Chapter 18: Problem 98
A factory wants to produce \(1.00 \times 10^{3} \mathrm{~kg}\) barium from the electrolysis of molten barium chloride. What current must be applied for \(4.00 \mathrm{~h}\) to accomplish this?
Chapter 18: Problem 98
A factory wants to produce \(1.00 \times 10^{3} \mathrm{~kg}\) barium from the electrolysis of molten barium chloride. What current must be applied for \(4.00 \mathrm{~h}\) to accomplish this?
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Get started for freeHydrazine is somewhat toxic. Use the half-reactions shown below to explain why household bleach (a highly alkaline solution of sodium hypochlorite) should not be mixed with household ammonia or glass cleansers that contain ammonia. \(\mathrm{ClO}^{-}+\mathrm{H}_{2} \mathrm{O}+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{OH}^{-}+\mathrm{Cl}^{-} \quad \mathscr{E}^{\circ}=0.90 \mathrm{~V}\) \(\mathrm{N}_{2} \mathrm{H}_{4}+2 \mathrm{H}_{2} \mathrm{O}+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{NH}_{3}+2 \mathrm{OH}^{-} \quad \mathscr{E}^{\circ}=-0.10 \mathrm{~V}\)
Nerve impulses are electrical "signals" that pass through neurons in the body. The electrical potential is created by the differences in the concentration of \(\mathrm{Na}^{+}\) and \(\mathrm{K}^{+}\) ions across the nerve cell membrane. We can think about this potential as being caused by a concentration gradient, similar to what we see in a concentration cell (keep in mind that this is a very simple explanation of how nerves work; there is much more involved in the true biologic process). A typical nerve cell has a resting potential of about \(-70 \mathrm{mV}\). Let's assume that this resting potential is due only to the \(\mathrm{K}^{+}\) ion concentration difference. In nerve cells, the \(\mathrm{K}^{+}\) concentration inside the cell is larger than the \(\mathrm{K}^{+}\) concentration outside the cell. Calculate the \(\mathrm{K}^{+}\) ion concentration ratio necessary to produce a resting potential of \(-70 . \mathrm{mV}\). $$\frac{\left[\mathrm{K}^{+}\right]_{\text {inside }}}{\left[\mathrm{K}^{+}\right]_{\text {outside }}}=?$$
A galvanic cell is based on the following half-reactions at \(25^{\circ} \mathrm{C}\) : $$\begin{aligned}\mathrm{Ag}^{+}+\mathrm{e}^{-} & \longrightarrow \mathrm{Ag} \\\ \mathrm{H}_{2} \mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} & \longrightarrow 2 \mathrm{H}_{2} \mathrm{O} \end{aligned}$$ Predict whether \(\mathscr{G}_{\text {cell }}\) is larger or smaller than \(\mathscr{C}_{\text {cell }}^{\circ}\) for the following cases. a. \(\left[\mathrm{Ag}^{+}\right]=1.0 \mathrm{M},\left[\mathrm{H}_{2} \mathrm{O}_{2}\right]=2.0 \mathrm{M},\left[\mathrm{H}^{+}\right]=2.0 \mathrm{M}\) b. \(\left[\mathrm{Ag}^{+}\right]=2.0 \mathrm{M},\left[\mathrm{H}_{2} \mathrm{O}_{2}\right]=1.0 M,\left[\mathrm{H}^{+}\right]=1.0 \times 10^{-7} M\)
Given the following two standard reduction potentials, $$\mathrm{M}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{M} \quad \mathscr{E}^{\circ}=-0.10 \mathrm{~V}$$ $$\mathrm{M}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{M} \quad \mathscr{E}^{\circ}=-0.50 \mathrm{~V}$$ solve for the standard reduction potential of the half- reaction$$\mathrm{M}^{3+}+\mathrm{e}^{-} \longrightarrow \mathrm{M}^{2+}$$
It takes \(15 \mathrm{kWh}\) (kilowatt-hours) of electrical energy to produce 1.0 kg aluminum metal from aluminum oxide by the HallHeroult process. Compare this to the amount of energy necessary to melt \(1.0 \mathrm{~kg}\) aluminum metal. Why is it economically feasible to recycle aluminum cans?
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