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Chapter 20: Problem 93

(a) Calculate the mass of Li formed by electrolysis of molten LiCl by a current of \(7.5 \times 10^{4}\) A flowing for a period of 24 \(\mathrm{h} .\) Assume the electrolytic cell is 85\(\%\) efficient. (b) What is the minimum voltage required to drive the reaction?

Short Answer

Expert verified
The mass of Li formed by the electrolysis of molten LiCl under these conditions is approximately \(395,937.61 g\), and the minimum voltage required to drive the reaction is \(3.04 V\).

Step by step solution

01

Calculate the total charge passed through the cell

Given the current, \(I=7.5 \times 10^4 A\), and the time period, \(t=24h\). The total charge passed through the cell, \(Q\), can be calculated using the formula: \[Q = It\] First, convert the given time to seconds since the unit of current is Amperes (A), which is coulombs per second (C/s). \[t=24h \times 60\frac{min}{h} \times 60\frac{s}{min} = 86400 s\] Now, calculate the total charge \(Q\): \[Q = (7.5 \times 10^4 A)(86400s) = 6.48 \times 10^9 C\]
02

Determine the number of moles of electrons

To find the number of moles of electrons passed through the cell, we will use Faraday's constant, \(F = 96485 C/mol\). The number of moles of electrons, \(n_e\), can be calculated as follows: \[n_e = \frac{Q}{F}\] So, \[n_e = \frac{6.48 \times 10^9 C}{96485 \frac{C}{mol}} \approx 67149.95 mol\]
03

Calculate the number of moles of Li produced

Given that the cell is 85% efficient, we can calculate the number of moles of Li produced, \(n_{Li}\), as follows: \[n_{Li} = efficiency \times n_e\] So, \[n_{Li} = (0.85)(67149.95 mol) \approx 57077.46 mol\]
04

Determine the mass of Li formed

Now that we know the number of moles of Li produced, we can find the mass, \(m_{Li}\), by multiplying it with the molar mass of Li, which is 6.94 g/mol: \[m_{Li} = n_{Li} \times M_{Li}\] So, \[m_{Li} = (57077.46 mol)(6.94 \frac{g}{mol}) \approx 395937.61 g\]
05

Calculate the minimum voltage required

To find the minimum voltage required to drive the reaction, we can use the standard potential of the Li/Li+ half-cell, \(E^0 = -3.04 V\). Because the reaction involves one mole of electrons, the minimum voltage required for the electrolysis of molten LiCl, \(V_{min}\), is equal to the magnitude of the standard potential: \[V_{min} = |-3.04 V| = 3.04 V\] So, the minimum voltage required to drive the reaction is 3.04 V. In summary, the mass of Li formed by the electrolysis of molten LiCl under these conditions is approximately 395,937.61 g, and the minimum voltage required to drive the reaction is 3.04 V.

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Most popular questions from this chapter

Using standard reduction potentials (Appendix E), calculate the standard emf for each of the following reactions: $$ \begin{array}{l}{\text { (a) } \mathrm{Cl}_{2}(g)+2 \mathrm{I}^{-}(a q) \longrightarrow 2 \mathrm{Cl}^{-}(a q)+\mathrm{I}_{2}(s)} \\ {\text { (b) } \mathrm{Ni}(s)+2 \mathrm{Ce}^{4+}(a q) \longrightarrow \mathrm{Ni}^{2+}(a q)+2 \mathrm{Ce}^{3+}(a q)} \\ {\text { (c) } \mathrm{Fe}(s)+2 \mathrm{Fe}^{3+}(a q) \longrightarrow 3 \mathrm{Fe}^{2+}(a q)} \\ {\text { (d) } 2 \mathrm{NO}_{3}^{-}(a q)+8 \mathrm{H}^{+}(a q)+3 \mathrm{Cu}(s) \longrightarrow 2 \mathrm{NO}(g)+} \\ \quad {4 \mathrm{H}_{2} \mathrm{O}(l)+3 \mathrm{Cu}^{2+}(a q)}\end{array} $$

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A voltaic cell is constructed that uses the following half-cell reactions: $$ \begin{array}{c}{\mathrm{Cu}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s)} \\ {\mathrm{I}_{2}(s)+2 \mathrm{e}^{-} \quad \longrightarrow 2 \mathrm{I}^{-}(a q)}\end{array} $$ The cell is operated at 298 \(\mathrm{K}\) with \(\left[\mathrm{Cu}^{+}\right]=0.25 M\) and \(\left[\mathrm{I}^{-}\right]=0.035 \mathrm{M}\) (a) Determine \(E\) for the cell at these concentrations. (b) Which electrode is the anode of the cell? (c) Is the answer to part (b) the same as it would be if the cell were operated under standard conditions? (d) If \(\left[\mathrm{Cu}^{+}\right]\) were equal to \(0.15 \mathrm{M},\) at what concentration of I \(\mathrm{I}^{-}\) would the cell have zero potential?

Iron corrodes to produce rust, \(\mathrm{Fe}_{2} \mathrm{O}_{3},\) but other corrosion products that can form are \(\mathrm{Fe}(\mathrm{O})(\mathrm{OH}),\) iron oxyhydroxide, and magnetite, \(\mathrm{Fe}_{3} \mathrm{O}_{4}\) . (a) What is the oxidation number of Fe in iron oxyhydroxide, assuming oxygen's oxidation number is \(-2 ?\) (b) The oxidation number for Fe in magnetite was controversial for a long time. If we assume that oxygen's oxidation number is \(-2,\) and Fe has a unique oxidation number, what is the oxidation number for Fe in magnetite? (c) It turns out that there are two different kinds of Fe in magnetite that have different oxidation numbers. Suggest what these oxidation numbers are and what their relative stoichiometry must be, assuming oxygen's oxidation number is - 2 .

A voltaic cell that uses the reaction $$ \operatorname{PdCl}_{4}^{2-}(a q)+\mathrm{Cd}(s) \longrightarrow \mathrm{Pd}(s)+4 \mathrm{Cl}^{-}(a q)+\mathrm{Cd}^{2+}(a q) $$ has a measured standard cell potential of \(+1.03 \mathrm{V}\) . (a) Write the two half-cell reactions. (b) By using data from Appendix E, determine \(E_{\text { red }}^{\circ}\) for the reaction involving Pd. (c) Sketch the voltaic cell, label the anode and cathode, and indicate the direction of electron flow.
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