Chapter 21: Problem 2
Why must an electrochemical process involve a redox reaction?
Short Answer
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Chapter 21: Problem 2
Why must an electrochemical process involve a redox reaction?
These are the key concepts you need to understand to accurately answer the question.
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Get started for freeHow many grams of aluminum can form by passing 305 C through an electrolytic cell containing a molten aluminum salt?
What does a negative \(E_{\mathrm{cell}}^{\circ}\) indicate about a redox reaction? What does it indicate about the reverse reaction?
To examine the effect of ion removal on cell voltage, a chemist constructs two voltaic cells, each with a standard hydrogen electrode in one compartment. One cell also contains a \(\mathrm{Pb} / \mathrm{Pb}^{2+}\) half-cell; the other contains a \(\mathrm{Cu} / \mathrm{Cu}^{2+}\) half-cell. (a) What is \(E^{\circ}\) of each cell at \(298 \mathrm{~K} ?\) (b) Which electrode in each cell is negative? (c) When \(\mathrm{Na}_{2} \mathrm{~S}\) solution is added to the \(\mathrm{Pb}^{2+}\) electrolyte, solid \(\mathrm{PbS}\) forms. What happens to the cell voltage? (d) When sufficient \(\mathrm{Na}_{2} \mathrm{~S}\) is added to the \(\mathrm{Cu}^{2+}\) electrolyte, CuS forms and \(\left[\mathrm{Cu}^{2+}\right]\) drops to \(1 \times 10^{-16} \mathrm{M} .\) Find the cell voltage.
Like any piece of apparatus, an electrolytic cell operates at less than \(100 \%\) efficiency. A cell depositing \(\mathrm{Cu}\) from a \(\mathrm{Cu}^{2+}\) bath operates for \(10 \mathrm{~h}\) with an average current of 5.8 A. If \(53.4 \mathrm{~g}\) of copper is deposited, at what efficiency is the cell operating?
Commercial electrolytic cells for producing aluminum operate at \(5.0 \mathrm{~V}\) and \(100,000 \mathrm{~A}\). (a) How long does it take to produce exactly 1 metric ton ( \(1000 \mathrm{~kg}\) ) of aluminum? (b) How much electrical power (in kilowatt-hours, \(\mathrm{kW} \cdot \mathrm{h}\) ) is used \(\left(1 \mathrm{~W}=1 \mathrm{~J} / \mathrm{s} ; 1 \mathrm{~kW} \cdot \mathrm{h}=3.6 \times 10^{3} \mathrm{~kJ}\right) ?\) (c) If electricity costs \(\$ 0.123\) per \(\mathrm{kW} \cdot \mathrm{h}\) and cell efficiency is \(90 . \%\) what is the cost of electricity to produce exactly 1 lb of aluminum?
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