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Chapter 18: Problem 133

An aqueous solution of an unknown salt of ruthenium is electrolyzed by a current of \(2.50\) A passing for \(50.0\) min. If \(2.618 \mathrm{~g}\) Ru is produced at the cathode, what is the charge on the ruthenium ions in solution?

Short Answer

Expert verified
The charge on the ruthenium ions in solution can be calculated by following these steps: 1. Calculate the total charge transferred during the electrolysis: Total charge (Q) = Current (I) × Time (t) = \(2.50 A \times 3000 s = 7500 C\). 2. Calculate the number of moles of electrons transferred: Number of moles of electrons (n) = Total charge (Q) / Faraday constant (F) = \(7500 C / 96485 C/mol = 0.0777 mol\). 3. Calculate the number of moles of ruthenium produced at the cathode: Number of moles of Ru = Mass of Ru produced (m) / Molar mass of Ru (M) = \(2.618 g / 101.07 g/mol = 0.0259 mol\). 4. Determine the charge on the ruthenium ions (z) in solution: Charge on ruthenium ions (z) = Number of moles of electrons (n) / Number of moles of Ru = \(0.0777 mol / 0.0259 mol = 3\). Thus, the charge on the ruthenium ions in solution is \(3+\).

Step by step solution

01

Calculate the number of electrons transferred in total

We need to find out how many electrons are transferred in total throughout the electrolysis process. We can do this by calculating the total charge transferred during the electrolysis using the formula: Total charge (Q) = Current (I) × Time (t) where I = 2.50 A and t = 50 min = 50 × 60 s = 3000 s Total charge (Q) = 2.50 A × 3000 s Calculate the value of Q.
02

Calculate the number of moles of electrons transferred

Now, let's find the number of moles of electrons (n) transferred by dividing the total charge by the Faraday constant (F), which is approximately \(96485\) coulombs per mole of electrons. Number of moles of electrons (n) = Total charge (Q) / F Calculate the value of n using Q from Step 1.
03

Calculate the number of moles of ruthenium

To calculate the number of moles of ruthenium (Ru) produced at the cathode, we use the formula: Number of moles of Ru = Mass of Ru produced (m) / Molar mass of Ru (M) where m = 2.618 g and M = 101.07 g/mol (molar mass of ruthenium) Number of moles of Ru = 2.618 g / 101.07 g/mol Calculate the value of number of moles of Ru.
04

Determine the charge on the ruthenium ions

Now that we have the number of moles of electrons and ruthenium, we can find the charge on the ruthenium ions (z) in solution using the following formula: Charge on ruthenium ions (z) = Number of moles of electrons (n) / Number of moles of Ru Calculate the value of z using n from Step 2 and the number of moles of Ru from Step 3. The result will give you the charge on the ruthenium ions in solution.

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Most popular questions from this chapter

Consider the following half-reactions: $$ \begin{aligned} \mathrm{Pt}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pt} & \mathscr{E}^{\circ} &=1.188 \mathrm{~V} \\ \mathrm{PtCl}_{4}^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pt}+4 \mathrm{Cl}^{-} & \mathscr{E}^{\circ} &=0.755 \mathrm{~V} \\ \mathrm{NO}_{3}^{-}+4 \mathrm{H}^{+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{NO}+2 \mathrm{H}_{2} \mathrm{O} & \mathscr{E}^{\circ} &=0.96 \mathrm{~V} \end{aligned} $$ Explain why platinum metal will dissolve in aqua regia (a mixture of hydrochloric and nitric acids) but not in either concentrated nitric or concentrated hydrochloric acid individually.

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A galvanic cell is based on the following half-reactions: $$ \begin{aligned} \mathrm{Cu}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s) & \mathscr{E}^{\circ} &=0.34 \mathrm{~V} \\ \mathrm{~V}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{V}(s) & \mathscr{E}^{\circ} &=-1.20 \mathrm{~V} \end{aligned} $$ In this cell, the copper compartment contains a copper electrode and \(\left[\mathrm{Cu}^{2+}\right]=1.00 M\), and the vanadium compartment contains a vanadium electrode and \(\mathrm{V}^{2+}\) at an unknown concentration. The compartment containing the vanadium \((1.00 \mathrm{~L}\) of solution) was titrated with \(0.0800 M \mathrm{H}_{2} \mathrm{EDTA}^{2-}\), resulting in the reaction $$ \begin{array}{r} \mathrm{H}_{2} \mathrm{EDTA}^{2-}(a q)+\mathrm{V}^{2+}(a q) \rightleftharpoons \mathrm{VEDTA}^{2-}(a q)+2 \mathrm{H}^{+}(a q) \\ K=? \end{array} $$ The potential of the cell was monitored to determine the stoichiometric point for the process, which occurred at a volume of \(500.0 \mathrm{~mL} \mathrm{H}_{2} \mathrm{EDTA}^{2-}\) solution added. At the stoichiometric point, \(\mathscr{E}_{\text {cell }}\) was observed to be \(1.98 \mathrm{~V}\). The solution was buffered at a pH of \(10.00\). a. Calculate \(\mathscr{E}_{\text {cell }}\) before the titration was carried out. b. Calculate the value of the equilibrium constant, \(K\), for the titration reaction. c. Calculate \(\mathscr{B}_{\text {cell }}\) at the halfway point in the titration.
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