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(a) What is meant by the term oxidation? (b) On which side of an oxidation half-reaction do the electrons appear? (c) What is meant by the term oxidant? (d) What is meant by the term oxidizing agent?

Short Answer

Expert verified
Oxidation involves electron loss; electrons appear on the right in the oxidation half-reaction. An oxidant or oxidizing agent accepts electrons, facilitating oxidation.

Step by step solution

01

Understanding Oxidation

Oxidation refers to the process in which an atom, ion, or molecule loses electrons, leading to an increase in its oxidation state. This is a key component of redox reactions, which involve both oxidation and reduction (the gain of electrons).
02

Electrons in Oxidation Half-Reaction

In an oxidation half-reaction, electrons are shown on the right side of the equation. This indicates that electrons are being released from the species undergoing oxidation. For example, in the half-reaction \( ext{A} ightarrow ext{A}^{+} + e^{-}\), the electron \(e^{-}\) appears on the right.
03

Defining an Oxidant

An oxidant (or oxidizing agent) is a substance that causes another substance to be oxidized. The oxidant itself gains electrons in the process, becoming reduced while facilitating the oxidation of another substance.
04

Understanding Oxidizing Agent

The term oxidizing agent is synonymous with oxidant. It refers to the substance that accepts electrons during an oxidation reaction, itself undergoing reduction. Hence, the oxidizing agent is instrumental in enabling the oxidation process.

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Key Concepts

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

Oxidizing agent
An oxidizing agent plays a crucial role in chemical reactions by accepting electrons from another substance. This interaction facilitates the process known as oxidation. When an oxidizing agent accepts electrons, it undergoes a complementary process called reduction.

This means the oxidizing agent is reduced while causing another substance to be oxidized. This dual action of losing and gaining electrons in reactions is typical of redox reactions.
  • An easy way to remember this is through the mnemonic 'OIL RIG' - Oxidation Is Loss, Reduction Is Gain (of electrons).
  • Oxidizing agents are often strong electron acceptors, such as oxygen or chlorine.
The role of the oxidizing agent extends to various practical applications, including batteries and rust formation, indicating its importance across different chemical processes.
Redox reactions
Redox reactions, short for reduction-oxidation reactions, are chemical processes involving the transfer of electrons between two substances. This transfer is essential as it drives the chemical changes needed in both natural and industrial reactions.

In a redox reaction, one substance undergoes oxidation by losing electrons, while another substance undergoes reduction by gaining those electrons. This electron transfer results in changes to the oxidation states of the reacting substances.
  • The substance that gets oxidized, losing electrons, is termed the reducing agent.
  • The oxidizing agent gains those electrons and is thereby reduced.
These reactions are critical in countless chemical and biological processes, such as cellular respiration and photosynthesis, illustrating their significant role in both living organisms and technology.
Half-reaction
Understanding half-reactions is fundamental to grasping how redox reactions function. A half-reaction focuses on either the oxidation or reduction part of a redox reaction separately, detailing the electron exchange process specifically.

For example, in the oxidation half-reaction, electrons are shown being released and written on the right side of the equation. A simple example of this would be \( \text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-\). Conversely, for a reduction half-reaction, electrons are added to the left side of the equation.
  • Half-reactions help balance overall chemical equations by keeping track of electron flow.
  • They also aid in identifying which species are oxidized and which are reduced.
By breaking down the entire redox reaction into half-reactions, one can isolate and visualize the specific processes that each reactant undergoes.

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Most popular questions from this chapter

A cell has a standard cell potential of \(+0.257 \mathrm{~V}\) at \(298 \mathrm{~K}\). What is the value of the equilibrium constant for the reaction \((\mathbf{a})\) if \(n=1 ?(\mathbf{b})\) if \(n=2 ?\) (c) if \(n=3 ?\)

A \(1 M\) solution of \(\mathrm{AgNO}_{3}\) is placed in a beaker with a strip of Ag metal. A \(1 \mathrm{M}\) solution of \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\) is placed in a second beaker with a strip of Cu metal. A salt bridge connects the two beakers, and wires to a voltmeter link the two metal electrodes. (a) Which electrode serves as the anode, and which as the cathode? (b) Which electrode gains mass, and which loses mass as the cell reaction proceeds? (c) Write the equation for the overall cell reaction. (d) What is the emf generated by the cell under standard conditions?

A voltaic cell utilizes the following reaction: $$ 2 \mathrm{Fe}^{3+}(a q)+\mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{Fe}^{2+}(a q)+2 \mathrm{H}^{+}(a q) $$ (a) What is the emf of this cell under standard conditions? (b) What is the emf for this cell when \(\left[\mathrm{Fe}^{3+}\right]=3.50 \mathrm{M}, P_{\mathrm{H}_{2}}=\) \(96.3 \mathrm{kPa},\left[\mathrm{Fe}^{2+}\right]=0.0010 \mathrm{M}\), and the \(\mathrm{pH}\) in both half-cells is \(4.00 ?\)

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?

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 \(\mathrm{CO}\) into \(\mathrm{CO}_{2}\).

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