Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 20: Problem 16

Indicate whether each of the following statements is true or false: (a) If something is reduced, it is formally losing electrons. (b) A reducing agent gets oxidized as it reacts. (c) An oxidizing agent is needed to convert CO into \(\mathrm{CO}_{2}\) .

Short Answer

Expert verified
(a) False - Reduction means a species is gaining electrons, not losing them. (b) True - A reducing agent gets oxidized as it reacts. (c) True - An oxidizing agent is needed to convert CO to \(\mathrm{CO}_{2}\).

Step by step solution

01

(a) Analyzing the statement about reduction and electron loss.

A reduction reaction is one in which a species gains electrons. It's important to remember the mnemonic "LEO says GER" or "Loss of Electrons is Oxidation; Gain of Electrons is Reduction." Thus, the statement is false since a reduction means an electron gain and not loss.
02

(b) Analyzing the statement about reducing agents.

A reducing agent is a substance that reduces other species, meaning it helps other species gain electrons. In doing so, the reducing agent itself loses electrons and gets oxidized. Therefore, the statement is true.
03

(c) Analyzing the statement about converting CO to CO2.

The conversion of CO to \(\mathrm{CO}_{2}\) involves adding an oxygen atom to the CO molecule. This process requires an oxidizing agent, which has the ability to remove electrons from another substance. Specifically, it's needed to transfer an oxygen atom (or a part of it) to CO. Thus, the statement is true. #Summary#: We have determined the truthfulness of the following statements: (a) False - Reduction means a species is gaining electrons, not losing them. (b) True - A reducing agent gets oxidized as it reacts. (c) True - An oxidizing agent is needed to convert CO to \(\mathrm{CO}_{2}\).

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.

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

(a) What is the difference between a battery and a fuel cell? (b) Can the "fuel" of a fuel cell be a solid?

(a) Write the reactions for the discharge and charge of a nickel-cadmium (nicad) rechargeable battery. (b) Given the following reduction potentials, calculate the standard emf of the cell: $$ \begin{array}{r}{\operatorname{Cd}(\mathrm{OH})_{2}(s)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cd}(s)+2 \mathrm{OH}^{-}(a q)} \\\ {E_{\mathrm{red}}^{\circ}=-0.76 \mathrm{V}}\end{array} $$ $$ \begin{array}{r}{\mathrm{NiO}(\mathrm{OH})(s)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{e}^{-} \longrightarrow \mathrm{Ni}(\mathrm{OH})_{2}(s)+\mathrm{OH}^{-}(a q)} \\\ {E_{\mathrm{red}}^{\circ}=+0.49 \mathrm{V}}\end{array} $$ (c) A typical nicad voltaic cell generates an emf of \(+1.30 \mathrm{V}\) . Why is there a difference between this value and the one you calculated in part (b)? (d) Calculate the equilibrium constant for the overall nicad reaction based on this typical emf value.

Given the following half-reactions and associated standard reduction potentials: $$ \begin{array}{c}{\text { AuBr }_{4}^{-}(a q)+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au}(s)+4 \mathrm{Br}^{-}(a q)} \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad {E_{\mathrm{red}}^{\circ}=-0.86 \mathrm{V}} \\ {\mathrm{Eu}^{3+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Eu}^{2+}(a q)} \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad {E_{\mathrm{red}}^{\circ}=-0.43 \mathrm{V}}\end{array} $$ $$ \begin{array}{r}{\mathrm{IO}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-} \longrightarrow \mathrm{I}^{-}(a q)+2 \mathrm{OH}^{-}(a q)} \\\ {E_{\mathrm{red}}^{\circ}=+0.49 \mathrm{V}}\end{array} $$ (a) Write the equation for the combination of these half-cell reactions that leads to the largest positive emf and calculate the value. (b) Write the equation for the combination of half-cell reactions that leads to the smallest positive emf and calculate that value.

For each of the following balanced oxidation-reduction reactions, (i) identify the oxidation numbers for all the elements in the reactants and products and (ii) state the total number of electrons transferred in each reaction. $$ \begin{array}{l}{\text { (a) } 2 \mathrm{MnO}_{4}^{-}(a q)+3 \mathrm{S}^{2-}(a q)+4 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 3 \mathrm{S}(s)+} \\ {\quad 2 \mathrm{MnO}_{2}(s)+8 \mathrm{OH}^{-}(a q)} \\ {\text { (b) } 4 \mathrm{H}_{2} \mathrm{O}_{2}(a q)+\mathrm{Cl}_{2} \mathrm{O}_{7}(g)+2 \mathrm{OH}^{-}(a q) \longrightarrow 2 \mathrm{ClO}_{2}^{-}(a q)+} \\ {\quad 5 \mathrm{H}_{2} \mathrm{O}(l)+4 \mathrm{O}_{2}(g)} \\\\{\text { (c) } \mathrm{Ba}^{2+}(a q)+2 \mathrm{OH}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}_{2}(a q)+2 \mathrm{ClO}_{2}(a q) \longrightarrow} \\ {\quad \mathrm{Ba}\left(\mathrm{ClO}_{2}\right)_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g)}\end{array} $$

Indicate whether each statement is true or false: (a) The cathode is the electrode at which oxidation takes place. (b) A galvanic cell is another name for a voltaic cell. (c) Electrons flow spontaneously from anode to cathode in a voltaic cell.
See all solutions

Recommended explanations on Chemistry Textbooks

Making Measurements

Read Explanation

Chemical Analysis

Read Explanation

Physical Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

The Earths Atmosphere

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