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Chapter 19: Problem 45

The half-reaction at an electrode is $$ \mathrm{Mg}^{2+}(\text { molten })+2 e^{-} \longrightarrow \mathrm{Mg}(s) $$ Calculate the number of grams of magnesium that can be produced by supplying \(1.00 \mathrm{~F}\) to the electrode.

Short Answer

Expert verified
The mass of magnesium that can be produced is approximately 12.15 grams.

Step by step solution

01

Understanding Faraday's Constant

Faraday's constant (\( F \)) denotes the amount of electric charge per one mole of electrons. It has a value of approximately \( 96,485 \) coulombs. In this problem, it is given that exactly \( 1.00 \) faraday of charge is supplied to the electrode, which means that \( 1.00 \) mole of electrons are supplied.
02

Calculating the Amount of Magnesium

According to the provided half-reaction equation, two electrons (\( 2e^{-} \)) are needed to produce one mole of magnesium (\( \mathrm{Mg} \)). Therefore, since \( 1.00 \) mole of electrons are supplied, this will produce \( 1.00/2 = 0.50 \) mole of magnesium.
03

Calculating the Mass of Magnesium

Knowing the molar amount of a substance allows us to easily calculate its mass. The molar mass of magnesium is approximately \( 24.3 \) grams per mole. So, the mass of magnesium that can be obtained is \( 0.50 ~ \mathrm{mole} \times 24.3 ~ \mathrm{g/mole} = 12.15 ~ \mathrm{g} \)

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Most popular questions from this chapter

To remove the tarnish \(\left(\mathrm{Ag}_{2} \mathrm{~S}\right)\) on a silver spoon, a student carried out the following steps. First, she placed the spoon in a large pan filled with water so the spoon was totally immersed. Next, she added a few tablespoonfuls of baking soda (sodium bicarbonate), which readily dissolved. Finally, she placed some aluminum foil at the bottom of the pan in contact with the spoon and then heated the solution to about \(80^{\circ} \mathrm{C}\). After a few minutes, the spoon was removed and rinsed with cold water. The tarnish was gone and the spoon regained its original shiny appearance. (a) Describe with equations the electrochemical basis for the procedure. (b) Adding \(\mathrm{NaCl}\) instead of \(\mathrm{NaHCO}_{3}\) would also work because both compounds are strong electrolytes. What is the added advantage of using \(\mathrm{NaHCO}_{3}\) ? (Hint: Consider the \(\mathrm{pH}\) of the solution.) (c) What is the purpose of heating the solution? (d) Some commercial tarnish removers containing a fluid (or paste) that is a dilute \(\mathrm{HCl}\) solution. Rubbing the spoon with the fluid will also remove the tarnish. Name two disadvantages of using this procedure compared to the one described here.

Which species in each pair is a better reducing agent under standard-state conditions? (a) \(\mathrm{Na}\) or \(\mathrm{Li},\) (b) \(\mathrm{H}_{2}\) or \(\mathrm{I}_{2},\) (c) \(\mathrm{Fe}^{2+}\) or \(\mathrm{Ag}\), (d) \(\mathrm{Br}^{-}\) or \(\mathrm{Co}^{2+}\).

What is Faraday's contribution to quantitative electrolysis?

From the following information, calculate the solubility product of AgBr: $$ \begin{array}{ll} \mathrm{Ag}^{+}(a q)+e^{-} \longrightarrow \mathrm{Ag}(s) & E^{\circ}=0.80 \mathrm{~V} \\ \operatorname{AgBr}(s)+e^{-} \longrightarrow \mathrm{Ag}(s)+\mathrm{Br}^{-}(a q) & E^{\circ}=0.07 \mathrm{~V} \end{array} $$

For a number of years it was not clear whether mercury(I) ions existed in solution as \(\mathrm{Hg}^{+}\) or as \(\mathrm{Hg}_{2}^{2+}\). To distinguish between these two possibilities, we could set up the following system: $$ \mathrm{Hg}(l) \mid \text { soln } \mathrm{A} \| \text { soln } \mathrm{B} \mid \mathrm{Hg}(l) $$ where soln A contained 0.263 g mercury(I) nitrate per liter and soln B contained 2.63 g mercury(I) nitrate per liter. If the measured emf of such a cell is \(0.0289 \mathrm{~V}\) at \(18^{\circ} \mathrm{C},\) what can you deduce about the nature of the mercury(I) ions?
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