= 25^{\circ} C = (25 + 273.15)K \)

Calculate the moles of hydrogen

We can now use the Ideal Gas Law to calculate the moles of hydrogen produced. The ideal gas law is given by: \( n_{H_{2}}=\frac{P V}{R T} \) Where P is the partial pressure of hydrogen, V is the volume, R is the ideal gas constant (0.08206 L atm/mol K), and T is the temperature. Plug in the values obtained in previous steps.

Calculate the number of faradays passed

Since 1 mole of H2 requires 1 faraday of electricity, the number of faradays passed can be calculated by multiplying the moles of hydrogen obtained in step 4 by 2. \(No.~of~ faradays = n_{H_{2}} × 2\)

Calculate the grams of metal lost per faraday passed

Now, we can calculate the grams of metal lost per faraday passed by dividing the mass lost by the number of faradays passed. \(g=\frac{no.~grams~lost}{no.~faradays~passed}=\frac{0.208 g}{No.~of~ faradays ~obtained~in~Step~5}\)

Determine the charge of the Fe ion

We know that the equivalent mass (EM) is equal to the molar mass (MM) of iron divided by the charge (n) on the iron ion. Rearrange the formula for n. \(n=\frac{MM~Fe}{EM}\) Substitute the values for MM Fe(55.845 g/mol) and the equivalent mass (EM) obtained in step 6: \(n=\frac{55.845~g/mol}{EM}\) Now we've successfully found the charge (n) of the Fe ion in the given scenario.

Key concepts

Electrochemistry

Electrochemistry is the study of the relationship between electricity and chemical reactions. It's a fascinating field that intersects chemistry and physics.
During electrolysis, electrical energy is used to drive a non-spontaneous chemical reaction. In the exercise, this process was utilized to produce hydrogen gas and oxidize the metal anode, causing it to lose mass.
• **Anode and Cathode:** - Electrolysis involves two electrodes: the anode (where oxidation occurs) and the cathode (where reduction occurs).
• **Movement of Ions:** - In an electrolysis cell, negatively charged ions move towards the anode and positively charged ions toward the cathode.
Understanding these basic ideas helps us see how electricity can trigger chemical transformations.

Ideal Gas Law

The Ideal Gas Law connects the pressure, volume, temperature, and moles of a gas. It is expressed by the equation: \[ PV = nRT \]
• **Variables:** - P = Pressure measured in atmospheres (atm) or mm Hg.
- V = Volume in liters (L).
- n = Number of moles of the gas.
- R = Ideal gas constant (0.08206 L atm/mol K).
- T = Temperature in Kelvin (K).
In this exercise, we used the Ideal Gas Law to calculate the number of moles of hydrogen produced. By knowing the partial pressure and applying these principles, we can easily assess the behavior of gases under given conditions.

Faraday's Law

Faraday's Law in electrochemistry relates the amount of material altered at an electrode during electrolysis to the quantity of electricity used.
The law states that \(1\) mole of a substance is produced or consumed at an electrode by \(1\) faraday of electricity. One faraday is equivalent to \(96,485\) coulombs (C).
In the problem, knowing that hydrogen production requires \(2\) moles of electrons per mole of hydrogen (\(H_2\)), we calculated the number of faradays through the moles of \(H_2\) formed.
This is essential in connecting the electrical current passed to the chemical yield in electrochemical reactions.

Partial Pressure

In a mixture of gases, each gas exerts its own pressure, known as partial pressure. The total pressure of the gas mixture equals the sum of the partial pressures of its constituents.
For our analysis:• **Using Dalton's Law:** - \[P_{total} = P_{H_2} + P_{H_2O}\]
We calculated the partial pressure of hydrogen \(P_{H_2}\) by subtracting the vapor pressure of water (\(23.8\) mm Hg) from the barometric pressure \(748\) mm Hg.
Converting this to atmospheres helped us in using the Ideal Gas Law effectively. Understanding partial pressures is vital in gas-related problems.

Molar Mass Calculation

Calculating the molar mass involves determining how many grams per mole a substance has. It is a crucial aspect of chemistry, enabling the understanding of chemical proportions.
In the exercise:• We calculated the equivalent mass (EM) from the metal's loss and used it to find the charge of the Fe ion by means of Faraday's law and molar mass relationship: - \[ n = \frac{MM_{Fe}}{EM} \]
This exercise exemplifies the use of experimental data to deduce fundamental properties. In electrolysis, knowing the molar mass of electrodes helps predict outcomes and balance equations.