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Chapter 20: Problem 91

(a) A \(\mathrm{Cr}^{3+}(a q)\) solution is electrolyzed, using a current of \(7.60 \mathrm{~A}\). What mass of \(\mathrm{Cr}(s)\) is plated out after 2.00 days? (b) What amperage is required to plate out $0.250 \mathrm{~mol} \mathrm{Cr}\( from a \)\mathrm{Cr}^{3+}\( solution in a period of \)8.00 \mathrm{~h}$ ?

Short Answer

Expert verified
(a) The mass of \(\mathrm{Cr}\) plated out after 2.00 days is 744.33 g. (b) The amperage required to plate out 0.250 mol \(\mathrm{Cr}\) from a \(\mathrm{Cr}^{3+}\) solution in a period of 8.00 h is 25.061 A.

Step by step solution

01

(a) Calculate the total charge

: First, we need to determine the total charge (in coulombs) that has passed through the solution. Since we have the current (7.60 A) and the time (2.00 days), we can calculate the charge using the formula: \[ Q = I \times t \] Convert the time to seconds: \[ 2.00\,\mathrm{days} \times \frac{24\,\mathrm{h}}{1\,\mathrm{day}} \times \frac{3600\,\mathrm{s}}{1\,\mathrm{h}} = 172800\,\mathrm{s} \] Now, calculate the total charge (Q): \[ Q = 7.60\,\mathrm{A} \times 172800\,\mathrm{s} = 1313280\,\mathrm{C} \]
02

(a) Apply Faraday's law

: Now, apply Faraday's law to calculate the amount of substance (moles of \(\mathrm{Cr}\)) produced during electrolysis: \[ n = \frac{Q}{z \times F} \] Where \(n\) is the amount (in moles) of substance (in this case, \(\mathrm{Cr}\)), \(Q\) is the total charge, \(z\) is the number of electrons transferred (3 for \(\mathrm{Cr}^{3+}\)), and \(F\) is Faraday's constant (\(96485\,\mathrm{C/mol}\)). \[ n = \frac{1313280\,\mathrm{C}}{3 \times 96485\,\mathrm{C/mol}} = 14.315\,\mathrm{mol} \]
03

(a) Convert moles to mass

: To find the mass of \(\mathrm{Cr}\), we need its molar mass, which is \(51.996\,\mathrm{g/mol}\): \[ m = n \times M \] Where \(m\) is the mass, \(n\) is the amount of substance and \(M\) is the molar mass. \[ m = 14.315\,\mathrm{mol} \times 51.996\,\mathrm{g/mol} = 744.33\,\mathrm{g} \]
04

(b) Calculate moles of Cr for given time

: For the second part of the problem, we are given the amount of substance (\(0.250\,\mathrm{mol}\,\mathrm{Cr}\)) and the desired time frame (\(8.00\,\mathrm{h}\)). Let's convert the time to seconds: \[ 8.00\,\mathrm{h} \times \frac{3600\,\mathrm{s}}{1\,\mathrm{h}} = 28800\,\mathrm{s} \]
05

(b) Use Faraday's law to find the required current

: We will once again use Faraday's law, but this time, we will rearrange the formula to find the required current (I): \[ I = \frac{n \times z \times F}{t} \] We already have the values for \(n\), \(z\), and \(t\), so plug in the values: \[ I = \frac{0.250\,\mathrm{mol} \times 3 \times 96485\,\mathrm{C/mol}}{28800\,\mathrm{s}} = 25.061\,\mathrm{A} \] #Answer#: (a) The mass of \(\mathrm{Cr}\) plated out after 2.00 days is 744.33 g. (b) The amperage required to plate out 0.250 mol \(\mathrm{Cr}\) from a \(\mathrm{Cr}^{3+}\) solution in a period of 8.00 h is 25.061 A.

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Most popular questions from this chapter

From each of the following pairs of substances, use data in Appendix \(\mathrm{E}\) to choose the one that is the stronger oxidizing agent: (a) \(\mathrm{Cl}_{2}(g)\) or \(\mathrm{Br}_{2}(l)\) (b) \(\mathrm{Zn}^{2+}(a q)\) or \(\mathrm{Cd}^{2+}(a q)\) (c) \(\mathrm{Cl}^{-}(a q)\) or \(\mathrm{ClO}_{3}^{-}(a q)\) (d) \(\mathrm{H}_{2} \mathrm{O}_{2}(a q)\) or \(\mathrm{O}_{3}(g)\)

(a) Which electrode of a voltaic cell, the cathode or the anode, corresponds to the higher potential energy for the electrons? (b) What are the units for electrical potential? How does this unit relate to energy expressed in joules?

Using the standard reduction potentials listed in Appendix E, calculate the equilibrium constant for each of the following reactions at \(298 \mathrm{~K}\) : $$ \begin{array}{l} \text { (a) } \mathrm{Fe}(s)+\mathrm{Ni}^{2+}(a q) \longrightarrow \mathrm{Fe}^{2+}(a q)+\mathrm{Ni}(s) \\ \text { (b) } \mathrm{Co}(s)+2 \mathrm{H}^{+}(a q) \longrightarrow \mathrm{Co}^{2+}(a q)+\mathrm{H}_{2}(g) \\ \text { (c) } 10 \mathrm{Br}^{-}(a q)+2 \mathrm{MnO}_{4}^{-}(a q)+16 \mathrm{H}^{+}(a q) \longrightarrow 2 \mathrm{Mn}^{2+}(a q)+8 \mathrm{H}_{2} \mathrm{O}(l)+5 \mathrm{Br}_{2}(l) \end{array} $$

For each of the following reactions, write a balanced equation, calculate the standard emf, calculate \(\Delta G^{\circ}\) at \(298 \mathrm{~K},\) and calculate the equilibrium constant \(K\) at \(298 \mathrm{~K}\). (a) Aqueous iodide ion is oxidized to \(\mathrm{I}_{2}(s)\) by \(\mathrm{Hg}_{2}^{2+}(a q) .\) (b) In acidic solution, copper(I) ion is oxidized to copper(II) ion by nitrate ion. (c) In basic solution, \(\mathrm{Cr}(\mathrm{OH})_{3}(s)\) is oxidized to \(\mathrm{CrO}_{4}^{2-}(a q)\) by \(\mathrm{ClO}^{-}(a q)\)

Predict whether the following reactions will be spontaneous in acidic solution under standard conditions: (a) oxidation of \(\mathrm{Cu}\) to \(\mathrm{Cu}^{2+}\) by \(\mathrm{I}_{2}\) (to form \(\mathrm{I}^{-}\) ), \((\mathbf{b})\) reduction of \(\mathrm{Fe}^{2+}\) to \(\mathrm{Fe}\) by \(\mathrm{H}_{2}\) (to form \(\mathrm{H}^{+}\) ), \(\left(\mathbf{c}\right.\) ) reduction of \(\mathrm{I}_{2}\) to \(\mathrm{I}^{-}\) by $\mathrm{H}_{2} \mathrm{O}_{2},(\mathbf{d})\( reduction of \)\mathrm{Ni}^{2+}\( to \)\mathrm{Ni}$ by \(\mathrm{Sn}^{2+}\left(\right.\) to form \(\left.\mathrm{Sn}^{4+}\right)\).
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