Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 20: Problem 43

From each of the following pairs of substances, use data in Appendix \(\mathrm{E}\) to choose the one that is the stronger reducing agent: (a) \(\mathrm{Al}(s)\) or \(\mathrm{Mg}(s)\) (b) \(\mathrm{Fe}(s)\) or \(\mathrm{Ni}(s)\) (c) \(\mathrm{H}_{2}(g\), acidic solution) or \(\operatorname{Sn}(s)\) (d) \(\mathrm{I}^{-}(a q)\) or \(\mathrm{Br}^{-}(a q)\)

Short Answer

Expert verified
Based on the reduction potentials in Appendix E, the stronger reducing agents among the given pairs are: (a) Mg(s) with \(E^{\circ}=-2.37V\), (b) Fe(s) with \(E^{\circ}=-0.44V\), (c) Sn(s) with \(E^{\circ}=-0.14V\), and (d) I⁻(aq) with \(E^{\circ}=0.54V\).

Step by step solution

01

(a): Al(s) or Mg(s)

: 1. The reduction potentials are: Al³⁺ + 3e⁻ → Al(s): \(E^{\circ}=-1.66V\) Mg²⁺ + 2e⁻ → Mg(s): \(E^{\circ}=-2.37V\) 2. Compare the reduction potentials. Since -2.37V < -1.66V, the magnesium reduction potential is more negative. 3. The stronger reducing agent is Mg(s).
02

(b): Fe(s) or Ni(s)

: 1. The reduction potentials are: Fe²⁺ + 2e⁻ → Fe(s): \(E^{\circ}=-0.44V\) Ni²⁺ + 2e⁻ → Ni(s): \(E^{\circ}=-0.25V\) 2. Compare the reduction potentials. Since -0.44V < -0.25V, the iron reduction potential is more negative. 3. The stronger reducing agent is Fe(s).
03

(c): H2(g, acidic solution) or Sn(s)

: 1. The reduction potentials are: 2H⁺ + 2e⁻ → H₂: \(E^{\circ}=0.00V\) Sn²⁺ + 2e⁻ → Sn(s): \(E^{\circ}=-0.14V\) 2. Compare the reduction potentials. Since -0.14V < 0.00V, the tin reduction potential is more negative. 3. The stronger reducing agent is Sn(s).
04

(d): I⁻(aq) or Br⁻(aq)

: 1. The reduction potentials are: I₂ + 2e⁻ → 2I⁻: \(E^{\circ}=0.54V\) Br₂ + 2e⁻ → 2Br⁻: \(E^{\circ}=1.07V\) 2. Compare the reduction potentials. Since 0.54V < 1.07V, the iodine reduction potential is more negative. 3. The stronger reducing agent is I⁻(aq).

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

Using the standard reduction potentials listed in Appendix E, calculate the equilibrium constant for each of the following reactions at \(298 \mathrm{~K}\) : (a) $\mathrm{Cu}(s)+2 \mathrm{Ag}^{+}(a q) \longrightarrow \mathrm{Cu}^{2+}(a q)+2 \mathrm{Ag}(s)$ (b) $3 \mathrm{Ce}^{4+}(a q)+\mathrm{Bi}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 3 \mathrm{Ce}^{3+}(a q)+ \mathrm{BiO}^{+}(a q)+2 \mathrm{H}^{+}(a q)$ (c) $\mathrm{N}_{2} \mathrm{H}_{5}^{+}(a q)+4 \mathrm{Fe}(\mathrm{CN})_{6}^{3-}(a q) \longrightarrow \mathrm{N}_{2}(g)+ 5 \mathrm{H}^{+}(a q)+4 \mathrm{Fe}(\mathrm{CN})_{6}^{4-}(a q)$

A voltaic cell similar to that shown in Figure 20.5 is constructed. One electrode half-cell consists of a magnesium strip placed in a solution of \(\mathrm{MgCl}_{2}\), and the other has a nickel strip placed in a solution of \(\mathrm{NiCl}_{2}\). The overall cell reaction is $$ \mathrm{Mg}(s)+\mathrm{Ni}^{2+}(a q) \longrightarrow \mathrm{Ni}(s)+\mathrm{Mg}^{2+}(a q) $$ (a) What is being oxidized, and what is being reduced? (b) Write the half- reactions that occur in the two half-cells. (c) Which electrode is the anode, and which is the cathode?(d) Indicate the signs of the electrodes. (e) Do electrons flow from the magnesium electrode to the nickel electrode or from the nickel to the magnesium? (f) In which directions do the cations and anions migrate through the solution?

Predict whether the following reactions will be spontaneous in acidic solution under standard conditions: (a) oxidation of \(\mathrm{Cu}\) to \(\mathrm{Cu}^{2+}\) by \(\mathrm{I}_{2}\) (to form \(\mathrm{I}^{-}\) ), \((\mathbf{b})\) reduction of \(\mathrm{Fe}^{2+}\) to \(\mathrm{Fe}\) by \(\mathrm{H}_{2}\) (to form \(\mathrm{H}^{+}\) ), \(\left(\mathbf{c}\right.\) ) reduction of \(\mathrm{I}_{2}\) to \(\mathrm{I}^{-}\) by $\mathrm{H}_{2} \mathrm{O}_{2},(\mathbf{d})\( reduction of \)\mathrm{Ni}^{2+}\( to \)\mathrm{Ni}$ by \(\mathrm{Sn}^{2+}\left(\right.\) to form \(\left.\mathrm{Sn}^{4+}\right)\).

Iron corrodes to produce rust, \(\mathrm{Fe}_{2} \mathrm{O}_{3},\) but other corrosion products that can form are \(\mathrm{Fe}(\mathrm{O})(\mathrm{OH})\), iron oxyhydroxide, and magnetite, \(\mathrm{Fe}_{3} \mathrm{O}_{4} .\) (a) What is the oxidation number of Fe in iron oxyhydroxide, assuming oxygen's oxidation number is \(-2 ?(\mathbf{b})\) The oxidation number for Fe in magnetite was controversial for a long time. If we assume that oxygen's oxidation number is \(-2,\) and Fe has a unique oxidation number, what is the oxidation number for Fe in magnetite? (c) It turns out that there are two different kinds of Fe in magnetite that have different oxidation numbers. Suggest what these oxidation numbers are and what their relative stoichiometry must be, assuming oxygen's oxidation number is -2 .

A voltaic cell is constructed with two \(\mathrm{Cu}^{2+}-\mathrm{Cu}\) electrodes. The two half-cells have $\left[\mathrm{Cu}^{2+}\right]=0.100 \mathrm{M}\( and \)\left[\mathrm{Cu}^{2+}\right]=1.00 \times 10^{-4} \mathrm{M}$, respectively. (a) Which electrode is the cathode of the cell? (b) What is the standard emf of the cell? (c) What is the cell emf for the concentrations given? (d) For each electrode, predict whether \(\left[\mathrm{Cu}^{2+}\right]\) will increase, decrease, or stay the same as the cell operates.
See all solutions

Recommended explanations on Chemistry Textbooks

Making Measurements

Read Explanation

Kinetics

Read Explanation

Chemical Reactions

Read Explanation

The Earths Atmosphere

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Physical Chemistry

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