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Chapter 18: Problem 131

An aqueous solution of an unknown salt of ruthenium is electrolyzed by a current of \(2.50\) A passing for \(50.0 \mathrm{~min}\). 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 the given aqueous solution is approximately +3.

Step by step solution

01

Identify the known values

In this problem, we are given: - Current (I) = 2.50 A - Time (t) = 50.0 minutes - Mass of Ru produced (m) = 2.618 g
02

Convert time to seconds

First, let's convert the time given in minutes to seconds, as it will make further calculations more straightforward. \(t (in~seconds) = 50.0 ~min \times 60~ \frac{seconds}{min} = 3000~ s\)
03

Calculate the total charge

Now we need to calculate the total charge passed during electrolysis using the formula: \(Q = I \times t\) where Q is the total charge in Coulombs, I is the current in Amperes, and t is the time in seconds. \(Q = 2.50~A \times 3000~s = 7500~C\)
04

Determine the moles of ruthenium

Using the molar mass of ruthenium (Ru), we can determine the moles of ruthenium produced at the cathode: Molar mass of Ru = 101.1 g/mol \(\text{moles of Ru} = \frac{\text{mass of Ru}}{\text{molar mass of Ru}} = \frac{2.618~g}{101.1~\frac{g}{mol}} = 0.02588~mol\)
05

Apply Faraday's Law of Electrolysis

According to Faraday's Law of Electrolysis, the total charge (Q) is equal to the number of moles of ions (n), the charge per ion (z), and Faraday's constant (F): \(Q = n \times z \times F\) Faraday's constant (F) = 96,485 C/mol We already calculated Q (total charge) and n (number of moles). Now, we will solve for z, the charge on the ruthenium ions. Rearranging the equation to solve for z: \(z = \frac{Q}{n \times F}\) Calculate z: \(z = \frac{7500~C}{0.02588~mol \times 96485~\frac{C}{mol}} = 2.998 \approx 3\) Since z must be a whole number, we can round it to the nearest integer.
06

Determine the charge on ruthenium ions

The calculated value of z is approximately 3, which corresponds to a charge of +3 on the ruthenium ions in solution. Therefore, the charge on the ruthenium ions is +3.

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Most popular questions from this chapter

Nerve impulses are electrical "signals" that pass through neurons in the body. The electrical potential is created by the differences in the concentration of \(\mathrm{Na}^{+}\) and \(\mathrm{K}^{+}\) ions across the nerve cell membrane. We can think about this potential as being caused by a concentration gradient, similar to what we see in a concentration cell (keep in mind that this is a very simple explanation of how nerves work; there is much more involved in the true biologic process). A typical nerve cell has a resting potential of about \(-70 \mathrm{mV}\). Let's assume that this resting potential is due only to the \(\mathrm{K}^{+}\) ion concentration difference. In nerve cells, the \(\mathrm{K}^{+}\) concentration inside the cell is larger than the \(\mathrm{K}^{+}\) concentration outside the cell. Calculate the \(\mathrm{K}^{+}\) ion concentration ratio necessary to produce a resting potential of \(-70 . \mathrm{mV}\). $$\frac{\left[\mathrm{K}^{+}\right]_{\text {inside }}}{\left[\mathrm{K}^{+}\right]_{\text {outside }}}=?$$

You have a concentration cell in which the cathode has a silver electrode with \(0.10 \mathrm{MAg}^{+}\). The anode also has a silver electrode with \(\mathrm{Ag}^{+}(a q), 0.050 \mathrm{M} \mathrm{S}_{2} \mathrm{O}_{3}{ }^{2-}\), and \(1.0 \times 10^{-3} \mathrm{M} \mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}{ }^{3-}\). You read the voltage to be \(0.76 \mathrm{~V}\). a. Calculate the concentration of \(\mathrm{Ag}^{+}\) at the anode. b. Determine the value of the equilibrium constant for the formation of \(\mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}{ }^{3-}\) \(\mathrm{Ag}^{+}(a q)+2 \mathrm{~S}_{2} \mathrm{O}_{3}{ }^{2-}(a q) \rightleftharpoons \mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}{ }^{3-}(a q) \quad K=?\)

Gold metal will not dissolve in either concentrated nitric acid on concentrated hydrochloric acid. It will dissolve, however, in aqua regia, a mixture of the two concentrated acids. The products of the reaction are the \(\mathrm{AuCl}_{4}^{-}\) ion and gaseous NO. Write a balanced equation for the dissolution of gold in aqua regia.

In making a specific galvanic cell, explain how one decides on the electrodes and the solutions to use in the cell.

Consider the galvanic cell based on the following half-reactions: $$\begin{array}{ll}\mathrm{Au}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au} & \mathscr{E}^{\circ}=1.50 \mathrm{~V} \\ \mathrm{Tl}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Tl} & \mathscr{E}^{\circ}=-0.34 \mathrm{~V} \end{array}$$ a. Determine the overall cell reaction and calculate \(\mathscr{E}_{\mathrm{ccll}}^{\circ}\) b. Calculate \(\Delta G^{\circ}\) and \(K\) for the cell reaction at \(25^{\circ} \mathrm{C}\). c. Calculate \(\mathscr{E}_{\text {cell }}\) at \(25^{\circ} \mathrm{C}\) when \(\left[\mathrm{Au}^{3+}\right]=1.0 \times 10^{-2} M\) and \(\left[\mathrm{Tl}^{+}\right]=1.0 \times 10^{-4} \mathrm{M}\)
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