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Chapter 18: Problem 94

Electrolysis of an alkaline earth metal chloride using a current of \(5.00 \mathrm{~A}\) for \(748 \mathrm{~s}\) deposits \(0.471 \mathrm{~g}\) of metal at the cathode. What is the identity of the alkaline earth metal chloride?

Short Answer

Expert verified
The identity of the alkaline earth metal chloride is magnesium chloride (MgCl2). This is determined by using Faraday's laws of electrolysis and calculating the molar mass of the deposited metal, which is approximately 24.32 g/mol, close to the molar mass of magnesium (24.305 g/mol).

Step by step solution

01

Gather information from the problem statement

We know the following information: 1. Current, I = 5.00 A 2. Time, t = 748 s 3. Mass of metal deposited, m = 0.471 g
02

Calculate the charge passed during the electrolysis

We can use the formula \(Q = It\) to find the charge passed during the electrolysis. So, \(Q = 5.00 \mathrm{~A} \times 748 \mathrm{~s} = 3740 \mathrm{~C}\)
03

Calculate the equivalent weight of the metal

According to Faraday's second law of electrolysis, the weight of a substance deposited during electrolysis is proportional to its equivalent weight, that is: \(W = Ze\) Where: - W is the weight of the substance deposited, - Z is the electrochemical equivalent (ECE), - e is the charge passed. We can calculate the electrochemical equivalent (ECE) using the formula: \(Z = \frac{W}{e}\) Substituting the given values, we get: \(Z = \frac{0.471 \mathrm{~g}}{3740 \mathrm{~C}} = 1.259 \times 10^{-4} \mathrm{\frac{g}{C}}\)
04

Calculate the molar mass of the metal

Since the alkaline earth metals have a valency of 2, the molar mass of the metal is twice the equivalent weight. So, Molar mass of metal = \(2 \times Z \times Faraday's \,\, constant\), where Faraday's constant (F) is \(96485 \mathrm{\frac{C}{mol}}\). Molar mass of metal = \(2 \times 1.259 \times 10^{-4} \mathrm{\frac{g}{C}} \times 96485 \mathrm{\frac{C}{mol}}\) Molar mass of metal \( = 24.32 \mathrm{~g/mol}\)
05

Identify the alkaline earth metal

The molar mass of the metal is approximately 24.32 g/mol. This value is very close to the molar mass of magnesium (24.305 g/mol). So, the identity of the alkaline earth metal chloride is magnesium chloride (MgCl2).

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Most popular questions from this chapter

Estimate \(\mathscr{C}^{\circ}\) for the half-reaction $$2 \mathrm{H}_{2} \mathrm{O}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_{2}+2 \mathrm{OH}^{-}$$ given the following values of \(\Delta G_{\mathrm{f}}^{\circ}\) : $$\begin{aligned}\mathrm{H}_{2} \mathrm{O}(l) &=-237 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{H}_{2}(g) &=0.0 \\\\\mathrm{OH}^{-}(a q) &=-157 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{e}^{-} &=0.0\end{aligned}$$ Compare this value of \(\mathscr{E}^{\circ}\) with the value of \(\mathscr{b}^{\circ}\) given in Table \(18.1\).

In making a specific galvanic cell, explain how one decides on the electrodes and the solutions to use in the cell.

A galvanic cell is based on the following half-reactions: $$\begin{array}{ll}\mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe}(s) & \mathscr{E}^{\circ}=-0.440 \mathrm{~V} \\ 2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_{2}(g) & \mathscr{E}^{\circ}=0.000 \mathrm{~V} \end{array}$$ where the iron compartment contains an iron electrode and \(\left[\mathrm{Fe}^{2+}\right]=1.00 \times 10^{-3} M\) and the hydrogen compartment contains a platinum electrode, \(P_{\mathrm{H}_{2}}=1.00 \mathrm{~atm}\), and a weak acid, \(\mathrm{HA}\), at an initial concentration of \(1.00 M .\) If the observed cell potential is \(0.333 \mathrm{~V}\) at \(25^{\circ} \mathrm{C}\), calculate the \(K_{\mathrm{a}}\) value for the weak acid HA.

A disproportionation reaction involves a substance that acts as both an oxidizing and a reducing agent, producing higher and lower oxidation states of the same element in the products. Which of the following disproportionation reactions are spontaneous under standard conditions? Calculate \(\Delta G^{\circ}\) and \(K\) at \(25^{\circ} \mathrm{C}\) for those reactions that are spontaneous under standard conditions. a. \(2 \mathrm{Cu}^{+}(a q) \rightarrow \mathrm{Cu}^{2+}(a q)+\mathrm{Cu}(s)\) b. \(3 \mathrm{Fe}^{2+}(a q) \rightarrow 2 \mathrm{Fe}^{3+}(a q)+\mathrm{Fe}(s)\) c. \(\mathrm{HClO}_{2}(a q) \rightarrow \mathrm{ClO}_{3}^{-}(a q)+\mathrm{HClO}(a q) \quad\) (unbalanced) Use the half-reactions: \(\mathrm{ClO}_{3}^{-}+3 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{HClO}_{2}+\mathrm{H}_{2} \mathrm{O} \quad \mathscr{E}^{\circ}=1.21 \mathrm{~V}\) \(\mathrm{HClO}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{HClO}+\mathrm{H}_{2} \mathrm{O} \quad \mathscr{E}^{\circ}=1.65 \mathrm{~V}\)

For the following half-reaction, \(\mathscr{E}^{\circ}=-2.07 \mathrm{~V}\) : $$\mathrm{AlF}_{6}^{3-}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Al}+6 \mathrm{~F}^{-}$$ Using data from Table \(18.1\), calculate the equilibrium constant at \(25^{\circ} \mathrm{C}\) for the reaction $$\mathrm{Al}^{3+}(a q)+6 \mathrm{~F}^{-}(a q) \rightleftharpoons \mathrm{AlF}_{6}^{3-}(a q) \quad K=?$$
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