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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 h. Assume the electrolytic cell is \(85 \%\) efficient. (b) What is the minimum voltage required to drive the reaction?

Short Answer

Expert verified
After applying Faraday's laws and accounting for the 85% efficiency, the mass of Li formed is calculated as \(1.98 \times 10^3\) g. The minimum voltage required to drive the reaction is 2.32 V.

Step by step solution

01

1. Calculate the total charge passed through the cell

To calculate the total charge passed through the cell, we will use the formula: Total charge (Q) = Current (I) x Time (t) The given current is \(7.5 \times 10^4\) A and the time is 24 hours. We need to convert the time into seconds before plugging the values into the formula. 1 hour = 3600 seconds So, 24 hours = 24 x 3600 = 86,400 seconds Now, we can calculate the total charge: Q = \(7.5 \times 10^4\) A x 86,400 s
02

2. Calculate the number of moles of electrons

Once we have the total charge, we can calculate the number of moles of electrons using Faraday's constant (F = 96,485 C/mol): Moles of electrons (n) = Total charge (Q) / Faraday's constant (F) n = Q / 96,485 C/mol
03

3. Apply the electrolytic cell efficiency

Given that the cell is 85% efficient, we now need to account for the efficiency in our calculations: Moles of electrons (n') = Moles of electrons (n) x 0.85
04

4. Calculate the moles of Lithium formed

Lithium is formed by the reduction of Li+ ions, which requires one mole of electrons for each mole of Li+ ions: Moles of Li formed = Moles of electrons (n')
05

5. Calculate the mass of Lithium formed

Now that we have the moles of Li formed, we can calculate the mass by using the molar mass of Li (6.94 g/mol): Mass of Li = Moles of Li x Molar mass of Li Mass of Li = Moles of Li x 6.94 g/mol
06

6. Calculate the minimum voltage required

For part (b), we're asked to find the minimum voltage required to drive the reaction. We can find this by using the standard electrode potentials of the half-reactions which are: Li+ + e- -> Li (E° = -3.04 V) 2Cl- -> Cl2 + 2e- (E° = +1.36 V) To calculate the minimum voltage required for the cell, we can use the formula: E°cell = E°cathode - E°anode E°cell = E°(Li) - E°(Cl) Now we can plug in the values to find the minimum voltage required for the electrolysis of molten LiCl.

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Most popular questions from this chapter

In a Li-ion battery the composition of the cathode is \(\mathrm{LiCoO}_{2}\) when completely discharged. On charging, approximately \(50 \%\) of the \(\mathrm{Li}^{+}\) ions can be extracted from the cathode and transported to the graphite anode where they intercalate between the layers. (a) What is the composition of the cathode when the battery is fully charged? (b) If the \(\mathrm{LiCoO}_{2}\) cathode has a mass of \(10 \mathrm{~g}\) (when fully discharged), how many coulombs of electricity can be delivered on completely discharging a fully charged battery?

Given the following half-reactions and associated standard reduction potentials: $$ \begin{aligned} \mathrm{AuBr}_{4}^{-}(a q)+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au}(s)+4 \mathrm{Br}^{-}(a q) & \\ E_{\mathrm{red}}^{\circ}=-0.86 \mathrm{~V} \\ \mathrm{Eu}^{3+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Eu}^{2+}(a q) & \\\ E_{\mathrm{red}}^{\circ}=-0.43 \mathrm{~V} \\ \mathrm{IO}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-} \longrightarrow \mathrm{I}^{-}(a q)+2 \mathrm{OH}^{-}(a q) & \\ E_{\mathrm{red}}^{\circ}=&+0.49 \mathrm{~V} \end{aligned} $$ (a) Write the equation for the combination of these halfcell reactions that leads to the largest positive emf and calculate the value. (b) Write the equation for the combination of half-cell reactions that leads to the smallest positive emf and calculate that value.

Mercuric oxide dry-cell batteries are often used where a flat discharge voltage and long life are required, such as in watches and cameras. The two half-cell reactions that occur in the battery are $$ \begin{array}{l} \mathrm{HgO}(s)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Hg}(l)+2 \mathrm{OH}^{-}(a q) \\ \mathrm{Zn}(s)+2 \mathrm{OH}^{-}(a q) \longrightarrow \mathrm{ZnO}(s)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-} \end{array} $$ (a) Write the overall cell reaction. (b) The value of $E_{\text {red }}^{\circ}\( for the cathode reaction is \)+0.098 \mathrm{~V}$. The overall cell potential is \(+1.35 \mathrm{~V}\). Assuming that both half-cells operate under standard conditions, what is the standard reduction potential for the anode reaction? (c) Why is the potential of the anode reaction different than would be expected if the reaction occurred in an acidic medium?

(a) Write the half-reaction that occurs at an oxygen electrode in acidic aqueous solution when it serves as the cathode of a voltaic cell. (b) Write the half-reaction that occurs at an oxygen electrode in acidic aqueous solution when it serves as the anode of a voltaic cell. (c) What is standard about the standard oxygen electrode?

(a) Which electrode of a voltaic cell, the cathode or the anode, corresponds to the higher potential energy for the electrons? (b) What are the units for electrical potential? How does this unit relate to energy expressed in joules?
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