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Look up the reduction potential for \(\mathrm{Fe}^{3+}\) to \(\mathrm{Fe}^{2+} .\) Look up the reduction potential for \(\mathrm{Fe}^{2+}\) to Fe. Finally, look up the reduction potential for \(\mathrm{Fe}^{3+}\) to Fe. You should notice that adding the reduction potentials for the first two does not give the potential for the third. Why not? Show how you can use the first two potentials to calculate the third potential.

Short Answer

Expert verified
The sum of the first two reduction potentials, \(0.771V\) and \(-0.447V\), does not equal the third potential \(-0.036V\) because it doesn't take into account the reaction quotients or concentrations. To obtain the correct potential, the Nernst equation should be used: \[E_3 = E_1 + E_2 - \frac{RT}{3F} \ln{\frac{1}{[Fe^3+]}}\] This equation includes the effects of ion concentrations on the reaction potentials, allowing accurate calculation of the third potential using the first two potentials.

Step by step solution

01

Look up the Reduction Potentials

First, we need to look up the reduction potentials of the ions: 1. Fe鲁鈦 to Fe虏鈦: \(E_1^\circ = 0.771V\) 2. Fe虏鈦 to Fe: \(E_2^\circ = -0.447V\) 3. Fe鲁鈦 to Fe: \(E_3^\circ = -0.036V\)
02

Attempt to Combine Reduction Potentials

Add the reduction potentials from step 1: \[E_1^\circ + E_2^\circ \ne E_3^\circ\] \[0.771V + (-0.447V) \ne -0.036V\] \[0.324V \ne -0.036V\] This shows that adding the first two potentials doesn't equal the third potential.
03

Analyze the Nernst Equation

To understand why merely adding the potentials doesn't give the correct result, we can look at the Nernst equation: \[E_{cell} = E_{cell}^\circ - \frac{RT}{nF} \ln{Q}\] The Nernst equation describes the relationship between the cell potential and the ion concentrations. The mismatch between the added potentials and the true potential indicates that the reaction quotient or concentrations are at play here.
04

Consider the Reaction Quotients

We need to consider the reaction quotients when adding the reactions. Here are the half-reactions: 1. Fe鲁鈦 + e鈦 鈫 Fe虏鈦 2. Fe虏鈦 + 2e鈦 鈫 Fe And the full reaction: Fe鲁鈦 + 3e鈦 鈫 Fe To find the correct reaction quotient for the full reaction, we have to multiply the reaction quotients for the half-reactions: For half-reaction 1: \[Q_1 = \frac{[Fe^2+]}{[Fe^3+]}\] For half-reaction 2: \[Q_2 = \frac{1}{[Fe^2+]}\] For the full reaction: \[Q = Q_1 * Q_2 = \frac{1}{[Fe^3+]}\]
05

Use the Concentration Term to Adjust Potentials

Now that we found the reaction quotient for the full reaction, we can use the Nernst equation to properly combine the half-cell potentials. \[E_{cell} = E_{cell}^\circ - \frac{RT}{nF} \ln{Q}\] For the full reaction, the Nernst equation should look like this: \[E_3 = E_1 + E_2 - \frac{RT}{3F} \ln{\frac{1}{[Fe^3+]}}\] Given that we know \(E_1\), \(E_2\), and \(E_3\), it is indeed possible to use the first two potentials to calculate the third potential. It is important to take into account the concentration term, as merely adding the standard reduction potentials does not always give an accurate result.

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Key Concepts

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

Electrochemistry
Electrochemistry is a branch of chemistry that deals with the interconversion of chemical and electrical energy. It is based on redox (reduction-oxidation) reactions. In these reactions, electrons are transferred between chemical species, leading to changes in their oxidation states.

Key components in electrochemistry include electrodes, electrolytes, and electric circuits. These components are used to study and harness the movement of electrons, which can generate electricity or drive chemical changes.
  • Electrodes are conductive materials that allow electrons to enter or exit a chemical system.
  • Electrolytes are substances that produce ions in solution, enabling the flow of electrical charge through the solution.
Understanding how these elements interact in an electrochemical cell provides insights into processes like corrosion, battery function, and electrolysis. In our example with iron ions, understanding the reduction potentials of different reactions helps predict how the electron transfer will occur, influencing the direction and magnitude of electron flow.
Nernst Equation
The Nernst equation allows us to calculate the potential of an electrochemical cell under non-standard conditions. Standard conditions involve 1 M concentrations of reactants and products, 1 atm pressure, and a temperature of 25掳C (298 K). When these standard conditions aren't met, the cell potential changes.

The Nernst equation is given by:\[ E_{cell} = E_{cell}^\circ - \frac{RT}{nF} \ln{Q} \]where:
  • \(E_{cell}^\circ\) is the standard cell potential,
  • \(R\) is the universal gas constant (8.314 J/mol K),
  • \(T\) is the temperature in Kelvin,
  • \(n\) is the number of moles of electrons transferred in the reaction,
  • \(F\) is Faraday's constant (96485 C/mol),
  • \(Q\) is the reaction quotient.
The Nernst equation shows that if the concentrations of the reacting species differ from their standard states, the cell potential will differ from the standard potential. It helps bridge the gap between theoretical predictions and real-world observations of electrochemical behaviors.
Half-Reaction
A half-reaction is a part of a redox reaction that shows either the oxidation or reduction process separately. In electrochemistry, redox reactions are often broken down into two half-reactions. Each half-reaction describes how electrons are either gained (reduction) or lost (oxidation).

In our example with iron:
  • Reduction half-reaction for Fe鲁鈦 to Fe虏鈦: \( \mathrm{Fe}^{3+} + e^- \rightarrow \mathrm{Fe}^{2+} \)
  • Reduction half-reaction for Fe虏鈦 to Fe: \( \mathrm{Fe}^{2+} + 2e^- \rightarrow \mathrm{Fe} \)
These half-reactions are crucial for calculating cell potentials. By understanding each component individually, we can better predict and manipulate the electrochemical reactions.
Breaking down redox processes into half-reactions also simplifies the use of formulae like the Nernst equation, which requires knowledge of the number of electrons transferred.
Reaction Quotient
The reaction quotient, denoted as \(Q\), is a measure of the relative amounts of products and reactants present during a reaction at any given time. It has the same form as the equilibrium constant \(K\), but differs because \(Q\) can be calculated at any point in time, not just at equilibrium.

In electrochemistry, \(Q\) plays a vital role in the Nernst equation, affecting the calculation of cell potentials especially under non-standard conditions. The formula for \(Q\) is derived from the concentrations of the chemical species involved in the reaction:
  • \(Q = \frac{[\text{products}]}{[\text{reactants}]}\)
For the specific case of our iron reduction example, the reaction quotient helps understand why the calculated potential differs when using non-standard conditions.
The link between \(Q\) and \(E_{cell}\) reflects dynamic shifts in the chemical environment, stressing the importance of \(Q\) in achieving accurate theoretical predictions and explaining why simple addition of standard reduction potentials may not always suffice.

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Most popular questions from this chapter

A galvanic cell is based on the following half-reactions at \(25^{\circ} \mathrm{C} :\) $$\begin{array}{c}{\mathrm{Ag}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}} \\\ {\mathrm{H}_{2} \mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}}\end{array}$$ Predict whether \(\mathscr{E}_{\text{cell}}\) is larger or smaller than \(\mathscr{E}^{\circ}_{\text{cell}}\) for the following cases. a. [Ag1] 5 1.0 a. \(\left[\mathrm{Ag}^{+}\right]=1.0 M,\left[\mathrm{H}_{2} \mathrm{O}_{2}\right]=2.0 M,\left[\mathrm{H}^{+}\right]=2.0 \mathrm{M}\) b. \(\left[\mathrm{Ag}^{+}\right]=2.0 \mathrm{M},\left[\mathrm{H}_{2} \mathrm{O}_{2}\right]=1.0 M,\left[\mathrm{H}^{+}\right]=1.0 \times 10^{-7} \mathrm{M}\)

When copper reacts with nitric acid, a mixture of \(\mathrm{NO}(g)\) and \(\mathrm{NO}_{2}(g)\) is evolved. The volume ratio of the two product gases depends on the concentration of the nitric acid according to the equilibrium $$2 \mathrm{H}^{+}(a q)+2 \mathrm{NO}_{3}^{-}(a q)+\mathrm{NO}(g) \rightleftharpoons 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)$$ Consider the following standard reduction potentials at \(25^{\circ} \mathrm{C} :\) $$3 \mathrm{e}^{-}+4 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$$ $$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \mathscr{E}^{\circ}=0.957 \mathrm{V}$$ $$\mathrm{e}^{-}+2 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$$ $$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \mathscr{E}^{\circ}=0.775 \mathrm{V}$$ a. Calculate the equilibrium constant for the above reaction. b. What concentration of nitric acid will produce a NO and NO \(_{2}\) mixture with only 0.20\(\% \mathrm{NO}_{2}\) (by moles) at \(25^{\circ} \mathrm{C}\) and 1.00 atm? Assume that no other gases are present and that the change in acid concentration can be neglected.

A galvanic cell is based on the following half-reactions: $$\begin{aligned} \mathrm{Ag}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s) & \mathscr{E}^{\circ}=0.80 \mathrm{V} \\ \mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s) & \mathscr{E}^{\circ}=0.34 \mathrm{V} \end{aligned}$$ In this cell, the silver compartment contains a silver electrode and excess AgCl(s) \(\left(K_{\mathrm{sp}}=1.6 \times 10^{-10}\right),\) and the copper compartment contains a copper electrode and \(\left[\mathrm{Cu}^{2+}\right]=2.0 \mathrm{M}\) . a. Calculate the potential for this cell at \(25^{\circ} \mathrm{C}\) . b. Assuming 1.0 \(\mathrm{L}\) of 2.0\(M \mathrm{Cu}^{2+}\) in the copper compartment, calculate the moles of \(\mathrm{NH}_{3}\) that would have to be added to give a cell potential of 0.52 \(\mathrm{V}\) at \(25^{\circ} \mathrm{C}\) (assume no volume change on addition of \(\mathrm{NH}_{3} )\) . $$\mathrm{Cu}^{2+}(a q)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q) \qquad K=1.0 \times 10^{13}$$

Consider a galvanic cell based on the following half-reactions: $$\begin{array}{ll}{\text {}} & { \mathscr{E}^{\circ}(\mathbf{V}) } \\ \hline {\mathrm{La}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{La}} & {-2.37} \\\ {\mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe}} & {-0.44}\end{array}$$ a. What is the expected cell potential with all components in their standard states? b. What is the oxidizing agent in the overall cell reaction? c. What substances make up the anode compartment? d. In the standard cell, in which direction do the electrons flow? e. How many electrons are transferred per unit of cell reaction? f. If this cell is set up at \(25^{\circ} \mathrm{C}\) with \(\left[\mathrm{Fe}^{2+}\right]=2.00 \times 10^{-4} M\) and \(\left[\mathrm{La}^{3+}\right]=3.00 \times 10^{-3} M,\) what is the expected cell potential?

An electrochemical cell consists of a nickel metal electrode immersed in a solution with \(\left[\mathrm{Ni}^{2+}\right]=1.0 M\) separated by a porous disk from an aluminum metal electrode. a. What is the potential of this cell at \(25^{\circ} \mathrm{C}\) if the aluminum electrode is placed in a solution in which \(\left[\mathrm{Al}^{3+}\right]=7.2 \times 10^{-3} M?\) b. When the aluminum electrode is placed in a certain solution in which \(\left[\mathrm{Al}^{3+}\right]\) is unknown, the measured cell potential at \(25^{\circ} \mathrm{C}\) is 1.62 \(\mathrm{V}\) . Calculate \(\left[\mathrm{Al}^{3+}\right]\) in the unknown solution. (Assume Al is oxidized.)

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