Question

In the electrolysis of a solution containing \(\mathrm{Ni}^{2+}\) (aq), metallic \(\mathrm{Ni}(\mathrm{s})\) deposits on the cathode. Using a current of 0.150 A for 12.2 min, what mass of nickel will form?

Short answer

Approximately 0.0334 grams of nickel will form.

Step-by-step solution

Calculate Total Charge Passed

To determine the total charge passed through the solution, use the formula: \( Q = I \times t \), where \( Q \) is the charge in coulombs, \( I \) is the current in amperes, and \( t \) is the time in seconds. First, convert the time from minutes to seconds: \( 12.2 \text{ min} = 12.2 \times 60 = 732 \text{ s} \). Then, calculate the total charge: \[ Q = 0.150 \text{ A} \times 732 \text{ s} = 109.8 \text{ C} \].

Relate Charge to Moles of Electrons

Use Faraday's constant to convert the charge to moles of electrons. Faraday's constant is approximately \( 96485 \text{ C/mol} \). The formula is \( n = \frac{Q}{F} \), where \( n \) is the moles of electrons. Calculate \( n \): \[ n = \frac{109.8 \text{ C}}{96485 \text{ C/mol}} \approx 0.001138 \text{ mol of electrons} \].

Convert Moles of Electrons to Moles of Ni

The reaction at the cathode is \( \mathrm{Ni}^{2+} + 2e^- \rightarrow \mathrm{Ni}(s) \). The stoichiometry tells us that 2 moles of electrons are needed to deposit 1 mole of Ni. Therefore, calculate moles of Ni: \[ \text{Moles of Ni} = \frac{0.001138}{2} \approx 0.000569 \text{ mol of Ni} \].

Convert Moles of Ni to Mass

Find the molar mass of Ni, which is approximately 58.69 g/mol. Use the formula \( \text{mass} = \text{moles} \times \text{molar mass} \) to find the mass of Ni deposited. Substitute the value: \[ \text{mass} = 0.000569 \text{ mol} \times 58.69 \text{ g/mol} \approx 0.0334 \text{ g} \].

Key concepts

Faraday's constant

Faraday's constant is named after the famous scientist Michael Faraday. It represents the amount of electric charge carried by one mole of electrons. This constant is crucial in electrochemistry, as it allows us to convert between charge in coulombs and moles of electrons.
In calculations involving electrolysis, Faraday's constant is used to determine how much substance is deposited during the process. Its value is approximately 96,485 coulombs per mole (\[ ext{C/mol} \]).
By knowing this value, you can break down complex electrochemical reactions into manageable steps, using the natural relationship between charge, electrons, and chemical changes. Thus, it serves as a bridge to connect electrical units with chemical quantities.

Current and charge calculation

Understanding current and charge is fundamental in electrolysis calculations. Current, measured in amperes (\[ ext{A} \]), describes the flow of electric charge through a conductor. In electrolysis, it determines how much electricity passes through the electrolyte solution.
The charge (\[ Q \]) transferred during electrolysis is calculated using the formula: \[ Q = I \times t \], where \[ I \] is the current in amperes and \[ t \] is the time in seconds.
First, convert any time given in minutes into seconds to use in the formula by multiplying the minutes by 60.

• For example: 12.2 minutes is 732 seconds.

Applying this formula gives the total charge, which then helps in further calculations related to the electrochemical reaction.

Nickel deposition

Nickel deposition refers to the process where a nickel (\[ ext{Ni} \]) coating is formed on a material, typically at the cathode, in an electrolytic cell.
This occurs when \[ ext{Ni}^{2+} \] ions in the solution gain electrons and form solid nickel. The half-reaction can be represented as:

• \[ ext{Ni}^{2+} + 2e^- \rightarrow ext{Ni}(s) \]

For every mole of nickel ions reduced, two moles of electrons are needed. Understanding this stoichiometry is key to calculating the amount of nickel deposited based on the electrons available, as derived from the total charge calculation.
Thus, knowing the moles of electrons allows us to determine the moles of nickel that will deposit, which can then be used to find the mass of nickel formed.

Stoichiometry in electrochemistry

Stoichiometry links chemical equations with real-world quantities. In electrochemistry, it plays a crucial role, especially in understanding the amount of substance that will be produced or consumed in a reaction.
In nickel deposition, the stoichiometry of \[ 2e^- \] needed for \[ ext{Ni}^{2+} \] to form \[ ext{Ni} \] helps determine the moles of nickel formed based on the moles of electrons calculated.
You use the stoichiometry from the balanced electrochemical equation to relate the charge to the substance deposited.

• The overall reaction for nickel deposition is:\[ ext{Ni}^{2+} + 2e^- \rightarrow ext{Ni}(s) \]

Once you calculate the moles of nickel formed using this stoichiometry, you can then convert it into mass by multiplying by the molar mass of nickel. This approach ensures that the calculations accurately reflect the quantities involved in the electrochemical process.