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Chapter 20: Problem 91

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

Short Answer

Expert verified
The mass of Chromium plated out after 2.00 days is 235.2 g. The amperage required to plate out 0.250 mol of Cr from a Cr^3+ solution in a period of 8.00 hours is 2.53 A.

Step by step solution

01

(a) Calculate the moles of electrons passed through the cell

First, we need to convert the time from days to seconds: \(2.00\,days × \frac{24\,hours}{1\,day} × \frac{60\,minutes}{1\,hour} × \frac{60\,seconds}{1\,minute} = 172,800\,s\). Then, we can use the equation for moles of electrons passed through the cell: moles of electrons = (7.60 A) × (172,800 s) / (96,485 C/mol) = 13.58 mol.
02

(a) Find moles of Chromium plated out

The stoichiometry of the reaction is as follows: \[Cr^{3+}(aq) + 3e^- \to Cr(s)\] For every 3 moles of electrons, we have 1 mole of Chromium plated out. Thus, we can find the moles of Chromium as follows: moles of Chromium = 13.58 mol electrons / 3 = 4.526 mol.
03

(a) Calculate the mass of Chromium plated out

Now, we can convert the moles of Chromium to mass using its molar mass (51.996 g/mol): mass of Chromium = (4.526 mol) × (51.996 g/mol) = 235.2 g. So the mass of Chromium plated out after 2.00 days is 235.2 g.
04

(b) Find the moles of electrons needed to plate out 0.250 mol Cr

From the stoichiometry of the reaction, we need 3 moles of electrons for 1 mole of Chromium. Thus, the moles of electrons needed = 3 × 0.250 mol = 0.750 mol.
05

(b) Calculate the time in seconds

We need to convert the given time from hours to seconds: \(8.00\,hours × \frac{60\,minutes}{1\,hour} × \frac{60\,seconds}{1\,minute} = 28,800\,s\).
06

(b) Calculate the amperage needed to plate out 0.250 mol Cr

Now, we can use the equation for moles of electrons passed through the cell to find the amperage: current (A) = (0.750 mol × 96,485 C/mol) / 28,800 s = 2.53 A. Thus, the amperage required to plate out 0.250 mol of Cr from a Cr^3+ solution in a period of 8.00 hours is 2.53 A.

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Most popular questions from this chapter

A voltaic cell consists of a strip of cadmium metal in a solution of \(\mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2}\) in one beaker, and in the other beaker a platinum electrode is immersed in a \(\mathrm{NaCl}\) solution, with \(\mathrm{Cl}_{2}\) gas bubbled around the electrode. A salt bridge connects the two beakers. (a) Which electrode serves as the anode and which as the cathode? (b) Does the Cd electrode gain or lose mass as the cell reaction proceeds? (c) Write the equation for the overall cell reaction. (d) What is the emf generated by the cell under standard conditions?

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

From each of the following pairs of substances, use data in Appendix \(\mathrm{E}\) to choose the one that is the stronger reducing agent: (a) \(\mathrm{Fe}(s)\) or \(\mathrm{Mg}(s)\) (b) \(\mathrm{Ca}(s)\) or \(\mathrm{Al}(s)\) (c) \(\mathrm{H}_{2}(g,\) acidic solution \()\) or \(\mathrm{H}_{2} \mathrm{~S}(g)\) (d) \(\mathrm{BrO}_{3}^{-}(a q)\) or \(\mathrm{IO}_{3}^{-}(a q)\)

A voltaic cell utilizes the following reaction: $$\mathrm{Al}(s)+3 \mathrm{Ag}^{+}(a q) \longrightarrow \mathrm{Al}^{3+}(a q)+3 \mathrm{Ag}(s)$$ What is the effect on the cell emf of each of the following changes? (a) Water is added to the anode half-cell, diluting the solution. (b) The size of the aluminum electrode is increased. (c) A solution of \(\mathrm{AgNO}_{3}\) is added to the cathode half-cell, increasing the quantity of \(\mathrm{Ag}^{+}\) but not changing its concentration. (d) \(\mathrm{HCl}\) is added to the \(\mathrm{AgNO}_{3}\) solution, precipitating some of the \(\mathrm{Ag}^{+}\) as \(\mathrm{AgCl}\).

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