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Chapter 17: Problem 25

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.

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Most popular questions from this chapter

A galvanic cell is based on the following half-reactions: $$ \begin{array}{ll} \mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe}(s) & \mathscr{E}^{\circ}=-0.440 \mathrm{V} \\ 2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_{2}(g) & \mathscr{E}^{\circ}=0.000 \mathrm{V} \end{array} $$ where the iron compartment contains an iron electrode and \(\left[\mathrm{Fe}^{2+}\right]=1.00 \times 10^{-3} \mathrm{M}\) and the hydrogen compartment contains a platinum electrode, \(P_{\mathrm{H}_{2}}=1.00\) atm, and a weak acid, HA, at an initial concentration of \(1.00 \mathrm{M}\). If the observed cell potential is \(0.333 \mathrm{V}\) at \(25^{\circ} \mathrm{C},\) calculate the \(K_{\mathrm{a}}\) value for the weak acid HA.

Consider the cell described below: $$ \mathrm{Al}\left|\mathrm{Al}^{3+}(1.00 M)\right|\left|\mathrm{Pb}^{2+}(1.00 M)\right| \mathrm{Pb} $$ Calculate the cell potential after the reaction has operated long enough for the \(\left[\mathrm{Al}^{3+}\right]\) to have changed by \(0.60 \mathrm{mol} / \mathrm{L}\). (Assume \(\left.T=25^{\circ} \mathrm{C} .\right)\)

Calculate \(\mathscr{E}^{\circ}\) values for the following cells. Which reactions are spontaneous as written (under standard conditions)? Balance the equations. Standard reduction potentials are found in Table \(17-1\) a. \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{I}^{-}(a q) \longrightarrow \mathrm{I}_{2}(a q)+\mathrm{Mn}^{2+}(a q)\) b. \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{F}^{-}(a q) \longrightarrow \mathrm{F}_{2}(g)+\mathrm{Mn}^{2+}(a q)\)

An experimental fuel cell has been designed that uses carbon monoxide as fuel. The overall reaction is $$ 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g) $$ The two half-cell reactions are $$ \begin{array}{c} \mathrm{CO}+\mathrm{O}^{2-} \longrightarrow \mathrm{CO}_{2}+2 \mathrm{e}^{-} \\\ \mathrm{O}_{2}+4 \mathrm{e}^{-} \longrightarrow 2 \mathrm{O}^{2-} \end{array} $$ The two half-reactions are carried out in separate compartments connected with a solid mixture of \(\mathrm{CeO}_{2}\) and \(\mathrm{Gd}_{2} \mathrm{O}_{3}\). \(\mathrm{Ox}\) ide ions can move through this solid at high temperatures (about \(800^{\circ} \mathrm{C}\) ). \(\Delta G\) for the overall reaction at \(800^{\circ} \mathrm{C}\) under certain concentration conditions is -380 kJ. Calculate the cell potential for this fuel cell at the same temperature and concentration conditions.

An aqueous solution of \(\mathrm{PdCl}_{2}\) is electrolyzed for 48.6 seconds, and during this time 0.1064 g of Pd is deposited on the cathode. What is the average current used in the electrolysis?
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