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Chapter 18: Problem 133

A \(9.00 \times 10^{2}-\mathrm{mL} 0.200 \mathrm{M} \mathrm{MgI}_{2}\) was electrolyzed. As a result, hydrogen gas was generated at the cathode and iodine was formed at the anode. The volume of hydrogen collected at \(26^{\circ} \mathrm{C}\) and \(779 \mathrm{mmHg}\) was \(1.22 \times 10^{3} \mathrm{~mL}\). (a) Calculate the charge in coulombs consumed in the process. (b) How long (in min) did the electrolysis last if a current of \(7.55 \mathrm{~A}\) was used? (c) A white precipitate was formed in the process. What was it and what was its mass in grams? Assume the volume of the solution was constant.

Short Answer

Expert verified
The charge consumed in the process was 4824.25 C. The electrolysis lasted for 10.65 min. The white precipitate was magnesium hydroxide (Mg(OH)2) and its mass was approximately 10.498 g.

Step by step solution

01

Determine the moles of Hydrogen gas

Since Hydrogen gas (H2) was produced at the cathode, first calculate the number of moles of H2 gas using the gas law \(PV=nRT\), where P is the pressure, V is the volume, n is the number of moles, R is the gas law constant and T is the absolute temperature. Convert the pressure to atm by dividing by 760, and convert the temperature to Kelvin by adding 273 to the Celsius temperature. Plug in the known values and solve for n. \(PV=nRT \Rightarrow n=PV/RT = (1.022 atm)(1.22L)/(0.0821 L atm mol^-1 K^-1)(299K) = 0.050 mol H2 \).
02

Calculate the charge consumed

In the electrolysis, H2 is produced at the cathode via the half-reaction \(2H+ + 2e^- \Rightarrow H2\). We calculate the charge consumed by considering that 1 mol of electrons (1 Faraday = 96485 C) corresponds to 1 mol of H2. Therefore the total charge used in the process is given by \(Charge = n*Faraday = 0.05 mol * 96485 C/mol = 4824.25 C \).
03

Find the duration of the process

Since the current (I) is the charge flow per unit time (t), we can rearrange the equation \(I=Q/t\) to solve for time as \( t = Q/I = 4824.25 C / 7.55 A = 639 s = 10.65 min \).
04

Identify the precipitate and its mass

During electrolysis, iodine is produced at the anode by the half-reaction \(2I^- \Rightarrow I2 + 2e^-\). Therefore, the white precipitate resulting from the reaction is magnesium hydroxide (Mg(OH)2), which forms when magnesium ions in the MgI2 solution react with water to produce Mg(OH)2 and H+. By stoichiometry, 1 mol of MgI2 produces 1 mol of Mg(OH)2, and since the initial solution was 0.200M, we have 0.2 mol/L * 0.9 L = 0.18 mol of Mg(OH)2. Consequently, the mass of the precipitate is given by \( mass = n*MolarMass = 0.18 mol * 58.319 g/mol = 10.498 g. \).

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Most popular questions from this chapter

The magnitudes (but not the signs) of the standard reduction potentials of two metals \(X\) and \(Y\) are $$ \begin{array}{ll} \mathrm{Y}^{2+}+2 e^{-} \longrightarrow \mathrm{Y} & \mid E^{\mathrm{O}} \mathrm{I}=0.34 \mathrm{~V} \\ \mathrm{X}^{2+}+2 e^{-} \longrightarrow \mathrm{X} & \mid E^{\circ} \mathrm{I}=0.25 \mathrm{~V} \end{array} $$ where the II notation denotes that only the magnitude (but not the sign) of the \(E^{\circ}\) value is shown. When the half-cells of \(\mathrm{X}\) and \(\mathrm{Y}\) are connected, electrons flow from \(X\) to \(Y\). When \(X\) is connected to a SHE, electrons flow from \(X\) to \(\mathrm{SHE}\). (a) Are the \(E^{\circ}\) values of the half-reactions positive or negative? (b) What is the standard emf of a cell made up of \(\mathrm{X}\) and \(\mathrm{Y} ?\)

A silver rod and a SHE are dipped into a saturated aqueous solution of silver oxalate, \(\mathrm{Ag}_{2} \mathrm{C}_{2} \mathrm{O}_{4},\) at \(25^{\circ} \mathrm{C}\). The measured potential difference between the rod and the \(\mathrm{SHE}\) is \(0.589 \mathrm{~V},\) the rod being positive. Calculate the solubility product constant for silver oxalate.

Industrially, copper is purified by electrolysis. The impure copper acts as the anode, and the cathode is made of pure copper. The electrodes are immersed in a CuSO \(_{4}\) solution. During electrolysis, copper at the anode enters the solution as \(\mathrm{Cu}^{2+}\) while \(\mathrm{Cu}^{2+}\) ions are reduced at the cathode. (a) Write halfcell reactions and the overall reaction for the electrolytic process. (b) Suppose the anode was contaminated with \(\mathrm{Zn}\) and \(\mathrm{Ag} .\) Explain what happens to these impurities during electrolysis. (c) How many hours will it take to obtain \(1.00 \mathrm{~kg}\) of \(\mathrm{Cu}\) at a current of \(18.9 \mathrm{~A} ?\)

An acidified solution was electrolyzed using copper electrodes. A constant current of 1.18 A caused the anode to lose \(0.584 \mathrm{~g}\) after \(1.52 \times 10^{3} \mathrm{~s}\). (a) What is the gas produced at the cathode and what is its volume at STP? (b) Given that the charge of an electron is \(1.6022 \times 10^{-19} \mathrm{C},\) calculate Avogadro's number. Assume that copper is oxidized to \(\mathrm{Cu}^{2+}\) ions.

A concentration cell is constructed having Cu electrodes in two CuSO \(_{4}\) solutions \(\mathrm{A}\) and \(\mathrm{B}\). At \(25^{\circ} \mathrm{C}\), the osmotic pressures of the two solutions are 48.9 atm and 4.89 atm, respectively. Calculate the \(E_{\text {cell }},\) assuming no ion-pair formation.
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