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Chapter 20: Problem 108

Magnesium is obtained by electrolysis of molten \(\mathrm{MgCl}_{2}\) . (a) Why is an aqueous solution of MgCl_ not used in the electrolysis? (b) Several cells are connected in parallel by very large copper bars that convey current to the cells. Assuming that the cells are 96\(\%\) efficient in producing the desired products in electrolysis, what mass of Mg is formed by passing a current of \(97,000\) A for a period of 24 \(\mathrm{h} ?\)

Short Answer

Expert verified
(a) An aqueous solution of \(\mathrm{MgCl}_{2}\) is not used in the electrolysis because water will undergo reactions instead of the Mg ions, producing hydrogen and chlorine gas. (b) The mass of Mg formed by passing a current of 97,000 A for a period of 24 hours with 96% efficiency is 1,018,000 g or 1,018 kg.

Step by step solution

01

Answer for (a)

In the electrolysis of an aqueous solution of \(\mathrm{MgCl}_{2}\), water will also undergo reactions. This is because water can be more easily reduced and oxidized than Mg ions. Therefore, the desired product (Mg metal) will not be formed in an aqueous solution. Instead, hydrogen gas and chlorine gas will be produced at the cathode and anode, respectively.
02

Formulae needed for calculations

To calculate the mass of Mg formed in part (b), we will use Faraday's law of electrolysis, which states that the mass of a substance formed during electrolysis is proportional to the charge passed through the electrolyte. \(m = \dfrac{Q\times M}{nF}\) Where: - \(m\) is the mass of the substance formed (in this case, Mg) - \(Q\) is the total charge passed through the electrolyte (Coulombs) - \(M\) is the molar mass of the substance (Mg) - \(n\) is the number of electrons involved in the reaction - \(F\) is Faraday's constant (96485 C/mol)
03

Conversion of current and time to total charge

First, we need to find the total charge passed through the electrolyte (Q). We can do this by multiplying the current (I) and the time (t): \(Q=I\times t\) Current (I) = 97,000 A Time (t) = 24 hours = 86,400 seconds (1 hour = 3600 seconds) \(Q = 97000 A \times 86400 s = 8,378,800,000 C\)
04

Calculation of mass formed

Now, we can use Faraday's law to find the mass of Mg formed: For Mg, the molar mass (M) is 24.3 g/mol and the number of electrons involved in the reaction (n) is 2 (as \(\mathrm{Mg^{2+}}\) gains 2 electrons to form Mg metal). \(m = \dfrac{Q\times M}{nF} = \dfrac{8,378,800,000 C\times 24.3 g/mol}{2\times 96485 C/mol} = 1060620 g \) Since the cells are 96% efficient, we must multiply the mass by the efficiency factor: Mass of Mg = \(1060620 g \times 0.96 = 1018,000 g\) or 1018 kg
05

Final answer

(a) An aqueous solution of \(\mathrm{MgCl}_{2}\) is not used in the electrolysis because water will undergo reactions instead of the Mg ions, producing hydrogen and chlorine gas. (b) The mass of Mg formed by passing a current of 97,000 A for a period of 24 hours with 96% efficiency is 1,018,000 g or 1,018 kg.

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Most popular questions from this chapter

Using the standard reduction potentials listed in Appendix E, calculate the equilibrium constant for each of the following reactions at \(298 \mathrm{K} :\) $$ \begin{array}{l}{\text { (a) } \mathrm{Fe}(s)+\mathrm{Ni}^{2+}(a q) \longrightarrow \mathrm{Fe}^{2+}(a q)+\mathrm{Ni}(s)} \\ {\text { (b) } \mathrm{Co}(s)+2 \mathrm{H}^{+}(a q) \longrightarrow \mathrm{Co}^{2+}(a q)+\mathrm{H}_{2}(g)} \\ {\text { (c) } 10 \mathrm{Br}^{-}(a q)+2 \mathrm{MnO}_{4}^{-}(a q)+16 \mathrm{H}^{+}(a q) \rightarrow} \\ {\quad 2 \mathrm{Mn}^{2+}(a q)+8 \mathrm{H}_{2} \mathrm{O}(l)+5 \mathrm{Br}_{2}(l)}\end{array} $$

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