Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Describe in words how you would calculate the standard potential of the \(\mathrm{Fe}^{2+} / \mathrm{Fe}(\mathrm{s})\) couple from those of \(\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}\) and \(\mathrm{Fe}^{3+} / \mathrm{Fe}(\mathrm{s})\).

Short Answer

Expert verified
To calculate the standard potential of the Fe2+/Fe(s) couple from those of the Fe3+/Fe2+ and Fe3+/Fe(s) couples, one must first write out the half-reactions for the given couples. Then, by using the Nernst equation, and considering that both reactions involve the Fe3+ ion, the standard potential for the Fe2+/Fe(s) couple can be found. This shows the link between the standard potential for different redox couples involving the same element, in this case iron.

Step by step solution

01

Expression of Redox Couples

Begin by expressing each of the two given redox couples as half-reactions. The first one is \(\mathrm{Fe}^{3+} + e^- \rightarrow \mathrm{Fe}^{2+}\) and the second is \(\mathrm{Fe}^{3+} + 3e^- \rightarrow \mathrm{Fe}(\mathrm{s})\).
02

Design the Required Half-Reaction

We want to obtain the half-reaction for the \(\mathrm{Fe}^{2+}/\mathrm{Fe}(s)\) couple, which is \(\mathrm{Fe}^{2+} + 2e^- \rightarrow \mathrm{Fe}(\mathrm{s})\). To obtain this, we can subtract the first half-reaction from the second, as they have the same oxidation state Fe3+ in common.
03

Apply the Nernst Equation

Use the Nernst equation to calculate the potential. In this situation, it's expressed as \(E = E_{\mathrm{Fe}^{3+}/\mathrm{Fe}(\mathrm{s})} - E_{\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}}\)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Redox Reactions
Redox reactions, or reduction-oxidation reactions, are processes in which electrons are transferred between substances. These reactions are fundamental in chemistry and involve two key components: reduction (gain of electrons) and oxidation (loss of electrons). In the context of electrode potentials, each half-reaction shows how electrons are exchanged.
For instance, in our example:
  • The half-reaction of \( \mathrm{Fe}^{3+} + e^- \rightarrow \mathrm{Fe}^{2+} \) involves reduction, as \( \mathrm{Fe}^{3+} \) gains an electron.
  • Conversely, \( \mathrm{Fe}^{3+} + 3e^- \rightarrow \mathrm{Fe}(\mathrm{s}) \) represents another reduction step where three electrons are gained.
Understanding these reactions helps in determining how to manipulate them to find the required reaction, such as the \( \mathrm{Fe}^{2+}/\mathrm{Fe}(s) \) couple. Matching and rearranging these half-reactions is crucial for solving the given problem.
Nernst Equation
The Nernst equation is a powerful tool in electrochemistry that relates the concentration of ions in solution to the electrode potential. It extends the idea of standard electrode potential to account for real-world conditions.
The standard form of the Nernst equation is:\[E = E^0 - \frac{RT}{nF} \ln Q\]where:
  • \( E \) is the electrode potential under non-standard conditions,
  • \( E^0 \) is the standard electrode 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 exchanged,
  • \( F \) is Faraday’s constant (96485 C/mol),
  • \( Q \) is the reaction quotient.
In the exercise, the Nernst equation helps compute the desired potential by adjusting the potentials of known reactions, like \( E_{\mathrm{Fe}^{3+}/\mathrm{Fe}(\mathrm{s})} \) and \( E_{\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}} \).
Students can master this concept by practicing how different variables influence the potential and how to manipulate them to solve for unknown values.
Electrochemistry Concepts
Electrochemistry covers the study of chemical processes that cause electrons to move, creating electric current. This subject involves key principles that explain how batteries, electroplating, and various sensors work. Here are some fundamental concepts:
  • **Electrode Potential:** This measures the tendancy of a chemical species to be reduced, commonly represented by \( E^0 \).
  • **Standard Electrode Potential:** It’s the potential of a half-cell under standard conditions (1 M concentration, 1 atm pressure, and 25°C).
  • **Galvanic Cells:** These are electrochemical cells that convert chemical energy into electrical energy through spontaneous redox reactions.
  • **Half-Cells:** Each part of a galvanic cell where oxidation or reduction occurs, often represented by half-reactions in calculations.
  • **Cell Potential:** The difference in electrode potential between two half-cells, measured in volts.
Knowing these concepts allows students to understand how to calculate potentials, set up electrochemical cells, and predict the direction of redox reactions. These foundational ideas are vital for exploring more advanced topics in electrochemistry.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Calculate the quantity indicated for each of the following electrolyses. (a) \(\left[\mathrm{Cu}^{2+}\right]\) remaining in \(425 \mathrm{mL}\) of a solution that was originally \(0.366 \mathrm{M} \mathrm{CuSO}_{4},\) after passage of \(2.68 \mathrm{A}\) for 282 s and the deposition of Cu at the cathode (b) the time required to reduce \(\left[\mathrm{Ag}^{+}\right]\) in \(255 \mathrm{mL}\) of \(\mathrm{AgNO}_{3}(\mathrm{aq})\) from 0.196 to \(0.175 \mathrm{M}\) by electrolyzing the solution between \(\mathrm{Pt}\) electrodes with a current of \(1.84 \mathrm{A}\)

Given that \(E_{\text {cell }}^{\circ}=3.20 \mathrm{V}\) for the reaction $$2 \mathrm{Na}(\mathrm{in} \mathrm{Hg})+\mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{Na}^{+}(\mathrm{aq})+2 \mathrm{Cl}^{-}(\mathrm{aq})$$ What is \(E^{\circ}\) for the reduction \(2 \mathrm{Na}^{+}(\mathrm{aq})+2 \mathrm{e}^{-} \longrightarrow\) \(2 \mathrm{Na}(\text { in } \mathrm{Hg}) ?\)

An aqueous solution of \(\mathrm{K}_{2} \mathrm{SO}_{4}\) is electrolyzed by means of Pt electrodes. (a) Which of the following gases should form at the anode: \(\mathrm{O}_{2}, \mathrm{H}_{2}, \mathrm{SO}_{2}, \mathrm{SO}_{3} ?\) Explain. (b) What product should form at the cathode? Explain. (c) What is the minimum voltage required? Why is the actual voltage needed likely to be higher than this value?

A voltaic cell, with \(E_{\text {cell }}=0.180 \mathrm{V},\) is constructed as follows: $$\mathrm{Ag}(\mathrm{s})\left|\mathrm{Ag}^{+}\left(\operatorname{satd} \mathrm{Ag}_{3} \mathrm{PO}_{4}\right) \| \mathrm{Ag}^{+}(0.140 \mathrm{M})\right| \mathrm{Ag}(\mathrm{s})$$ What is the \(K_{\mathrm{sp}}\) of \(\mathrm{Ag}_{3} \mathrm{PO}_{4} ?\)

For the voltaic cell, $$\begin{array}{l} \mathrm{Ag}(\mathrm{s}) | \mathrm{Ag}^{+}(0.015 \mathrm{M}) \| \mathrm{Fe}^{3+}(0.055 \mathrm{M}) \\ \quad \mathrm{Fe}^{2+}(0.045 \mathrm{M}) | \mathrm{Pt}(\mathrm{s}) \end{array}$$ (a) what is \(E_{\text {cell initially? }}\) (b) As the cell operates, will \(E_{\text {cell increase }}\) decrease, or remain constant with time? Explain. (c) What will be \(E_{\text {cell }}\) when \(\left[\mathrm{Ag}^{+}\right]\) has increased to \(0.020 \mathrm{M} ?\) (d) What will be \(\left[\mathrm{Ag}^{+}\right]\) when \(E_{\text {cell }}=0.010 \mathrm{V} ?\) (e) What are the ion concentrations when \(E_{\text {cell }}=0 ?\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free