When jump-starting a car with a dead battery, the ground jumper should be attached to a remote part of the engine block. Why?

Short Answer

Expert verified
When jump-starting a car with a dead battery, the ground jumper should be attached to a remote part of the engine block because it helps prevent sparking near the battery, which can be dangerous due to the potential explosion of hydrogen gas released from the battery. Additionally, it ensures a stable ground connection, reducing the risk of damage to electrical components and improving current transfer between vehicles for effective charging of the dead battery.

Step by step solution

01

Connect the positive jumper cable

First, connect the positive (red) jumper cable to the positive terminal on the dead battery. Make sure the cable is securely attached.
02

Connect the other end of the positive jumper cable

Connect the other end of the positive (red) jumper cable to the positive terminal on the working battery of the assisting vehicle.
03

Connect the negative jumper cable

Connect the negative (black) jumper cable to the negative terminal on the working battery of the assisting vehicle. Ensure the cable is securely attached.
04

Attach the ground jumper to the engine block

Instead of connecting the other end of the negative (black) jumper cable to the negative terminal on the dead battery, attach it to a remote part of the engine block (an unpainted metal surface). Doing this helps in preventing sparking near the battery, as hydrogen gas released from a battery might be explosive. Also, it ensures a stable ground connection, reducing the risk of damage to electrical components and improving current transfer between vehicles.
05

Start the assisting vehicle

Start the engine of the assisting vehicle and let it run for a few minutes to charge the dead battery.
06

Start the vehicle with the dead battery

Attempt to start the vehicle with the dead battery. If it starts, let it run for a few minutes to further charge the battery. If it does not start, wait a few more minutes and try again.
07

Disconnect the jumper cables

Once the dead-battery vehicle is running, disconnect the jumper cables in the reverse order of how you connected them. Start with the ground (black) cable from the engine block, then the black cable from the assisting vehicle, followed by the red cable from the assisting vehicle, and finally the red cable from the previously dead battery. Be cautious not to let the cable clamps touch each other while they are still connected to vehicles.

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!

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

What is electrochemistry? What are redox reactions? Explain the difference between a galvanic and an electrolytic cell.

A solution containing \(\mathrm{Pt}^{4+}\) is electrolyzed with a current of 4.00 A. How long will it take to plate out \(99 \%\) of the platinum in 0.50 L of a \(0.010-M\) solution of \(\mathrm{Pt}^{4+} ?\)

Which of the following statement(s) is/are true? a. Copper metal can be oxidized by \(\mathrm{Ag}^{+}\) (at standard conditions). b. In a galvanic cell the oxidizing agent in the cell reaction is present at the anode. c. In a cell using the half reactions \(A l^{3+}+3 e^{-} \longrightarrow A l\) and \(\mathrm{Mg}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Mg},\) aluminum functions as the anode. d. In a concentration cell electrons always flow from the compartment with the lower ion concentration to the compartment with the higher ion concentration. e. In a galvanic cell the negative ions in the salt bridge flow in the same direction as the electrons.

What is the maximum work that can be obtained from a hydrogen-oxygen fuel cell at standard conditions that produces \(1.00 \mathrm{kg}\) water at \(25^{\circ} \mathrm{C} ?\) Why do we say that this is the maximum work that can be obtained? What are the advantages and disadvantages in using fuel cells rather than the corresponding combustion reactions to produce electricity?

An electrochemical cell consists of a nickel metal electrode immersed in a solution with \(\left[\mathrm{Ni}^{2+}\right]=1.0 \mathrm{M}\) separated by a porous disk from an aluminum metal electrode immersed in a solution with \(\left[\mathrm{Al}^{3+}\right]=1.0 \mathrm{M} .\) Sodium hydroxide is added to the aluminum compartment, causing \(\mathrm{Al}(\mathrm{OH})_{3}(s)\) to precipitate. After precipitation of Al(OH) \(_{3}\) has ceased, the concentration of \(\mathrm{OH}^{-}\) is \(1.0 \times 10^{-4} \mathrm{M}\) and the measured cell potential is \(1.82 \mathrm{V}\). Calculate the \(K_{\mathrm{sp}}\) value for \(\mathrm{Al}(\mathrm{OH})_{3}\) $$ \mathrm{Al}(\mathrm{OH})_{3}(s) \rightleftharpoons \mathrm{Al}^{3+}(a q)+3 \mathrm{OH}^{-}(a q) \quad K_{\mathrm{sp}}=? $$

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