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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 min. If 2.618 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 unknown salt of ruthenium ions in solution is approximately +3.

Step by step solution

01

Calculate the charge passed during electrolysis

First, let's calculate the charge passed during electrolysis using the formula Q = I * t. I = 2.50 A t = 50.0 min * (60 s/min) = 3000 s Now, plug the values into the formula. Q = I * t Q = 2.50 A * 3000 s Q = 7500 C So, the charge passed during the electrolysis process is 7500 C.
02

Calculate moles of electrons involved in the electrolysis process

Now, we will use the charge passed during the electrolysis and Faraday's constant to find the moles of electrons involved in the electrolysis process using the formula Q = n * F. F = 96,485 C/mol Rearrange the formula to solve for n n = Q / F Now, plug in the values. n = 7500 C / 96,485 C/mol n = 0.0777 mol So, about 0.0777 moles of electrons were involved in the electrolysis process.
03

Calculate the moles of ruthenium produced

Given that Ru has a molar mass of 101.07 g/mol, we can calculate the moles of ruthenium produced at the cathode using the mass of Ru. moles of Ru = mass of Ru / molar mass of Ru moles of Ru = 2.618 g Ru / 101.07 g/mol Ru moles of Ru = 0.0259 mol
04

Determine the charge on the ruthenium ions

Since the electrolysis process involves the exchange of electrons between cathode and ruthenium ions, the moles of electrons should be equal to the product of moles of Ru and the charge on the Ru ions. Let's denote the charge on the ruthenium ions as x. moles of electrons = moles of Ru * x Now, plug in the values and solve for x. 0.0777 mol = 0.0259 mol * x x = 0.0777 mol / 0.0259 mol x ≈ 3 Hence, the charge on the unknown salt of ruthenium ions in solution is approximately +3.

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Most popular questions from this chapter

The general rule for salt bridges is that anions flow to the anode and cations flow to the cathode. Explain why this is true.

A solution containing \(\mathrm{Pt}^{4+}\) is electrolyzed with a current of 4.00 \(\mathrm{A} .\) How long will it take to plate out 99\(\%\) of the platinum in 0.50 \(\mathrm{L}\) of a \(0.010-M\) solution of \(\mathrm{Pt}^{4+} ?\)

Consider the following electrochemical cell: a. If silver metal is a product of the reaction, is the cell a galvanic cell or electrolytic cell? Label the cathode and anode, and describe the direction of the electron flow. b. If copper metal is a product of the reaction, is the cell a galvanic cell or electrolytic cell? Label the cathode and anode, and describe the direction of the electron flow. c. If the above cell is a galvanic cell, determine the standard cell potential. d. If the above cell is an electrolytic cell, determine the minimum external potential that must be applied to cause the reaction to occur.

A galvanic cell consists of a standard hydrogen electrode and a copper electrode immersed in a Cu(NO \(_{3} )_{2}(a q)\) solution. If you wish to construct a calibration curve to show how the cell potential varies with \(\left[\mathrm{Cu}^{2+}\right],\) what should you plot to obtain a straight line? What will be the slope of this line?

Sketch the galvanic cells based on the following half- reactions. Show the direction of electron flow, show the direction of ion migration through the salt bridge, and identify the cathode and anode. Give the overall balanced equation, and determine \(\mathscr{E}^{\circ}\) for the galvanic cells. Assume that all concentrations are 1.0 \(M\) and that all partial pressures are 1.0 atm. a. $\mathrm{Cl}_{2}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cl}^{-} \quad \mathscr{E}^{\circ}=1.36 \mathrm{V}$ $\mathrm{Br}_{2}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Br}^{-} \quad \mathscr{E}^{\circ}=1.09 \mathrm{V}$ b. $\mathrm{MnO}_{4}^{-}+8 \mathrm{H}^{+}+5 \mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O} \quad \mathscr{E}^{\circ}=1.51 \mathrm{V}$ $\mathrm{IO}_{4}^{-}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{IO}_{3}^{-}+\mathrm{H}_{2} \mathrm{O} \quad \quad \mathscr{E}^{\circ}=1.60 \mathrm{V}$
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