Redox Reactions Half-Reactions
If a zinc rod is dipped into a beaker containing a solution of copper sulfate, after some time the blue color (due to copper ions) of the solution starts fading. The zinc rod starts dissolving partially and its surface gets coated with metallic copper. This happens because the following redox reaction takes place.
⇒ \(\mathrm{Zn}(\mathrm{s})+\mathrm{CuSO}_4(\mathrm{aq}) \longrightarrow \mathrm{ZnSO}_4(\mathrm{aq})+\mathrm{Cu}(\mathrm{s})\)
In an aqueous solution, copper sulfate dissociates to form Cu2+ (aq) and SO4+(aq).
So we can write the reaction as:
⇒ \(\mathrm{Zn}(\mathrm{s})+\mathrm{Cu}^{2+}(\mathrm{aq})+\mathrm{SO}_4^{2-}(\mathrm{aq}) \longrightarrow \mathrm{Cu}(\mathrm{s})^{+} \mathrm{Zn}^2+(\mathrm{aq})+\mathrm{SO}_4^{2-}(\mathrm{aq})\)
The sulfate ions do not participate in the reaction. Zinc loses electrons and gets oxidized to Zn2+ while Cu2+ gains electrons and gets reduced to metallic copper.
The overall reaction consists of two parts—oxidation and reduction, which can be represented separately as:
⇒ \(\mathrm{Zn}-2 \mathrm{e}^{-} \longrightarrow \mathrm{Zn}^{2+}\)
⇒ \(\mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}\)
Both (a) and (b) represent half-reactions*—(a) represents an oxidation half-reaction and (b) a reduction half- reaction
“Redox reactions, definition, types, examples, and applications”
It is interesting to note that if we dip a copper rod in a zinc sulfate solution, no reaction occurs.
The solution does not turn blue as it would have if copper had dissolved giving Cu2+(aq).
The attempt to detect Cu2+ in the solution by adding H2S also fails as no black color of cupric sulfide (CuS) is observed.
In the redox reaction discussed above zinc acts as a reducing agent, reducing Cu2+ to copper while copper ion (Cu2+) acts as an oxidizing agent, oxidizing Zn to Zn2+.
Whether a substance acts as an oxidant or a reductant depends upon its ability to accept or donate electrons.
This will be dear to you by studying another reaction occurring within the same experimental set-up as shown here.
As you, a copper rod is dipped in a solution of silver nitrate. After some time the copper rod dissolves partially, being oxidized to Cu2+.
The initially colorless solution turns blue. Metallic silver gets deposited on the copper rod from the silver nitrate solution.
⇒ \(\mathrm{Cu}(\mathrm{s})+2 \mathrm{Ag}^{+}(\mathrm{aq})+2 \mathrm{NO}_3^{-}(\mathrm{aq}) \longrightarrow \mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s})+2 \mathrm{NO}_3^{-}(\mathrm{aq})\)
or
⇒ \(\mathrm{Cu}(\mathrm{s})+2 \mathrm{Ag}^{+}(\mathrm{aq}) \longrightarrow \mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s})\)
The nitrate ion is passive in the above reaction. The silver ion accepts electrons and gets reduced while copper metal loses electrons and gets oxidized.
Thus, the silver ion is the oxidant and copper metal is the reductant. However, the reverse reaction does not take place, that is, silver does not dissolve in copper nitrate.
“Oxidation-reduction reactions, types, examples, and real-world applications”
Now consider the two reactions observed in the above two experiments.
As you can see from the above representations, the role of copper is reversed in the two reactions.
In the reaction with zinc, Cu2+ is reduced to Cu while in the reaction with Ag+, Cu is oxidized to Cu2+.
These reactions proceed in this manner as in the state of equilibrium, the formation of products is greatly favored over that of the reactants.
Now let us consider the reaction occurring between metallic iron and a cobalt sulfate solution. The expected reaction will be
At equilibrium, it is found that the solution contains both Fe2+ and Co2+, i.e., the reaction does not proceed to completion as neither the formation of the reactants nor that of the products is greatly favored.
Thus, we find that some metals readily donate electrons whilst some do not. Thus, the metals may be arranged in order of ease of donation of electrons.
“Redox reactions, chemical process, examples, and industrial applications”
If we consider the three metals Zn, Cu, and Ag, the electron-releasing tendency would be Zn > Cu > Ag.
This kind of series in which metals are arranged in their decreasing order of reactivity (i.e., electron-releasing tendency) is called the electrochemical series.
Redox reactions are of great significance not only because they account for many biochemical and industrial processes but they are also used to generate electricity, for instance, in electrochemical cells such as galvanic cells.
The chemical reactions taking place in such cells are the sources of electricity.
Types Of Redox Reactions
We have just seen that any reaction accompanied by a change in the oxidation number of any species is considered to be a redox reaction.
Most chemical changes can be classified as a combination, decomposition, and replacement or substitution reactions, and many such reactions are also redox reactions as shown by the examples cited here.
1. Combination reactions
A combination reaction may be denoted as
X + Y→ Z
In this case, if any one of the reactants or both are in elemental form then the reaction is a redox reaction. When an element forms a compound, its oxidation number changes.
⇒ \(\stackrel{0}{\mathrm{C}}+\stackrel{0}{\mathrm{O}}_2 \longrightarrow \stackrel{+4}{\mathrm{C}} \stackrel{-2}{\mathrm{O}}_2\)
⇒ \(2 \stackrel{0}{\mathrm{~A}}+\stackrel{0}{\mathrm{~N}}_2 \longrightarrow 2 \stackrel{+3}{2} \stackrel{-3}{\mathrm{~N}}\)
⇒ \(2 \stackrel{\mathrm{Al}}{+}+\stackrel{0}{\mathrm{~N}}_2 \longrightarrow 2 \mathrm{Al}^{+3} \stackrel{-3}{\mathrm{~N}}\)
⇒ \(2 \stackrel{0}{\mathrm{Mg}}+\stackrel{0}{\mathrm{O}_2} \longrightarrow 2 \stackrel{+2}{\mathrm{Mg} \mathrm{O}}^{-2}\)
“Types of redox reactions, oxidation, reduction, and their significance”
⇒ \(\stackrel{0}{\mathrm{~N}}_2+3 \stackrel{0}{\mathrm{H}}{ }_2 \longrightarrow 2 \stackrel{+3}{\mathrm{~N}} \stackrel{-1}{\mathrm{H}}_3\)
⇒ \(\stackrel{0}{\mathrm{Fe}}+\stackrel{0}{\mathrm{~S}} \longrightarrow \stackrel{+2}{\mathrm{Fe}}-\stackrel{-}{\mathrm{S}}^2\)
⇒ \(2 \stackrel{-}{\mathrm{C}}_4 \stackrel{+1}{\mathrm{H}}_{10}+13 \stackrel{0}{\mathrm{O}}_2 \longrightarrow 8 \stackrel{4}{\mathrm{C}}^{-2} \mathrm{O}_2+10 \mathrm{H}_2 \mathrm{O}^{-2}\)
When elements combine to form compounds the oxidation number of the more electronegative element decreases while that of the other increases.
2. Decomposition reactions
Decomposition reactions are the opposite of combination reactions. In such a reaction, the reactant (a compound) decomposes to give two or more products, which may be either elements or compounds. Such reactions may also be redox reactions. A few examples are shown as follows.
⇒ \(2 \stackrel{+2}{\mathrm{H}} \stackrel{-2}{\mathrm{O}} \longrightarrow 2 \stackrel{0}{\mathrm{H}} \longrightarrow \stackrel{0}{\mathrm{O}}\)
⇒ \(2 \stackrel{+1}{2} \stackrel{-}{O}_2 \longrightarrow 2 \stackrel{0}{\mathrm{H}}_2+\stackrel{0}{\mathrm{O}_2}\)
⇒ \(2 \mathrm{~K}^{+1} \stackrel{+5}{\mathrm{C}}-20, \longrightarrow 2 \mathrm{O}^{+1} \mathrm{~K}^{-1} \mathrm{Cl}+3 \mathrm{O}_2\)
It may be noted that the oxidation number of K in potassium chlorate remains unchanged but that of Cl decreases while that of oxygen increases.
3. Displacement reactions
A displacement or substitution reaction is one in which an atom or a molecule is replaced by another atom or a molecule.
A+BC→AC+B
The above equation shows that A replaces B. We may come across reactions where a metal displaces another metal or reactions where a nonmetal displaces another nonmetal. Let us discuss both types in brief.
Metal displacement In such reactions a metal in a compound is displaced by another metal in the uncombined state. We have already seen the case of Zn replacing Cu in CuS04,
⇒ \(\stackrel{0}{\mathrm{Zn}}+\stackrel{+2}{\mathrm{CuSO}_4} \longrightarrow \stackrel{+2}{\mathrm{ZnSO}_4}+\stackrel{0}{\mathrm{Cu}}\)
There are many other examples
⇒ \(\stackrel{+4}{\mathrm{TiCl}}{ }_4+2 \stackrel{0}{\mathrm{Mg}} \longrightarrow 2 \stackrel{+2}{\mathrm{MgCl}_2}+\stackrel{0}{\mathrm{~T} i}\)
⇒ \(\stackrel{+3}{\mathrm{Fe}} \mathrm{O}_3+2 \stackrel{0}{\mathrm{Al}} \longrightarrow \stackrel{+3}{\mathrm{Al}_2} \mathrm{O}_3+2 \stackrel{0}{\mathrm{Fe}}\)
⇒ \(\stackrel{+3}{\mathrm{Cr}_2} \mathrm{O}_3+2 \stackrel{0}{\mathrm{Al}} \longrightarrow \stackrel{+3}{\mathrm{Al}_2 \mathrm{O}_3}+\stackrel{0}{2 \mathrm{Cr}}\)
Many such reactions are used in metallurgical operations to obtain a metal from its oxide. Reduction with aluminum is called a thermite reaction.
“Redox reactions, everyday examples, industrial uses, and biological importance”
Nonmetal displacement A nonmetal may be displaced by a metal or another nonmetal. A metal may displace hydrogen from water or acids.
⇒ \(2 \stackrel{0}{\mathrm{Na}}+2 \stackrel{+1}{2} \mathrm{H}_2 \mathrm{O} \longrightarrow 2 \stackrel{+1}{\mathrm{NaOH}}+\stackrel{0}{\mathrm{H}_2}\)
⇒ \(\stackrel{0}{\mathrm{Mg}}+\stackrel{+1}{2} \mathrm{H}_2 \mathrm{O} \longrightarrow \stackrel{+2}{\mathrm{M}}(\mathrm{OH})_2+\stackrel{0}{\mathrm{H}_2}\)
⇒ \(\stackrel{0}{\mathrm{Zn}}+\stackrel{+1}{2} \mathrm{HCl} \longrightarrow \stackrel{+2}{\mathrm{Z}} \mathrm{nCl}_2+\stackrel{0}{\mathrm{H}}_2\)
⇒ \(\stackrel{0}{\mathrm{Fe}}+\stackrel{+1}{\mathrm{H}_2} \mathrm{SO}_4 \longrightarrow \stackrel{+2}{\mathrm{FeSO}} \mathrm{SO}_4+\stackrel{0}{\mathrm{H}_2}\)
The reactivity of a metal may be adjudged by these reactions. Highly reactive metals like sodium and potassium react violently with cold water.
Magnesium reacts with warm water while iron reacts with steam. Also, metals like iron and zinc liberate hydrogen readily from ads whereas less reactive metals like copper, silver, and gold do not.
A classical example of one nonmetal displacing another is the case of a more reactive halogen displacing a less reactive halogen from halides. The reactivity of halogens follows the order:
F>Cl>Br>I
⇒ \(\stackrel{0}{\mathrm{C}} \mathrm{l}_2+2 \stackrel{+1}{\mathrm{Na}} \mathrm{-}^{-1} \longrightarrow 2 \mathrm{Na} \stackrel{-1}{\mathrm{Cl}}^{-1} \stackrel{0}{\mathrm{X}}_2(\mathrm{X}=\mathrm{Br}, \mathrm{I})\)
⇒ \(\stackrel{0}{\mathrm{Br}_2}+2 \mathrm{NaI} \longrightarrow 2 \mathrm{Na} \stackrel{-1}{\mathrm{~N}} \longrightarrow \mathrm{I}_2\)
In all the cases shown above the higher halogen is reduced while the lower one is oxidised.
Halogens are obtained by the oxidation of halides. Huorine is the most reactive halogen and the strongest oxidizing agent.
Thus F- cannot be chemically oxidized to F-, this is achieved by electrolysis. In fact, fluorine is so reactive that when it comes in contact with water it displaces the oxygen of the water vigorously
⇒ \(2 \stackrel{0}{\mathrm{~F}}_2+2 \stackrel{+1}{\mathrm{H}}_2-\stackrel{2}{\mathrm{O}} \longrightarrow 4 \stackrel{+1}{\mathrm{H}} \stackrel{-1}{\mathrm{~F}}+\stackrel{0}{\mathrm{O}}{ }_2\)
“Redox reactions, electrochemical cells, examples, and practical applications”
Thus, fluorine cannot be used to displace chlorine, bromine, and iodine in their aqueous solutions.
4. Disproportionation reactions
In these reactions, a substance is both oxidized and reduced. For example, when a metal superoxide.
⇒ \(\begin{aligned}
& 2 \mathrm{KO}_2^{-1 / 2}(\mathrm{~s})+\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \longrightarrow 0^0(\mathrm{~g})+2 \mathrm{~K}^{+}(\mathrm{aq})+\mathrm{HO}_2^{-1}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \\
& \text { Potassium } \\
& \text { superoxid } \\
&
\end{aligned}\)
dissolves in water it decomposes with the evolution of oxygen. In the given reaction, the oxygen in K02 is simultaneously oxidized from the)/1 oxidation state into the 0 oxidation state in 02 and reduced from the}2 oxidation state in 02 to the -1 oxidation state in HO-
2. Some other disproportionation reactions are as follows.
⇒ \(2 \stackrel{+1}{\mathrm{H}}_2 \stackrel{-1}{\mathrm{O}}_2 \longrightarrow 2 \stackrel{+1}{\mathrm{H}}_2 \stackrel{-2}{\mathrm{O}}+\stackrel{0}{\mathrm{O}_2}\)
⇒ \(2 \mathrm{Cu}^{+1} \mathrm{Cl}^{-1} \longrightarrow \stackrel{+2}{\mathrm{C} u} \mathrm{Cl}_2+\stackrel{0}{\mathrm{C} u}\)
⇒ \(3 \stackrel{+1}{\mathrm{Na}} \stackrel{+1}{\mathrm{C}}-2 \mathrm{O}^{-} \longrightarrow \stackrel{+1}{\mathrm{Na}} \stackrel{+5}{\mathrm{C}}-2 \mathrm{O}_3+2 \stackrel{+1}{\mathrm{Na}} \mathrm{Cl}^{-1}\)
Phosphorus, sulfur, and halogens also undergo disproportionation reactions in alkaline media.
⇒ \(\stackrel{0}{\mathrm{P}}_4(\mathrm{~s})+3 \mathrm{NaOH}(\mathrm{aq})+3 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \longrightarrow \stackrel{-3}{\mathrm{P}}_3(\mathrm{~g})+3 \mathrm{NaH}_2 \stackrel{+1}{\mathrm{P}}_2(\mathrm{aq})\)
⇒ \(\stackrel{0}{\mathrm{X}_2}(\mathrm{~g})+2 \mathrm{NaOH}(\mathrm{aq}) \longrightarrow \underset{(\mathrm{X}=\mathrm{Cl}, \mathrm{Br}, \mathrm{l})}{\mathrm{NaX}} \stackrel{+1}{(\mathrm{aq})}+\mathrm{NaX}^{-1}(\mathrm{aq})+\mathrm{H}_2 \mathrm{O}(\mathrm{l})\)
⇒ \(\stackrel{0}{\mathrm{~S}}_8(\mathrm{~s})+12 \mathrm{NaOH}(\mathrm{aq}) \longrightarrow 4 \mathrm{Na}_2 \stackrel{-2}{\mathrm{~S}}(\mathrm{aq})+2 \mathrm{Na}_2 \stackrel{+2}{\mathrm{~S}}_2{ }_2^{-2} \mathrm{O}_3(\mathrm{aq})+6 \mathrm{H}_2 \mathrm{O}(\mathrm{I})\)
“Oxidation and reduction reactions, examples, and their role in chemistry”
- Fluorine does not undergo such a disproportionation reaction as it does not display a positive oxidation state.
- Thus, it is obvious that in a disproportionation reaction, the reactant is present in an intermediate oxidation state and decomposes to give one product showing a higher oxidation state and another showing a lower oxidation state.
- A compound containing an element in the highest oxidation state will not be disproportionate as a higher oxidation state is not attainable.
Balancing Redox Reactions
For any chemical reaction to be balanced, the number of atoms of each element on both sides of the chemical equation representing it should be the same (law’ of conservation of mass).
Simple reactions can be balanced by the hit-and-trial method but for complex ones, a systematic approach has to be followed.
“Redox reactions, balancing methods, examples, and their importance”
Redox reactions can be balanced in two systematic ways. The first method is based on the change in the oxidation number of reductant and oxidant and is called the oxidation-number method.
In the second method, the reaction is split into two parts—one denoting reduction and the other oxidation. This is known as the half-reaction method.