Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 19: Problem 90

An aqueous solution of a platinum salt is electrolyzed at a current of 2.50 A for \(2.00 \mathrm{~h}\). As a result, \(9.09 \mathrm{~g}\) of metallic \(\mathrm{Pt}\) are formed at the cathode. Calculate the charge on the Pt ions in this solution.

Short Answer

Expert verified
The charge on the Pt ions in this solution is \(4+\).

Step by step solution

01

Calculate the total charge transferred

The total charge can be found using the formula for charge which is \(Q = I \cdot t\). Where \(I\) is current given as 2.50 A and time \(t\) is given in hours, but we need to convert it into seconds. We know that \(1 \mathrm{~h} = 3600 \mathrm{~s}\), so \(2.00 \mathrm{~h} = 7200 \mathrm{~s}\). Therefore, \( Q = 2.5 A \cdot 7200 s = 18000 C (coulombs) \)
02

Calculate the amount of Pt formed

We can calculate the amount of Pt in moles by using the formula \(n = m/M\), where \(m = 9.09 g\) is the mass of Pt formed and \(M = 195.08 g/mol\) is the molar mass of Pt (found in the Periodic Table). Therefore, \( n = 9.09g / 195.08 g/mol = 0.0466 mol \)
03

Faraday’s Law of Electrolysis and calculating charge on Pt ions

According to Faraday’s Law, \(1 \mathrm{~mol}\) of electrons carries a charge: \(Q = n \cdot F\), where \(F = 96485 C/mol\). So in our case, the charge transferred is equivalent to \(0.0466 \cdot F\) Hence, we calculate \(x = Q/F\), where \(x\) is the charge on Pt ions that we look for and \(Q\) is the total charge transferred (18000 C). Therefore, \(x = Q/ (n \cdot F)\). Plugging in values we find that \(x = 18000 C / (0.0466 \cdot 96485) \approx 4 \) Therefore, the charge on the Pt ions is 4.

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.

Start your free trial

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

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

A galvanic cell is constructed as follows. One halfcell consists of a platinum wire immersed in a solution containing \(1.0 M \mathrm{Sn}^{2+}\) and \(1.0 M \mathrm{Sn}^{4+} ;\) the other half-cell has a thallium rod immersed in a solution of \(1.0 M \mathrm{TI}^{+}\). (a) Write the half-cell reactions and the overall reaction. (b) What is the equilibrium constant at \(25^{\circ} \mathrm{C} ?\) (c) What is the cell voltage if the \(\mathrm{TI}^{+}\) concentration is increased tenfold? \(\left(E_{\mathrm{T} 1^{+} / \mathrm{T} 1}^{\circ}=-0.34 \mathrm{~V} .\right)\)

When \(25.0 \mathrm{~mL}\) of a solution containing both \(\mathrm{Fe}^{2+}\) and \(\mathrm{Fe}^{3+}\) ions is titrated with \(23.0 \mathrm{~mL}\) of \(0.0200 \mathrm{M}\) \(\mathrm{KMnO}_{4}\) (in dilute sulfuric acid), all of the \(\mathrm{Fe}^{2+}\) ions are oxidized to \(\mathrm{Fe}^{3+}\) ions. Next, the solution is treated with \(Z\) n metal to convert all of the \(\mathrm{Fe}^{3+}\) ions to \(\mathrm{Fe}^{2+}\) ions. Finally, \(40.0 \mathrm{~mL}\) of the same \(\mathrm{KMnO}_{4}\) solution are added to the solution in order to oxidize the \(\mathrm{Fe}^{2+}\) ions to \(\mathrm{Fe}^{3+}\). Calculate the molar concentrations of \(\mathrm{Fe}^{2+}\) and \(\mathrm{Fe}^{3+}\) in the original solution.

A \(9.00 \times 10^{2}-\mathrm{mL} 0.200 \mathrm{M} \mathrm{MgI}_{2}\) was electrolyzed. As a result, hydrogen gas was generated at the cathode and iodine was formed at the anode. The volume of hydrogen collected at \(26^{\circ} \mathrm{C}\) and \(779 \mathrm{mmHg}\) was \(1.22 \times 10^{3} \mathrm{~mL}\). (a) Calculate the charge in coulombs consumed in the process. (b) How long (in min) did the electrolysis last if a current of 7.55 A was used? (c) A white precipitate was formed in the process. What was it and what was its mass in grams? Assume the volume of the solution was constant.

In recent years there has been much interest in electric cars. List some advantages and disadvantages of electric cars compared to automobiles with internal combustion engines.

A galvanic cell using \(\mathrm{Mg} / \mathrm{Mg}^{2+}\) and \(\mathrm{Cu} / \mathrm{Cu}^{2+}\) half-cell's operates under standard-state conditions at \(25^{\circ} \mathrm{C}\) and each compartment has a volume of \(218 \mathrm{~mL}\). The cell delivers 0.22 A for \(31.6 \mathrm{~h}\). (a) How many grams of \(\mathrm{Cu}\) are deposited? (b) What is the \(\left[\mathrm{Cu}^{2+}\right]\) remaining?
See all solutions

Recommended explanations on Chemistry Textbooks

Making Measurements

Read Explanation

Chemistry Branches

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Kinetics

Read Explanation

Chemical Reactions

Read Explanation

The Earths Atmosphere

Read Explanation
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