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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

Consider a galvanic cell consisting of a magnesium electrode in contact with \(1.0 \mathrm{MMg}\left(\mathrm{NO}_{3}\right)_{2}\) and a cadmium electrode in contact with \(1.0 \mathrm{M} \mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2}\) Calculate \(E^{\circ}\) for the cell, and draw a diagram showing the cathode, anode, and direction of electron flow.

Use the standard reduction potentials to find the equilibrium constant for each of the following reactions at \(25^{\circ} \mathrm{C}\): (a) \(\operatorname{Br}_{2}(l)+2 \mathrm{I}^{-}(a q) \rightleftharpoons 2 \mathrm{Br}^{-}(a q)+\mathrm{I}_{2}(s)\) (b) \(2 \mathrm{Ce}^{4+}(a q)+2 \mathrm{Cl}^{-}(a q) \rightleftharpoons$$\mathrm{Cl}_{2}(g)+2 \mathrm{Ce}^{3+}(a q)\) (c) \(5 \mathrm{Fe}^{2+}(a q)+\mathrm{MnO}_{4}^{-}(a q)+8 \mathrm{H}^{+}(a q) \rightleftharpoons\) \(\mathrm{Mn}^{2+}(a q)+4 \mathrm{H}_{2} \mathrm{O}(l)+5 \mathrm{Fe}^{3+}(a q)\)

What is the difference between a galvanic cell (such as a Daniell cell) and an electrolytic cell?

The \(\mathrm{SO}_{2}\) present in air is mainly responsible for the phenomenon of acid rain. The concentration of \(\mathrm{SO}_{2}\) can be determined by titrating against a standard permanganate solution as follows: $$ 5 \mathrm{SO}_{2}+2 \mathrm{MnO}_{4}^{-}+2 \mathrm{H}_{2} \mathrm{O} \longrightarrow{\mathrm{SSO}_{4}^{2-}}+2 \mathrm{Mn}^{2+}+4 \mathrm{H}^{+} $$ Calculate the number of grams of \(\mathrm{SO}_{2}\) in a sample of air if \(7.37 \mathrm{~mL}\) of \(0.00800 \mathrm{M} \mathrm{KMnO}_{4}\) solution are required for the titration.

Consider a galvanic cell composed of the SHE and a half-cell using the reaction \(\mathrm{Ag}^{+}(a q)+e^{-} \rightarrow \operatorname{Ag}(s)\). (a) Calculate the standard cell potential. (b) What is the spontaneous cell reaction under standard-state conditions? (c) Calculate the cell potential when \(\left[\mathrm{H}^{+}\right]\) in the hydrogen electrode is changed to (i) \(1.0 \times\) \(10^{-2} M\) and (ii) \(1.0 \times 10^{-5} M,\) all other reagents being held at standard-state conditions. (d) Based on this cell arrangement, suggest a design for a pH meter.
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