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Chapter 19: Problem 51

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
Therefore, the amounts of \(Cu\) and \(Br_{2}\) produced are 52.7 g and 126 g, respectively.

Step by step solution

01

Write Down Faraday's Law of Electrolysis

Faraday's law of electrolysis states: \[m = (I \cdot t \cdot M) / (n \cdot F)\] where \(m\) is the mass of the substance produced at the electrode, \(I\) is the electric current, \(t\) is the time, \(M\) is the molar mass of the substance, \(n\) is the number of electrons involved in the redox reaction, and \(F\) is Faraday's constant.
02

Identify the Variables in the Equation

We know that \(M_{Cu} = 63.5 g/mol\), \(M_{Br_{2}} = 2 \cdot 79.9 g/mol\), \(n_{Cu} = 2\), \(n_{Br_{2}} = 2\), \(I = 4.5 A\), \(t = 1.0 h = 3600 s\), and \(F = 96500 C/mol\).
03

Calculate the Mass of Cu

Substitute the known values into Faraday's law to calculate the mass of Cu: \(m_{Cu} = (4.5 A \cdot 3600 s \cdot 63.5 g/mol) / (2 \cdot 96500 C/mol) = 52.7 g\).
04

Calculate the Mass of Br_{2}

Similarly, substitute the known values into Faraday's law to calculate the mass of \(Br_{2}\): \(m_{Br_{2}} = (4.5 A \cdot 3600 s \cdot 2 \cdot 79.9 g/mol) / (2 \cdot 96500 C/mol) = 126 g\).

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Most popular questions from this chapter

A constant electric current flows for \(3.75 \mathrm{~h}\) through two electrolytic cells connected in series. One contains a solution of \(\mathrm{AgNO}_{3}\) and the second a solution of \(\mathrm{CuCl}_{2}\). During this time \(2.00 \mathrm{~g}\) of silver are deposited in the first cell. (a) How many grams of copper are deposited in the second cell? (b) What is the current flowing, in amperes?

Explain the differences between a primary galvanic cell-one that is not rechargeable- and a storage cell (for example, the lead storage battery), which is rechargeable.

Balance the following redox equations by the halfreaction method: (a) \(\mathrm{H}_{2} \mathrm{O}_{2}+\mathrm{Fe}^{2+} \longrightarrow \mathrm{Fe}^{3+}+\mathrm{H}_{2} \mathrm{O}\) (in acidic solution) (b) \(\mathrm{Cu}+\mathrm{HNO}_{3} \longrightarrow \mathrm{Cu}^{2+}+\mathrm{NO}+\mathrm{H}_{2} \mathrm{O}\) (in acidic solution) (c) \(\mathrm{CN}^{-}+\mathrm{MnO}_{4}^{-} \longrightarrow \mathrm{CNO}^{-}+\mathrm{MnO}_{2}\) (in basic solution) (d) \(\mathrm{Br}_{2} \longrightarrow \mathrm{BrO}_{3}^{-}+\mathrm{Br}^{-}\) (in basic solution) (e) \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}+\mathrm{I}_{2} \longrightarrow \mathrm{I}^{-}+\mathrm{S}_{4} \mathrm{O}_{6}^{2-}\) (in acidic solution)

Explain why chlorine gas can be prepared by electrolyzing an aqueous solution of \(\mathrm{NaCl}\) but fluorine gas cannot be prepared by electrolyzing an aqueous solution of NaF.

When \(25.0 \mathrm{~mL}\) of a solution containing both \(\mathrm{Fe}^{2+}\) and \(\mathrm{Fe}^{3+}\) ions is titrated with \(23.0 \mathrm{~mL}\) of \(0.0200 \mathrm{M}\) \(\mathrm{KMnO}_{4}\) (in dilute sulfuric acid), all of the \(\mathrm{Fe}^{2+}\) ions are oxidized to \(\mathrm{Fe}^{3+}\) ions. Next, the solution is treated with \(Z\) n metal to convert all of the \(\mathrm{Fe}^{3+}\) ions to \(\mathrm{Fe}^{2+}\) ions. Finally, \(40.0 \mathrm{~mL}\) of the same \(\mathrm{KMnO}_{4}\) solution are added to the solution in order to oxidize the \(\mathrm{Fe}^{2+}\) ions to \(\mathrm{Fe}^{3+}\). Calculate the molar concentrations of \(\mathrm{Fe}^{2+}\) and \(\mathrm{Fe}^{3+}\) in the original solution.
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