Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Electrolysis of a solution of \(\mathrm{CuSO}_{4}\) (aq) to give copper metal is carried out using a current of 0.66 A. How long should electrolysis continue to produce \(0.50 \mathrm{g}\) of copper?

Short Answer

Expert verified
Electrolysis should continue for about 38.34 minutes to produce 0.50 g of copper.

Step by step solution

01

Determine Molar Mass of Copper

The molar mass of copper (Cu) is required. Copper has a molar mass of 63.55 g/mol. This value is essential to convert the mass of copper to moles.
02

Calculate Moles of Copper

Use the mass of copper to find the number of moles. Use the formula: \(\text{moles of Cu} = \frac{\text{mass of Cu}}{\text{molar mass of Cu}}\). Thus, the calculation is \(\frac{0.50 \text{ g}}{63.55 \text{ g/mol}} = 0.00787 \text{ mol Cu}\).
03

Determine Moles of Electrons Needed

Copper deposits in the reaction \(\mathrm{Cu}^{2+} + 2e^- \rightarrow \mathrm{Cu}\). Therefore, 1 mole of copper requires 2 moles of electrons. Calculate moles of electrons needed by multiplying the moles of copper by 2: \(0.00787 \times 2 = 0.01574 \text{ mol of e}^-=\).
04

Use Faraday's Constant

Faraday's constant (approximately 96500 C/mol) relates charge to moles of electrons. Calculate the total charge in coulombs needed using \(\text{charge} = 0.01574 \text{ mol e}^- \times 96500 \text{ C/mol} = 1518.31 \text{ C}\).
05

Find Time Using Current

Use the relationship between charge, current, and time: \(\text{charge} = \text{current} \times \text{time}\). Rearranging gives \(\text{time} = \frac{\text{charge}}{\text{current}} = \frac{1518.31}{0.66} \approx 2300.47 \text{ seconds}\).
06

Convert Time to Minutes

Convert the time from seconds into minutes. Since 1 minute = 60 seconds, you use \(\frac{2300.47 \text{ s}}{60} \approx 38.34 \text{ minutes}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Copper deposition
When copper is deposited in the process of electrolysis, it involves the reduction reaction of copper ions (\(\mathrm{Cu}^{2+}\)) into copper metal. This happens when an electric current is passed through an aqueous solution of copper sulfate (\(\mathrm{CuSO}_4\)).
At the cathode, copper ions gain electrons and get deposited as copper metal. The balanced chemical reaction for this process is:
  • \(\mathrm{Cu}^{2+} + 2e^- \rightarrow \mathrm{Cu}\)
Here, \(\mathrm{Cu}^{2+}\) represents copper ions and \(e^-\) represents electrons. Due to this reaction, one mole of copper ions accepts two moles of electrons to form one mole of copper metal. This principle is key because the number of electrons involved directly relates to the amount of copper deposited.
This understanding helps in calculating how much time and current is needed to achieve a specific amount of copper deposit.
Faraday's constant
Faraday's constant is an important value in electrochemistry, which quantifies the charge carried by one mole of electrons. The commonly used approximate value of Faraday's constant is 96500 Coulombs per mole (C/mol).
It plays a crucial role in determining the total electric charge needed in electrolysis calculations.
  • In the calculation situation: \(\text{charge} = \text{moles of } e^- \times \text{Faraday's constant}\).
For instance, if you determine the moles of electrons needed for your copper deposition process, multiplying it by Faraday's constant gives the total charge in Coulombs. Using Faraday's constant helps to bridge the gap between the electrochemical reaction and the electrical current needed to carry out the reaction.
By knowing the total charge, you can then use the relationship with current to find out how long you need the current to run.
Molar mass of copper
Copper is an element with an atomic symbol Cu and an atomic number of 29. In chemistry, understanding the molar mass of elements like copper is essential in stoichiometric calculations, such as converting mass to moles.
Molar mass is a measure of the mass of one mole of a substance. The molar mass of copper is 63.55 grams per mole (g/mol).
  • This is important for converting mass to moles using the formula: \(\text{moles of Cu} = \frac{\text{mass of Cu}}{\text{molar mass of Cu}}\).
In practice, if you have 0.50 grams of copper and you need to find out how many moles this represents, you would divide the mass by the molar mass, resulting in approximately 0.00787 moles. This conversion from mass to moles is crucial when calculating how much of a substance reacts or is produced in chemical reactions, especially in electrolysis where quantities of charge and reaction stoichiometry need to be precise.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Write equations for the half-reactions that occur at the anode and cathode in the electrolysis of molten KBr. What are the products formed at the anode and cathode in the electrolysis of aqueous KBr?

An aqueous solution of KBr is placed in a beaker with two inert platinum electrodes. When the cell is attached to an external source of electrical energy, electrolysis occurs. (a) Hydrogen gas and hydroxide ion form at the cathode Write an equation for the halfreaction that occurs at this electrode. (b) Bromine is the primary product at the anode. Write an equation for its formation.

Write balanced equations for the following halfreactions. Specify whether each is an oxidation or reduction. (a) \(\mathrm{Cr}(\mathrm{s}) \rightarrow \mathrm{Cr}^{3+}(\text { aq })\) (in acid) (b) \(\mathrm{AsH}_{3}(\mathrm{g}) \rightarrow \mathrm{As}(\mathrm{s})\) (in acid) (c) \(\mathrm{VO}_{3}^{-}(\mathrm{aq}) \rightarrow \mathrm{V}^{2+}(\mathrm{aq})\) (in acid) (d) \(\mathrm{Ag}(\mathrm{s}) \rightarrow \mathrm{Ag}_{2} \mathrm{O}(\mathrm{s})\) (in base)

The metallurgy of aluminum involves electrolysis of \(\mathrm{Al}_{2} \mathrm{O}_{3}\) dissolved in molten cryolite \(\left(\mathrm{Na}_{3} \mathrm{AlF}_{6}\right)\) at about \(950^{\circ} \mathrm{C}\). Aluminum metal is produced at the cathode. Predict the anode product and write equations for the reactions occurring at both electrodes.

Balance the following redox equations. All occur in acid solution. (a) \(\mathrm{Ag}(\mathrm{s})+\mathrm{NO}_{3}^{-}(\mathrm{aq}) \rightarrow \mathrm{NO}_{2}(\mathrm{g})+\mathrm{Ag}^{+}(\mathrm{aq})\) (b) \(\mathrm{MnO}_{4}^{-}(\mathrm{aq})+\mathrm{HSO}_{3}^{-}(\mathrm{aq}) \rightarrow\) \(\mathrm{Mn}^{2+}(\mathrm{aq})+\mathrm{SO}_{4}^{2-}(\mathrm{aq}\) (c) \(\mathrm{Zn}(\mathrm{s})+\mathrm{NO}_{3}^{-}(\mathrm{aq}) \rightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{N}_{2} \mathrm{O}(\mathrm{g})\) (d) \(\mathrm{Cr}(\mathrm{s})+\mathrm{NO}_{3}^{-}(\mathrm{aq}) \rightarrow \mathrm{Cr}^{3+}(\mathrm{aq})+\mathrm{NO}(\mathrm{g})\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free