Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 20: Problem 107

(a) How many coulombs are required to plate a layer of chromium metal 0.25 \(\mathrm{mm}\) thick on an auto bumper with a total area of 0.32 \(\mathrm{m}^{2}\) from a solution containing \(\mathrm{CrO}_{4}^{2-}\) ? The density of chromium metal is 7.20 \(\mathrm{g} / \mathrm{cm}^{3} .\) (b) What current flow is required for this electroplating if the bumper is to be plated in 10.0 s? (c) If the external source has an emf of \(+6.0 \mathrm{V}\) and the electrolytic cell is 65\(\%\) efficient, how much electrical power is expended to electroplate the bumper?

Short Answer

Expert verified
The number of coulombs required to plate a layer of chromium metal 0.25 mm thick on an auto bumper with a total area of 0.32 m² from a solution containing \(\mathrm{CrO}_{4}^{2-}\) is 6,407,281 C. The current flow required for this electroplating, if the bumper is to be plated in 10.0 s, is 640.728 kA. If the electrolytic cell is 65% efficient and the external source has an emf of +6.0 V, the electrical power expended to electroplate the bumper is approximately 5,910 kW.

Step by step solution

01

Calculate the volume and mass of Chromium metal required for electroplating

Given the thickness and total area of the required chromium layer, we can determine its volume. Use the density of chromium to calculate the mass of the metal required. Volume = (Total area) × (Thickness) Total area = \(0.32 m^2\) Thickness = \(0.25 mm\) = \(0.025 cm\) Volume = \((0.32 × 10^4) \text{cm}^2 × 0.025 \text{cm}\) = \(80 \text{cm}^3\) Now, let's calculate the mass. Mass = Volume × Density Density of Chromium = \(7.20 \frac{\text{g}}{\text{cm}^3}\) Mass = \(80 \text{cm}^3 × 7.20 \frac{\text{g}}{\text{cm}^3}\) = \(576 \,\text{g}\)
02

Calculate the moles of Chromium metal required for electroplating

To calculate the number of coulombs needed, we first need to find the moles of Chromium metal involved in the reaction. We can use the molar mass of Chromium to determine the number of moles. Moles = \(\frac{\text{Mass}}{\text{Molar Mass}}\) Molar mass of Chromium = \(51.996 \,\text{g/mol}\) Moles = \(\frac{576 \,\text{g}}{51.996 \,\text{g/mol}}\) = \(11.078 \, \text{mol}\)
03

Calculate the number of coulombs required

Now we can calculate the number of coulombs needed for electroplating using Faraday's law of electrolysis. Coulombs = Moles × Faraday's Constant × Number of electrons involved in the reaction In this case, the reaction is: \(\mathrm{CrO}_{4}^{2-} + 6e^{-} \rightarrow \mathrm{Cr} + 2 \, \mathrm{O}_{2(g)}\) Faraday's constant (F) = \(96,485 \, \frac{\text{C}}{\text{mol}}\) Number of electrons (n) = 6 Coulombs = \(11.078 \, \text{mol} × 96,485 \,\frac{\text{C}}{\text{mol}} × 6\) = \(6,407,281 \,\text{C}\)
04

Calculate the required current flow for electroplating

We are given the required time for plating the bumper as 10.0 s. We can calculate the current required for the process using Ohm's Law. Current (I) = \(\frac{\text{Coulombs}}{\text{Time}}\) Current = \(\frac{6,407,281 \, \text{C}}{10.0 \, \text{s}}\) = \(640,728 \, \text{A}\) or \(640.728 \, \text{kA}\)
05

Calculate the electrical power expended for electroplating

We are given the cell efficiency as 65%. Using the emf, efficiency, and current flow, we can calculate the electrical power expended for electroplating. Efficiency = \(\frac{\text{Output Power}}{\text{Input Power}}\) Output Power = 6.0 V × 640.728 kA Input Power = \(\frac{\text{Output Power}}{\text{Efficiency}}\) Input Power = \(\frac{6.0 \, \text{V}\ × 640.728 \, \text{kA}}{0.65}\) = \(6.0 \times \frac{640.728}{0.65} \, \text{kW}\) ≈ \(5,910 \, \text{kW}\) The electrical power expended to electroplate the bumper is approximately 5,910 kW.

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) Write the reactions for the discharge and charge of a nickel-cadmium (nicad) rechargeable battery. (b) Given the following reduction potentials, calculate the standard emf of the cell: $$ \begin{array}{r}{\operatorname{Cd}(\mathrm{OH})_{2}(s)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cd}(s)+2 \mathrm{OH}^{-}(a q)} \\\ {E_{\mathrm{red}}^{\circ}=-0.76 \mathrm{V}}\end{array} $$ $$ \begin{array}{r}{\mathrm{NiO}(\mathrm{OH})(s)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{e}^{-} \longrightarrow \mathrm{Ni}(\mathrm{OH})_{2}(s)+\mathrm{OH}^{-}(a q)} \\\ {E_{\mathrm{red}}^{\circ}=+0.49 \mathrm{V}}\end{array} $$ (c) A typical nicad voltaic cell generates an emf of \(+1.30 \mathrm{V}\) . Why is there a difference between this value and the one you calculated in part (b)? (d) Calculate the equilibrium constant for the overall nicad reaction based on this typical emf value.

(a) What is the difference between a battery and a fuel cell? (b) Can the "fuel" of a fuel cell be a solid?

(a) What is meant by the term reduction? (b) On which side of a reduction half-reaction do the electrons appear? (c) What is meant by the term reductant? (d) What is meant by the term reducing agent?

If the equilibrium constant for a one-electron redox reaction at 298 \(\mathrm{K}\) is \(8.7 \times 10^{4}\) , calculate the corresponding \(\Delta G^{\circ}\) and \(E^{\circ} .\)

A voltaic cell that uses the reaction $$ \operatorname{PdCl}_{4}^{2-}(a q)+\mathrm{Cd}(s) \longrightarrow \mathrm{Pd}(s)+4 \mathrm{Cl}^{-}(a q)+\mathrm{Cd}^{2+}(a q) $$ has a measured standard cell potential of \(+1.03 \mathrm{V}\) . (a) Write the two half-cell reactions. (b) By using data from Appendix E, determine \(E_{\text { red }}^{\circ}\) for the reaction involving Pd. (c) Sketch the voltaic cell, label the anode and cathode, and indicate the direction of electron flow.
See all solutions

Recommended explanations on Chemistry Textbooks

Nuclear Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemistry Branches

Read Explanation

Organic Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Making Measurements

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