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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 electrolysis of molten LiCl by a current of \(7.5 \times 10^{4} \mathrm{A}\) flowing for a period of 24 hours, assuming the electrolytic cell is 85% efficient, is approximately \(3.96 \times 10^{5} \mathrm{g}\). The minimum voltage required to drive the reaction is \(3.04 \mathrm{V}\).

Step by step solution

01

Find the total charge passed during electrolysis

The total charge (Q) passed during electrolysis can be found using the formula: \[ Q = I \times t \] where \( I \) is the current in amperes and \( t \) is the time in seconds. Given current \( I = 7.5 \times 10^{4} \mathrm{~A} \) and time \( t = 24 \mathrm{~hours} \). We need to convert the time into seconds: \[ t = 24 \mathrm{~h} \times \frac{3600 \mathrm{~s}}{1 \mathrm{~h}} = 86400 \mathrm{~s} \] Now, calculate the total charge: \[ Q = (7.5 \times 10^{4} \mathrm{~A}) \times (86400 \mathrm{~s}) = 6.48 \times 10^{9} \mathrm{~C} \]
02

Calculate the charge passed during Li formation considering cell efficiency

The cell is 85% efficient, so we need to find the effective charge (Q') that is used in Li formation by multiplying the total charge by the cell efficiency: \[Q' = Q \times \frac{85}{100} \] \[Q' = (6.48 \times 10^{9} \mathrm{~C}) \times 0.85 = 5.508 \times 10^{9} \mathrm{~C} \]
03

Use Faraday's laws of electrolysis to calculate the amount of substance of Li formed

From Faraday's laws of electrolysis, the amount of substance (in moles) of Li formed can be calculated using: \[ n = \frac{Q'}{zF} \] Where: - \( n \) is the amount of substance of lithium formed (in moles) - \( Q' \) is the effective charge in coulombs - \( z \) is the number of electrons transferred in the reduction half-reaction (for Li, \( z = 1 \) since the reaction is \( \mathrm{Li^{+} + e^{-} \rightarrow Li} \)) - \( F \) is the Faraday constant, which is \( 9.6485 \times 10^{4} \mathrm{~C/mol} \) Now, calculate the amount of substance of Li formed: \[ n = \frac{5.508 \times 10^{9} \mathrm{~C}}{(1)(9.6485 \times 10^{4} \mathrm{~C/mol})} = 5.703 \times 10^{4} \mathrm{~mol} \]
04

Calculate the mass of Li formed

The mass of lithium (m) formed can be calculated using: \[ m = n \times M \] Where: - \( m \) is the mass of lithium in grams - \( n \) is the amount of substance of lithium formed (from Step 3) - \( M \) is the molar mass of lithium, which is \( 6.94 \mathrm{~g/mol} \) Now, calculate the mass of Li formed: \[ m = (5.703 \times 10^{4} \mathrm{~mol}) \times (6.94 \mathrm{~g/mol}) = 3.96 \times 10^{5} \mathrm{~g} \] Thus, the mass of Li formed by electrolysis is \( 3.96 \times 10^{5} \mathrm{~g} \).
05

Calculate the minimum voltage required

The reduction half-reaction of Li is: \[ \mathrm{Li^{+} + e^{-} \rightarrow Li} \] The standard electrode potential for the reduction of Li is \( -3.04 \mathrm{~V} \) (data from the standard reduction potential table). However, since the given reaction is an electrolysis (non-spontaneous) reaction, the minimum voltage required is at least the positive value of the standard electrode potential, which is: \[ V_{min} = |-3.04 \mathrm{~V}| = 3.04 \mathrm{~V} \] So, the minimum voltage required to drive the reaction is \( 3.04 \mathrm{~V} \).

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Most popular questions from this chapter

An iron object is plated with a coating of cobalt to protect against corrosion. Does the cobalt protect iron by cathodic protection? Explain.

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(a) What is meant by the term reduction? (b) On which side of a reduction half-reaction do the electrons appear? (c) What is meant by the term reductant? (d) What is meant by the term reducing agent?

A voltaic cell is constructed that uses the following reaction and operates at \(298 \mathrm{~K}\) : $$\mathrm{Zn}(s)+\mathrm{Ni}^{2+}(a q) \longrightarrow \mathrm{Zn}^{2+}(a q)+\mathrm{Ni}(s)$$ (a) What is the emf of this cell under standard conditions? (b) What is the emf of this cell when \(\left[\mathrm{Ni}^{2+}\right]=3.00 \mathrm{M}\) and \(\left[\mathrm{Zn}^{2+}\right]=0.100 \mathrm{M} ?\) (c) What is the emf of the cell when \(\left[\mathrm{Ni}^{2+}\right]=0.200 \mathrm{M}\) and \(\left[\mathrm{Zn}^{2+}\right]=0.900 \mathrm{M} ?\)
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