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

Short Answer

Expert verified
8 electrons are transferred in reaction (c), 2 in (a) and (b).

Step by step solution

01

Reaction (a) Determine Oxidation Numbers

For the reaction \( \mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) \longrightarrow 2 \mathrm{HF}(g) \), we assign oxidation numbers as follows:- \( \mathrm{H}_{2} \): Hydrogen has an oxidation number of 0 (elemental state).- \( \mathrm{F}_{2} \): Fluorine has an oxidation number of 0 (elemental state).- \( \mathrm{HF} \): Hydrogen has an oxidation number of +1, and fluorine has an oxidation number of -1.
02

Reaction (a) Count Electrons Transferred

In the reaction, each \( \mathrm{H} \) atom goes from an oxidation state of 0 to +1, losing 1 electron, while each \( \mathrm{F} \) atom goes from an oxidation state of 0 to -1, gaining 1 electron.Therefore, the total number of electrons transferred is 2 (one from each \( \mathrm{H} \) atom to a \( \mathrm{F} \) atom).
03

Reaction (b) Determine Oxidation Numbers

For the reaction \( 2 \mathrm{Fe}^{2+}(aq) + \mathrm{H}_{2}\mathrm{O}_{2}(aq) + 2\mathrm{H}^{+}(aq) \longrightarrow 2 \mathrm{Fe}^{3+}(aq) + \mathrm{H}_{2}\mathrm{O}(l) \):- \( \mathrm{Fe}^{2+} \): Iron has an oxidation number of +2.- \( \mathrm{Fe}^{3+} \): Iron has an oxidation number of +3.- \( \mathrm{H}_{2}\mathrm{O}_{2} \): Each oxygen usually has an oxidation number of -1.- \( \mathrm{H}_{2}\mathrm{O} \): Hydrogen is +1, and oxygen is -2.
04

Reaction (b) Count Electrons Transferred

From \( \mathrm{Fe}^{2+} \) to \( \mathrm{Fe}^{3+} \), 1 electron is lost per iron ion. Since there are 2 \( \mathrm{Fe}^{2+} \) ions, a total of 2 electrons are transferred in this reaction.
05

Reaction (c) Determine Oxidation Numbers

For the reaction \( \mathrm{CH}_{4}(g) + 2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) + 2 \mathrm{H}_{2}\mathrm{O}(l) \):- In \( \mathrm{CH}_{4} \): Carbon is -4, hydrogen is +1.- In \( \mathrm{O}_{2} \): Oxygen is 0 (elemental state).- In \( \mathrm{CO}_{2} \): Carbon is +4, oxygen is -2.- In \( \mathrm{H}_{2}\mathrm{O} \): Hydrogen is +1, oxygen is -2.
06

Reaction (c) Count Electrons Transferred

Carbon goes from -4 in \( \mathrm{CH}_{4} \) to +4 in \( \mathrm{CO}_{2} \), losing a total of 8 electrons. Each oxygen in \( \mathrm{O}_{2} \) goes from 0 to -2 (each gaining 2 electrons), totaling 8 electrons gained by all oxygen atoms in the products.

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Key Concepts

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

Oxidation Numbers
Oxidation numbers are essential for understanding oxidation-reduction (redox) reactions. An oxidation number is a theoretical charge that an atom would have if all bonds it forms were ionic. This concept helps chemists identify which atoms are oxidized and which are reduced in a reaction.

To assign oxidation numbers, follow these basic guidelines:
  • Elements in their elemental form have an oxidation number of 0, such as Hydrogen in \( \mathrm{H}_{2} \) and Fluorine in \( \mathrm{F}_{2} \).
  • For ions, the oxidation number equals the charge of the ion. For instance, the oxidation number of \( \mathrm{Fe}^{2+} \) is +2, and for \( \mathrm{Fe}^{3+} \), it is +3.
  • Oxygen usually has an oxidation number of -2, except in peroxides like \( \mathrm{H}_2\mathrm{O}_2 \), where it is -1.
  • Hydrogen is typically +1, except when bonded to metals in binary compounds (like metal hydrides), where it is -1.
Identifying these oxidation numbers helps further understand electron transfer and how substances in a reaction change.
Electron Transfer
The heart of an oxidation-reduction reaction is electron transfer. This transfer is what causes changes in oxidation states of the elements involved in a reaction. Oxidation refers to the loss of electrons, while reduction refers to the gain of electrons. This interplay is often remembered using the mnemonic "OIL RIG" — "Oxidation Is Loss" and "Reduction Is Gain."
  • In the reaction \( \mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) \longrightarrow 2 \mathrm{HF}(g) \), hydrogen is oxidized as it loses electrons transitioning from an oxidation state of 0 to +1.
  • On the same note, each fluorine atom is reduced by accepting electrons, going from an oxidation state of 0 to -1.
  • This specific reaction results in a total transfer of 2 electrons — one electron per hydrogen transferred to each fluorine atom.
Understanding electron transfer is imperative. It not only aids in comprehending the oxidation-reduction reaction but also helps in predicting the products formed.
Balancing Chemical Equations
Balancing chemical equations can seem daunting, but it's a crucial skill in chemistry to ensure the law of conservation of mass is satisfied. Each reaction must have the same number of each type of atom on both sides of the equation. This step is vital in ensuring that the amount of matter remains constant throughout the process.

For example, take the equation:
\(2 \mathrm{Fe}^{2+}(aq) + \mathrm{H}_{2}\mathrm{O}_{2}(aq) + 2\mathrm{H}^{+}(aq) \longrightarrow 2 \mathrm{Fe}^{3+}(aq) + \mathrm{H}_{2}\mathrm{O}(l)\)
It's balanced because:
  • There are equal numbers of iron, hydrogen, and oxygen atoms on both sides.
  • Furthermore, charge is balanced; the total charge on the left equals the total charge on the right.
Balancing equations allows chemists to precisely understand the stoichiometry of a reaction, predicting the amounts of reactants needed and products formed. It also assists in calculating the number of electrons transferred, crucial for redox reactions. Balancing helps ensure the reaction is depicted accurately, aligning with basic chemical principles.

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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. $$ \text { (a) } 2 \mathrm{AgNO}_{3}(a q)+\mathrm{CoCl}_{2}(a q) \longrightarrow 2 \mathrm{AgCl}(s)+ $$ \(\mathrm{Co}\left(\mathrm{NO}_{3}\right)_{2}(a q)\) $$ \begin{array}{l} \text { (b) } 2 \mathrm{PbO}_{2}(s) \longrightarrow 2 \mathrm{PbO}(s)+\mathrm{O}_{2}(g) \\ \text { (c) } 2 \mathrm{H}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{NaBr}(s) \longrightarrow \mathrm{Br}_{2}(l)+\mathrm{SO}_{2}(g)+ \end{array} $$ \(\mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)\)

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. \((\mathbf{a})\) 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.

(a) What is the definition of the volt? (b) Do all voltaic cells produce a positive cell potential?

Disulfides are compounds that have \(\mathrm{S}-\mathrm{S}\) bonds, like peroxides have \(\mathrm{O}-\mathrm{O}\) bonds. Thiols are organic compounds that have the general formula \(\mathrm{R}-\mathrm{SH},\) where \(\mathrm{R}\) is a generic hydrocarbon. The \(\mathrm{SH}^{-}\) ion is the sulfur counterpart of hydroxide, \(\mathrm{OH}^{-}\). Two thiols can react to make a disulfide, \(\mathrm{R}-\mathrm{S}-\mathrm{S}-\mathrm{R} .\) (a) What is the oxidation state of sulfur in a thiol? (b) What is the oxidation state of sulfur in a disulfide? (c) If you react two thiols to make a disulfide, are you oxidizing or reducing the thiols? (d) If you wanted to convert a disulfide to two thiols, should you add a reducing agent or oxidizing agent to the solution? (e) Suggest what happens to the H's in the thiols when they form disulfides.

A voltaic cell is constructed with two silver-silver chloride electrodes, each of which is based on the following half-reaction: $$ \mathrm{AgCl}(s)+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s)+\mathrm{Cl}^{-}(a q) $$ The two half-cells have \(\left[\mathrm{Cl}^{-}\right]=0.0150 \mathrm{M}\) and \(\left[\mathrm{Cl}^{-}\right]=\) \(2.55 \mathrm{M},\) 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 [Cl^{-} ] will increase, decrease, or stay the same as the cell operates.

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