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?
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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?
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It took \(150 .\) s for a current of 1.25 A to plate out 0.109 g of a metal from a solution containing its cations. Show that it is not possible for the cations to have a charge of \(1+.\)
The overall reaction and standard cell potential at \(25^{\circ} \mathrm{C}\) for the rechargeable nickel-cadmium alkaline battery is $$\mathrm{Cd}(s)+\mathrm{NiO}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ni}(\mathrm{OH})_{2}(s)+\mathrm{Cd}(\mathrm{OH})_{2}(s) \quad \mathscr{E}^{\circ}=1.10 \mathrm{V} $$ For every mole of Cd consumed in the cell, what is the maximum useful work that can be obtained at standard conditions?
What reaction will take place at the cathode and the anode when each of the following is electrolyzed? a. molten KF b. molten \(\mathrm{CuCl}_{2}\) c. molten \(\mathrm{MgI}_{2}\)
The solubility product for \(\operatorname{CuI}(s)\) is \(1.1 \times 10^{-12} .\) Calculate the value of \(\mathscr{E}^{\circ}\) for the half-reaction $$ \operatorname{CuI}(s)+\mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s)+\mathrm{I}^{-}(a q) $$
You have a concentration cell with Cu electrodes and \(\left[\mathrm{Cu}^{2+}\right]\) \(=1.00 M\) (right side) and \(1.0 \times 10^{-4} M\) (left side). a. Calculate the potential for this cell at \(25^{\circ} \mathrm{C}\) b. The \(\mathrm{Cu}^{2+}\) ion reacts with \(\mathrm{NH}_{3}\) to form \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\) by the following equation: $$\begin{aligned} &\mathrm{Cu}^{2+}(a q)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q) & K=1.0 \times 10^{13} \end{aligned}$$ Calculate the new cell potential after enough \(\mathrm{NH}_{3}\) is added to the left cell compartment such that at equilibrium \(\left[\mathrm{NH}_{3}\right]=2.0 \mathrm{M}\)
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