Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 17: Problem 5

Sketch a cell that forms iron metal from iron(II) while changing chromium metal to chromium(III). Calculate the voltage, show the electron flow, label the anode and cathode, and balance the overall cell equation.

Short Answer

Expert verified
The cell consists of two half-cells connected by a salt bridge. At the anode (left), oxidation occurs with the reaction Cr -> Cr^3+ + 3e^-, while at the cathode (right), reduction occurs with the reaction Fe^2+ + 2e^- -> Fe. The overall balanced cell equation is 2Cr + 6Fe^2+ -> 2Cr^3+ + 6Fe, and the cell voltage is -1.18 V. Electrons flow from the anode (chromium) to the cathode (iron).

Step by step solution

01

Write the half-reactions for both iron and chromium

We need to convert iron(II) to iron metal and change chromium metal to chromium(III). For iron: Fe^2+ + 2e^- -> Fe (reduction half-reaction) For chromium: Cr -> Cr^3+ + 3e^- (oxidation half-reaction)
02

Determine the standard reduction potentials

Look up the standard reduction potentials for both half-reactions: Fe^2+ + 2e^- -> Fe (E° = -0.44 V) Cr^3+ + 3e^- -> Cr (E° = -0.74 V) Since we want to convert chromium-metal to chromium(III), we need to reverse the chromium half-reaction: Cr -> Cr^3+ + 3e^- (E° = 0.74 V)
03

Calculate the cell potential

The cell potential (E°(cell)) can be calculated as: E°(cell) = E°(cathode) - E°(anode) Since the iron half-reaction has a higher reduction potential, it will be the cathode (reduction occurs at the cathode): E°(cell) = (-0.44 V) - (0.74 V) = -1.18 V
04

Sketch the cell, show electron flow, and label the anode and cathode

- Left: Anode (Chromium): Write the oxidation half-reaction: Cr -> Cr^3+ + 3e^- Label chromium as the anode (A) Electrons flow from anode to cathode - Right: Cathode (Iron): Write the reduction half-reaction: Fe^2+ + 2e^- -> Fe Label iron as the cathode (C) - Connect the two half-cells with a salt bridge. A salt bridge is essential to maintain electrical neutrality in both half-cells by allowing the exchange of ions from one half-cell to the other.
05

Balance the overall cell equation and calculate the voltage

To balance the overall cell equation, multiply each half-reaction by the appropriate number to equalize the number of electrons exchanged: Oxidation: Cr -> Cr^3+ + 3e^- (x2) Reduction: Fe^2+ + 2e^- -> Fe (x3) Now add them up: 2Cr + 6Fe^2+ -> 2Cr^3+ + 6Fe The voltage of the cell is -1.18 V.

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.

Start your free trial

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

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

The table below lists the cell potentials for the 10 possible galvanic cells assembled from the metals \(\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D},\) and \(\mathrm{E},\) and their respective \(1.00 \space \mathrm{M} \space 2+\) ions in solution. Using the data in the table, establish a standard reduction potential table similar to Table \(17-1\) in the text. Assign a reduction potential of \(0.00 \mathrm{V}\) to the half-reaction that falls in the middle of the series. You should get two different tables. Explain why, and discuss what you could do to determine which table is correct. $$\begin{array}{|lcccc|} \hline & \begin{array}{c} \mathrm{A}(s) \text { in } \\ \mathrm{A}^{2+}(a q) \end{array} & \begin{array}{c} \mathrm{B}(s) \text { in } \\ \mathrm{B}^{2+}(a q) \end{array} & \begin{array}{c} \mathrm{c}(s) \text { in } \\ \mathrm{c}^{2+}(a q) \end{array} & \begin{array}{c} \mathrm{D}(s) \text { in } \\ \mathrm{D}^{2+}(a q) \end{array} \\ \hline \mathrm{E}(s) \text { in } \mathrm{E}^{2+}(a q) & 0.28 \mathrm{V} & 0.81 \mathrm{V} & 0.13 \mathrm{V} & 1.00 \mathrm{V} \\ \mathrm{D}(s) \text { in } \mathrm{D}^{2+}(a q) & 0.72 \mathrm{V} & 0.19 \mathrm{V} & 1.13 \mathrm{V} & \- \\ \mathrm{C}(s) \text { in } \mathrm{C}^{2+}(a q) & 0.41 \mathrm{V} & 0.94 \mathrm{V} & \- & \- \\ \mathrm{B}(s) \text { in } \mathrm{B}^{2+}(a q) & 0.53 \mathrm{V} & \- & \- & \- \\ \hline \end{array}$$

Define oxidation and reduction in terms of both change in oxidation number and electron loss or gain.

Is the following statement true or false? Concentration cells work because standard reduction potentials are dependent on concentration. Explain.

How can one construct a galvanic cell from two substances, each having a negative standard reduction potential?

The overall reaction in the lead storage battery is $$ \begin{array}{r} \mathrm{Pb}(s)+\mathrm{PbO}_{2}(s)+2 \mathrm{H}^{+}(a q)+2 \mathrm{HSO}_{4}^{-}(a q) \longrightarrow \\ 2 \mathrm{PbSO}_{4}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \end{array} $$ Calculate \(\mathscr{E}\) at \(25^{\circ} \mathrm{C}\) for this battery when \(\left[\mathrm{H}_{2} \mathrm{SO}_{4}\right]=4.5 \mathrm{M}\) that is, \(\left[\mathrm{H}^{+}\right]=\left[\mathrm{HSO}_{4}^{-}\right]=4.5 \mathrm{M} .\) At \(25^{\circ} \mathrm{C}, 8^{\circ}=2.04 \mathrm{V}\) for the lead storage battery.
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Reactions

Read Explanation

Inorganic Chemistry

Read Explanation

Making Measurements

Read Explanation

Kinetics

Read Explanation

Nuclear Chemistry

Read Explanation

Chemistry Branches

Read Explanation
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