Question

A steady current was passed through molten ${\mathrm{CoSO}}_4$ until $2.35\mathrm{g}$ of metallic cobalt was produced. Calculate the number of coulombs of electricity used.

Short answer

About 7708 Coulombs of electricity were used.

Step-by-step solution

Determine the Molar Mass of Cobalt

The periodic table shows that the atomic mass of cobalt (Co) is approximately 58.93 g/mol. This will help us convert the mass of the produced cobalt to moles."

Calculate the Moles of Cobalt Produced

Convert the given mass of cobalt to moles using its molar mass. $$\text{Moles of Co}=\frac{\text{mass}}{\text{molar mass}}=\frac{2.35\text{ g}}{58.93\text{ g/mol}}\approx 0.0399\text{ mol}$$

Use Faraday's Law of Electrolysis

According to Faraday's law, the amount of substance liberated at an electrode is directly proportional to the quantity of electricity passed through the electrolyte. For cobalt, which forms as Co, each mole of cobalt requires 2 moles of electrons (since it goes from Co^{2+} to Co).

Calculate the Charge in Coulombs

Use the formula:$$Q=n\times F\times z$$where $Q$ is the charge in coulombs, $n$ is the number of moles of cobalt, $F$ is Faraday's constant (approximately 96485 C/mol), and $z$ is the number of electrons per atom of cobalt (which is 2 here, for Co^{2+} to Co). So,$$Q=0.0399\times 96485\times 2\approx 7708\text{ C}$$

Conclusion

The amount of electricity used to produce the metallic cobalt is approximately 7708 coulombs.

Key concepts

Moles of Cobalt Calculation

To find out how many moles of cobalt are produced, we use the formula: $$\text{Moles of Co}=\frac{\text{mass of Co}}{\text{molar mass of Co}}$$ In this exercise, we are given 2.35 grams of cobalt. By using its molar mass, we can convert this mass to moles. The equation becomes: $$\text{Moles of Co}=\frac{2.35\text{ g}}{58.93\text{ g/mol}}\approx 0.0399\text{ mol}$$ This calculation tells us that about 0.0399 moles of cobalt have been formed. This conversion is crucial as it sets the foundation for further calculations in electrolysis.

Molar Mass

The molar mass of an element is key in converting grams to moles, a vital step in many chemistry calculations. - **Definition**: The molar mass is the mass of one mole of a given substance, typically measured in grams per mole (g/mol). - **For Cobalt**: Using the periodic table, we find cobalt's atomic mass to be approximately 58.93 g/mol. When solving problems related to masses and moles, always remember: - Look up the atomic/molar mass. - Ensure units are consistent for reliable conversions. By understanding molar mass, you can unlock the relationships between quantities of substances in chemical reactions.

Coulombs Calculation

Calculating the charge in coulombs involves using Faraday's law, which relates the amount of a substance deposited during electrolysis to the electricity used. Let's break this down:- **Formula**: $$Q=n\times F\times z$$ where $Q$ is the charge in coulombs, $n$ is the number of moles of the substance, $F$ is Faraday's constant, and $z$ is the number of electrons transferred.- **In this exercise**: We know from our previous calculation that $n=0.0399$ moles of cobalt. Cobalt forms ${\text{Co}}^{2+}$ requiring 2 electrons (thus, $z=2$). - **Calculation**: $$Q=0.0399\times 96485\times 2\approx 7708\text{ C}$$This amount, 7708 coulombs, represents the total electrical charge needed for the reaction.

Faraday's Constant

Faraday's constant is a critical value in electrochemistry that connects charge and amount of substance. - **Definition**: This constant is approximately 96485 coulombs per mole (C/mol). It represents the charge of one mole of electrons. - **Why It Matters**: Faraday’s constant allows us to calculate the charge needed to deposit or dissolve a given amount of ions during electrolysis. For example, in our calculation with cobalt, using Faraday's constant helps determine how many coulombs are needed to produce the cobalt metal. Utilizing Faraday's constant simplifies the link between electricity and chemical change in various scientific and industrial applications, emphasizing its importance in understanding electrolysis.