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Chapter 20: Problem 14

(a) What is meant by the term reduction? (b) On which side of a reduction half-reaction do the electrons appear? (c) What is meant by the term reductant? (d) What is meant by the term reducing agent?

Short Answer

Expert verified
(a) Reduction is the gain of electrons. (b) Electrons appear on the reactant side in reduction half-reactions. (c) A reductant donates electrons. (d) A reducing agent causes reduction by donating electrons.

Step by step solution

01

Understanding Reduction

Reduction is a chemical reaction that involves the gain of electrons by a molecule, atom, or ion. This process decreases the oxidation state of the chemical species involved.
02

Electrons in Reduction Half-Reaction

In a reduction half-reaction, electrons appear on the reactant side. This signifies that the chemical species involved is gaining electrons.
03

Defining a Reductant

The term reductant refers to a substance that donates electrons to another species; in other words, it gets oxidized itself in the process of reducing another substance.
04

Defining a Reducing Agent

A reducing agent is the same as a reductant鈥攊t's a substance that causes reduction by giving away electrons to another substance. This causes the reducing agent to be oxidized in the process.

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Key Concepts

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

Reduction Half-Reaction
In the world of chemistry, reduction is a significant concept that involves the gain of electrons. When we talk about a reduction half-reaction, it specifically describes one side of a redox (reduction-oxidation) reaction. Redox reactions always involve two complementary processes: one being oxidation and the other being reduction. In the reduction half-reaction, electrons appear on the left-hand or the reactant side of the chemical equation. This means that those electrons are added to the species undergoing reduction.
As a species gains these electrons, its oxidation state decreases, signaling that it has been reduced. An easy way to remember this is by using the phrase "OIL RIG": Oxidation Is Loss, Reduction Is Gain (of electrons). Understanding the position of electrons in these reactions and the changes in oxidation states is critical to mastering redox chemistry.
Reductant
The concept of a reductant is pivotal when discussing redox reactions. A reductant is a substance that donates electrons to another substance. By donating electrons, the reductant itself undergoes oxidation because it loses electrons in the process. As it gets oxidized, it helps the other species in the reaction to get reduced, hence acting as an agent of reduction.
Think of the reductant as a kind of electrons' donor. It's willing to give away its electrons to partner species that need them. This electron-supplying trait is essential in driving not just chemical reactions but also many biological processes.
Reducing Agent
The terms "reducing agent" and "reductant" are often used interchangeably in redox chemistry. A reducing agent is simply a substance that causes another substance to undergo reduction by donating electrons. As a result, the reducing agent itself gets oxidized.
  • Acts as an electron donor to the recipient species.
  • Becomes oxidized while reducing another element or compound.
It might be helpful to visualize reducing agents as helpful teammates in a reaction relay race. They hand off electrons, enabling their partners to finish the race, that is, reach a lower oxidation state. Understanding the role of reducing agents allows for a clearer comprehension of how many chemical reactions, including those crucial to life processes, occur.

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Most popular questions from this chapter

Indicate whether the following balanced equations involve oxidation-reduction. If they do, identify the elements that undergo changes in oxidation number. $$ \text { (a) } 2 \mathrm{AgNO}_{3}(a q)+\mathrm{CoCl}_{2}(a q) \longrightarrow 2 \mathrm{AgCl}(s)+ $$ \(\mathrm{Co}\left(\mathrm{NO}_{3}\right)_{2}(a q)\) $$ \begin{array}{l} \text { (b) } 2 \mathrm{PbO}_{2}(s) \longrightarrow 2 \mathrm{PbO}(s)+\mathrm{O}_{2}(g) \\ \text { (c) } 2 \mathrm{H}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{NaBr}(s) \longrightarrow \mathrm{Br}_{2}(l)+\mathrm{SO}_{2}(g)+ \end{array} $$ \(\mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)\)

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. (a) \(\mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) \longrightarrow 2 \mathrm{HF}(g)\) $$ \begin{array}{l} \text { (b) } 2 \mathrm{Fe}^{2+}(a q)+\mathrm{H}_{2} \mathrm{O}_{2}(a q)+2 \mathrm{H}^{+}(a q) \\ \longrightarrow 2 \mathrm{Fe}^{3+}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \\ \text { (c) } \mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) \end{array} $$

Complete and balance the following equations, and identify the oxidizing and reducing agents: (a) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{I}^{-}(a q) \longrightarrow \mathrm{Cr}^{3+}(a q)+\mathrm{IO}_{3}^{-}(a q)\) (acidic solution) (b) \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{CH}_{3} \mathrm{OH}(a q) \longrightarrow \mathrm{Mn}^{2+}(a q)+\) HCOOH \((a q)\) (acidic solution) (c) \(\mathrm{I}_{2}(s)+\mathrm{OCl}^{-}(a q) \longrightarrow \mathrm{IO}_{3}^{-}(a q)+\mathrm{Cl}^{-}(a q)\) (acidic solution) (d) \(\mathrm{As}_{2} \mathrm{O}_{3}(s)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{H}_{3} \mathrm{AsO}_{4}(a q)+\mathrm{N}_{2} \mathrm{O}_{3}(a q)\) (acidic solution) (e) \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{Br}^{-}(a q) \longrightarrow \mathrm{MnO}_{2}(s)+\mathrm{BrO}_{3}^{-}(a q)\) (basic solution) (f) \(\mathrm{Pb}(\mathrm{OH})_{4}^{2-}(a q)+\mathrm{ClO}^{-}(a q) \longrightarrow \mathrm{PbO}_{2}(s)+\mathrm{Cl}^{-}(a q)\) (basic solution)

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.

Gold exists in two common positive oxidation states, +1 and +3 . The standard reduction potentials for these oxidation states are $$ \begin{array}{l} \mathrm{Au}^{+}(a q)+\mathrm{e}^{-} \quad \longrightarrow \mathrm{Au}(s) \quad E_{\mathrm{red}}^{\circ}=+1.69 \mathrm{~V} \\ \mathrm{Au}^{3+}(a q)+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au}(s) \quad E_{\mathrm{red}}^{\circ}=+1.50 \mathrm{~V} \end{array} $$ (a) Can you use these data to explain why gold does not tarnish in the air? (b) Suggest several substances that should be strong enough oxidizing agents to oxidize gold metal. (c) Miners obtain gold by soaking gold-containing ores in an aqueous solution of sodium cyanide. A very soluble complex ion of gold forms in the aqueous solution because of the redox reaction $$ \begin{aligned} 4 \mathrm{Au}(s)+8 \mathrm{NaCN}(a q) &+2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g) \\ & \longrightarrow 4 \mathrm{Na}\left[\mathrm{Au}(\mathrm{CN})_{2}\right](a q)+4 \mathrm{NaOH}(a q) \end{aligned} $$ What is being oxidized, and what is being reduced in this reaction? (d) Gold miners then react the basic aqueous product solution from part (c) with \(\mathrm{Zn}\) dust to get gold metal. Write a balanced redox reaction for this process. What is being oxidized, and what is being reduced?
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