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Chapter 18: Problem 54

Calculate the amounts of \(\mathrm{Cu}\) and \(\mathrm{Br}_{2}\) produced in \(1.0 \mathrm{~h}\) at inert electrodes in a solution of \(\mathrm{CuBr}_{2}\) by a current of 4.50 A.

Short Answer

Expert verified
The amount of Cu and Br2 produced will be 5.34 g and 13.44 g respectively.

Step by step solution

01

Identify Fundamental Information from Exercise

The current (I) is 4.5A, the time (t) is 1 hour. Also, the one mole of Cu and Br2 are deposited or liberated respectively when 2 moles of electrons are supplied to CuBr2 solution. The atomic masses of Cu and Br this instance are, Cu=63.5 g/mole, Br=80g/mole.
02

Convert Time to Seconds

The time for electrolysis is given in hours. It's convenient to convert this to seconds for the calculations because the unit of electric current (amperes) is coulombs/second. So, 1 hour = 3600 s.
03

Calculate the Total Charge (Q)

The total charge (Q, in coulombs) passed through the cell can be obtained using the formula Q = It, where I is the current and t is the time. Substituting the given values, we get Q = 4.5A * 3600s = 16200 C.
04

Calculate the Amount of Substance Electrolyzed

According to Faraday’s law, the amount of substance produced is directly proportional to the amount of electrical charge passed. We use the formula n = Q/(z*F), where z is the number of electrons transferred per formula unit during the reaction (z=2 for CuBr2), F is Faraday’s constant (96500 C/mol). For Cu and Br2, n = 16200 C / (2 * 96500 C/mol) = 0.084 mol.
05

Compute the Mass of Cu and Br2

We can calculate the masses of Cu and Br2 from the number of moles. For Cu, mass = n * atomic mass = 0.084 mol * 63.5 g/mol = 5.34 g. For Br2, since it is diatomic, molar mass = 80g/mol * 2 = 160 g/mol. So, mass of Br2 = n * molar mass = 0.084 mol * 160 g/mol = 13.44 g.

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Most popular questions from this chapter

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.

Consider two electrolytic cells A and B. Cell A contains a \(0.20 \mathrm{M} \mathrm{CoSO}_{4}\) solution and platinum electrodes. Cell B differs from cell A only in that cobalt metals are used as electrodes. In each case, a current of \(0.20 \mathrm{~A}\) is passed through the cell for \(1.0 \mathrm{~h}\). (a) Write equations for the half-cell and overall cell reactions for these cells. (b) Calculate the products formed (in grams) at the anode and cathode in each case.

Comment on whether \(\mathrm{F}_{2}\) will become a stronger oxidizing agent with increasing \(\mathrm{H}^{+}\) concentration.

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