It took \(150 .\) s for a current of 1.25 A to plate out 0.109 g of a metal from a solution containing its cations. Show that it is not possible for the cations to have a charge of \(1+.\)

Short Answer

Expert verified
In this problem, we calculated the total charge transferred (Q = 187.5 C) using the given current and time. Then, we determined the number of moles of electrons transferred (n_e ≈ 0.00194 mol) using the Faraday constant. Assuming a 1+ charge on the cations, we equated the number of moles of metal ions plated out (n_M) to n_e. This led to an apparent molar mass of the metal (M ≈ 56 g/mol), which is too small for most metals and matches iron, which cannot form a 1+ ion. Therefore, it is not possible for the cations to have a charge of 1+.

Step by step solution

01

Calculate the total charge transferred

The total charge Q transferred can be calculated using the formula Q = I × t, where I is the current (1.25 A) and t is the time (150 s). Q = I × t Q = 1.25 A × 150 s Q = 187.5 C
02

Calculate the number of moles of electrons transferred

Using the Faraday constant, we can determine the number of moles of electrons transferred (n_e). The Faraday constant F = 96485 C/mol. We have: n_e = Q / F n_e = 187.5 C / 96485 C/mol n_e ≈ 0.00194 mol
03

Calculate the number of moles of metal ions plated out assuming a 1+ charge

Assume the charge of the cations is 1+, which means one mole of electrons is needed to plate one mole of metal ions. The number of moles of metal ions plated out (n_M) is equal to the number of moles of electrons transferred (n_e): n_M = n_e n_M ≈ 0.00194 mol
04

Calculate the mass of metal plated out assuming a 1+ charge

Use the molar mass of the metal (M) to find the mass of the metal plated out. Since we don't know the molar mass of the metal, we'll express the mass as a function of M: mass = n_M × M
05

Compare with the actual mass of metal plated out

We are given that the actual mass of the metal plated out is 0.109 g. Assuming a 1+ charge, we have: 0.109 g = 0.00194 mol × M Solving for M: M ≈ 56 g/mol This molar mass is smaller than most of the metals. Moreover, it is the same molar mass as iron, which can never form a 1+ ion. Thus, we have shown that it is not possible for the cations to have a charge of 1+.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Faraday's Constant
Faraday's constant plays a crucial role in the field of electrochemistry, providing a bridge between the macroscopic observables that we can measure (like current and time) and the microscopic events at the atomic level (like the transfer of electrons).

It is defined as the amount of electric charge per mole of electrons. To put it simply, Faraday's constant, denoted as F, tells us how many coulombs of charge are needed to move one mole of electrons through a circuit. The accepted value of F is approximately 96,485 coulombs per mole (C/mol).

Understanding Faraday's constant is essential for calculating the amount of a substance that undergoes oxidation or reduction at the electrodes during electrolysis. By dividing the total electric charge by Faraday's constant, we obtain the number of moles of electrons that have been transferred in the process. This calculation is fundamental to solving problems related to the mass of substances deposited or dissolved in electrochemical reactions, as seen in the textbook exercise solution provided.
Molar Mass
The molar mass of a substance is another pivotal concept in chemistry, particularly when examining reactions that involve the transfer of mass. It represents the mass of one mole of a substance and is typically expressed in grams per mole (g/mol).

The importance of molar mass in electrochemistry arises when we need to convert between the amount of substance in moles and the corresponding mass in grams. With a known molar mass, we can determine how much mass corresponds to a specific number of moles.

For the electroplating exercise, the molar mass is the missing link that, once known, allows us to conclude whether the given metal could indeed have a charge of +1 based on the mass of metal deposited. Given that the theoretical mass calculated under this assumption did not correspond to a known metal capable of forming a +1 ion, we can infer that the assumed charge was incorrect.
Metal Cations
Metal cations are positively charged ions that form when metal atoms lose electrons. The charge on a metal cation is indicative of the number of electrons a metal atom has donated to achieve a stable electron configuration. Metal ions are ubiquitous in electrochemical reactions, where they often undergo reduction to form solid metal deposits.

In the context of our exercise, cations in solution were reduced to form a metal during the electroplating process. The charge on these cations is directly related to the stoichiometry of the reaction—specifically, the number of electrons needed to neutralize one ion. Common metal cation charges are +1, +2, or +3, but can be higher for some metals. The incorrect assumption of a 1+ charge for the metal ion in the exercise would imply a certain behavior during electrolysis, which, as deduced from the molar mass and the practical outcome, was not consistent.
Charge and Mass Relationship
The relationship between charge and mass in electrochemical processes is dictated by the stoichiometry of the reaction taking place at the electrodes. When we talk about this relationship, we are referring to how the charge carried by electrons relates to the mass of the substance being deposited or dissolved.

For a given amount of charge passed through an electrolyte, Faraday's laws of electrolysis help us understand the quantity of a substance transformed at an electrode. Specifically, it states that the amount of substance reacted is directly proportional to the total charge that flows through the system. By incorporating Faraday's constant, we bridge the gap between the macroscopic flow of charge and its microscopic effect in terms of moles of electrons influenced.

In the solution to the exercise we examined, it was crucial to understand this relationship to determine if the assumption of a 1+ charge for the cations was plausible. The final calculation, using the mass of metal deposited and the determined number of moles of electrons transferred, was key in concluding that this charge did not align with the known behaviors of metal cations.

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Most popular questions from this chapter

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

Consider the standard galvanic cell based on the following half-reactions: $$\begin{array}{c} \mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu} \\ \mathrm{Ag}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Ag} \end{array}$$ The electrodes in this cell are \(\mathrm{Ag}(s)\) and \(\mathrm{Cu}(s) .\) Does the cell potential increase, decrease, or remain the same when the following changes occur to the standard cell? a. \(\mathrm{CuSO}_{4}(s)\) is added to the copper half-cell compartment (assume no volume change). b. \(\mathrm{NH}_{3}(a q)\) is added to the copper half-cell compartment. [Hint: \(\left.\mathrm{Cu}^{2+} \text { reacts with } \mathrm{NH}_{3} \text { to form } \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q) .\right]\) c. \(\mathrm{NaCl}(s)\) is added to the silver half-cell compartment. [Hint: \(\left.\mathrm{Ag}^{+} \text {reacts with } \mathrm{Cl}^{-} \text {to form } \mathrm{AgCl}(s) .\right]\) d. Water is added to both half-cell compartments until the volume of solution is doubled. e. The silver electrode is replaced with a platinum electrode. $$ \mathrm{Pt}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pt} \quad \mathscr{E}^{\circ}=1.19 \mathrm{V} $$

A chemist wishes to determine the concentration of \(\mathrm{CrO}_{4}^{2-}\) electrochemically. A cell is constructed consisting of a saturated calomel electrode (SCE; see Exercise 115 ) and a silver wire coated with \(\mathrm{Ag}_{2} \mathrm{CrO}_{4} .\) The \(8^{\circ}\) value for the following half-reaction is \(0.446 \mathrm{V}\) relative to the standard hydrogen electrode: $$\mathrm{Ag}_{2} \mathrm{CrO}_{4}+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Ag}+\mathrm{CrO}_{4}^{2-}$$ a. Calculate \(\mathscr{C}_{\text {cell }}\) and \(\Delta G\) at \(25^{\circ} \mathrm{C}\) for the cell reaction when \(\left[\mathrm{CrO}_{4}^{2-}\right]=1.00 \mathrm{mol} / \mathrm{L}\) b. Write the Nernst equation for the cell. Assume that the SCE concentrations are constant. c. If the coated silver wire is placed in a solution (at \(25^{\circ} \mathrm{C}\) ) in which \(\left[\mathrm{CrO}_{4}^{2-}\right]=1.00 \times 10^{-5} \mathrm{M},\) what is the expected cell potential? d. The measured cell potential at \(25^{\circ} \mathrm{C}\) is \(0.504 \mathrm{V}\) when the coated wire is dipped into a solution of unknown \(\left[\mathrm{CrO}_{4}^{2-}\right] .\) What is \(\left[\mathrm{CrO}_{4}^{2-}\right]\) for this solution? e. Using data from this problem and from Table \(17-1,\) calculate the solubility product \(\left(K_{\mathrm{sp}}\right)\) for \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\)

An electrochemical cell consists of a standard hydrogen electrode and a copper metal electrode. a. What is the potential of the cell at \(25^{\circ} \mathrm{C}\) if the copper electrode is placed in a solution in which \(\left[\mathrm{Cu}^{2+}\right]=\) \(2.5 \times 10^{-4} \mathrm{M} ?\) b. The copper electrode is placed in a solution of unknown \(\left[\mathrm{Cu}^{2+}\right] .\) The measured potential at \(25^{\circ} \mathrm{C}\) is \(0.195 \mathrm{V}\). What is \(\left[\mathrm{Cu}^{2+}\right] ?\) (Assume \(\mathrm{Cu}^{2+}\) is reduced.)

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