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Redox reactions


Oxidation and reduction

The key chemical event in a redox reaction is the net movement of electrons from one reactant to the other. As a result of electron movement, some reactants gain electrons (reduction), and others lose electrons (oxidation).

Oxidation is the loss of electron(s) by an atom or ion.

Reduction is the gain of electron(s) by an atom or ion.

Example 1: Overall reaction: \[ \ce{CuO}\left(s\right)+\ce{H}_{2}\left(g\right)\rightarrow\ce{Cu}\left(s\right)+\ce{H}_{2}\ce{O}\left(g\right) \]

Where reduction is \[ \ce{Cu}^{2+}+2e\rightarrow\ce{Cu} \]

and oxidation is \[ \ce{H}_{2}\rightarrow2\ce{H}^{+}+2e \]

Example 2: The following chemical redox reaction: \[ 2\ce{Ag}^{+}+\ce{Zn}\rightarrow2\ce{Ag}+\ce{Zn}^{2+} \]

Can be described by the two so-called electrochemical “half reactions”: \(\ce{Ag}^{+}\) is reduced \[ 2\ce{Ag}^{+}+2e\rightarrow2\ce{Ag} \]

and \(\ce{Zn}\) is oxidised \[ \ce{Zn}\rightarrow\ce{Zn}^{2+}+2e \]

Oxidising agents and reducing agents

Oxidising agents can oxidise another reactant by accepting electrons. Therefore, the oxidising agent becomes reduced (by gaining electrons) while oxidising another reactant (causing it to lose electrons).

Reducing agents reduce another reactant by offering electrons to the other reactant. Thus, the reducing agent becomes oxidised (by losing electrons) while reducing another reactant (causing it to gain electrons).

Oxidation number

In any discussion of redox reactions it is useful to be able to assign an oxidation number to any element in a given compound in order to keep track of the electrons during a reaction. The oxidation number is defined as the apparent electrical charge on the atom according to a set of rules like those following.

We need to recognize that:

Oxidation leads to a loss of electrons, and an increase in the oxidation number.

Reduction is a gain of electrons, and so leads to a decrease in the oxidation number.

Oxidation number rules
  1. The oxidation number of an atom as an element is zero. Examples: \(\ce{O}_{2},\ce{P}_{4},\ce{Fe},\ce{Br}_{2}\) are the elemental forms at room temperature, and they all have an oxidation number of zero.

  2. The oxidation number of oxygen is usually \(-2\) (except when bonded to \(\ce{F}\)). Remember that in the element, oxygen \(\left(\ce{O}_{2}\right)\) has an oxidation number of zero, whilst in peroxides \(\left(\ce{O}_{2}^{2-}\right)\) each oxygen atom has an oxidation number of \(-1\).

  3. The oxidation number of a monoatomic ion is equal to the ion’s charge. Examples: \(\ce{Na}^{+}=\ce{Na}\left(+1\right),\ce{Cl}^{-}=\ce{Cl}\left(-1\right),\ce{S}^{2-}=\ce{S}\left(-2\right)\).

  4. The algebraic sum of the oxidation numbers in a neutral polyatomic compound is zero; in a polyatomic ion the sum of the oxidation number is equal to the ion’s overall charge.

  5. In a combination of elements, the more electronegative element has its characteristic negative oxidation number (for example, \(\left(-3\right)\) for nitrogen, \(\left(-2\right)\) for oxygen, \(\left(-1\right)\) for chlorine), and the more electropositive element has a positive oxidation number (for example, \(\left(+1\right)\) for hydrogen, \(\left(+2\right)\) for calcium, \(\left(+3\right)\) for aluminium).

  6. The oxidation number of hydrogen is usually \(+1\) due to its electronegativity being lower than that of the main-group elements to which it is normally bonded. With more electropositive elements, hydrogen has an oxidation number of \(\left(-1\right)\). For instance, \(\ce{NaH}=\ce{Na}^{+}\ce{H}^{-}\).

Example 1: Find the oxidation number of \(\ce{S}\) in \(\ce{H}_{2}\ce{SO}_{4}\left(aq\right)\).

Using oxidation number rule \(4\), we can write:

\(2\times\) the oxidation number of \(\ce{H}+\) the oxidation number of \(\ce{S}+4\times\) the oxidation number of \(\ce{O}=0\)

Rule \(2\) states the oxidation number of \(\ce{O}\) is \(-2\) and rule \(6\) states the oxidation number of \(\ce{H}\) is \(+1\)

Lets consider the oxidation number of \(\ce{S}\) as “\(x\)”

Therefore, \[\begin{align*} 2\times\left(+1\right)+x+4\times\left(-2\right) & =0\\ 2+x-8 & =0\\ x-6 & =0\\ x & =+6 \end{align*}\] Example 2: Find the oxidation number of \(\ce{I}\), in \(\ce{ICl}_{4}^{-}\).

We use rule \(4\) to write:

\(\left(\textrm{oxidation number of }\ce{I}\right)+\left(4\times\textrm{oxidation number of }\ce{Cl}\right)=-1\)

According to rule \(3,\) since \(\ce{Cl}\) is more electronegative than \(\ce{I}\), so \(\ce{Cl}\) has the conventional oxidation number of \(-1\). Let’s consider the oxidation number of \(\ce{I}\) as \(x\).

Thus, \[\begin{align*} x+4\times\left(-1\right) & =-1\\ x-4 & =-1\\ x & =-1+4\\ x & =+3 \end{align*}\]

Predicting the products of an oxidation or reduction reaction

We can predict the outcome of the reaction between two redox couples such as \(\ce{MnO}_{4}^{-}/\ce{Mn}^{2+}\) and \(\ce{H}_{2}\ce{SO}_{3}/\ce{SO}_{4}^{2-}\) in the following way:

Because each redox reaction involves the movement of electrons, it also a voltage associated with it. This is called the standard reduction potential \(\left(E^{0}\right)\) which is defined according to \[ Ox^{n+}+ne^{-}\rightarrow Red \]

Where \(Ox^{n+}=\) the oxidised form of the element,

\(Red=\) the reduced form of the element,

\(n=\) the number of electrons transferred in the redox half-reactions

\(E^{0}\) range:

\(E^{0}=\) Standard reduction potential, this is the electron driving force for the reaction as it proceeds from left to right when both \(Ox^{n+}\) and \(Red\) are present at \(1M\) concentration (or one atmosphere partial pressure for gases), and at \(25^{0}C.\)

NOTE: The greater (more positive) the value of \(E^{0}\), the greater the tendency for: \(Ox^{n+}+ne^{-}\rightarrow Red\)

So, from the data in the table below we can decide on the spontaneous (favourable) reaction:

\(Ox^{n+}/Red\) couple \(E^{0}\)(volt) \(e^{-}\) donor/acceptor Oxidising or Reducing Agents
\(\ce{MnO}_{4}^{-}/\ce{Mn}^{2+}\) \(+1.51\) \(e^{-}\) acceptors Oxidising agent
\(\ce{SO}_{4}^{2-}/\ce{H}_{2}\ce{SO}_{3}\left(\ce{SO}_{2}\right)\) \(+0.17\) \(e^{-}\) donor Reducing agent

It is crucial to note that the redox couple with the more positive \(E^{0}\) value will remove electrons from a redox couple with a less positive \(E^{0}\). The oxidising agent is the species which causes the oxidation to occur, i.e. It is the agent (redox couple) which removes electrons from the other couple.

In the example above, \(\ce{MnO}_{4}^{-}\) has the more positive \(E^{0}\) and hence the greater affinity (liking) for electrons: therefore \(\ce{MnO}_{4}^{-}\) will be as the electron acceptor (i.e. Oxidant= oxidising agent). The sulphur system must donate electrons: hence \(\ce{H}_{2}\ce{SO}_{3}\) will be the electron donor (ie Reducing agent).

So we predict: \(\ce{MnO}_{4}^{-}\rightarrow\ce{Mn}^{2+}\) and \(\ce{H}_{2}\ce{SO}_{3}\rightarrow\ce{SO}_{4}^{2-}\)

Reduction and oxidation half-reactions and the balancing of redox equations

Redox equations are usually written as balanced net ionic equations, with spectator ions (ions that do not take part in the overall redox process) being omitted.

The determination of the stoichiometry of a redox reaction can be difficult to deduce even when the products of such a reaction are known. For this reason it is much more convenient and methodical to concentrate upon the stoichiometry of each redox “half-reaction” or “redox couple” individually.

Balancing net ionic equations for redox reactions

A five step method for balancing redox equations, called the “half-reaction” method, is used, and after mastering these operations, any chemical redox equation can be balanced by combining the five steps.

  1. Divide the equation into the appropriate “half-reactions” which can be balanced separately, for example:
    Overall reaction \[ \ce{H}_{2}\ce{SO}_{3}+\ce{MnO}_{4}^{-}\rightarrow\ce{SO}_{4}^{2-}+\ce{Mn}^{2+} \] Assign an oxidation number to each element in the ions or molecules of reactants and products:
    The oxidation number of \(\ce{Mn}\) in \(\ce{MnO}_{4}^{-}\) is \(+7\) and the \(\ce{Mn}\) oxidation number in \(\ce{Mn}^{2+}\) is \(+2\)
    Thus, \(\ce{Mn}^{7+}\)is being reduced to \(\ce{Mn}^{2+}\).
    So, the reduction half-reaction is \[ \ce{MnO}_{4}^{-}\rightarrow\ce{Mn}^{2+} \] The oxidation number of \(\ce{S}\) in \(\ce{H}_{2}\ce{SO}_{3}\) is \(+4\), and the oxidation number of \(\ce{S}\) in \(\ce{SO}_{4}^{2-}\) ion is \(+6\)
    So, \(\ce{S}\) is being oxidised from \(+4\rightarrow+6\).
    So, the oxidation half-reaction \[ \ce{H}_{2}\ce{SO}_{3}\rightarrow\ce{SO}_{4}^{2-} \]

  2. Write a separate balanced chemical equation for each redox-active species. This is the mass balance step:
    Add water to balance oxygen.
    \[ \ce{MnO}_{4}^{-}+8\ce{H}^{+}\rightleftarrows\ce{Mn}^{2+}+4\ce{H}_{2}\ce{O} \] Add \(\ce{H}^{+}\) to balance hydrogens. \[ \ce{H}_{2}\ce{SO}_{3}+\ce{H}_{2}\ce{O}\rightleftarrows\ce{SO}_{4}^{2-}+4\ce{H}^{+} \]

  3. Although we have now achieved mass balance, we also need to balance the charge, by adding electrons (negative charge) as needed. This generates the balanced electrochemical half-reaction. The total charge on each side of a balanced reaction must be the same. \[ \ce{MnO}_{4}^{-}+8\ce{H}^{+}+5e\rightleftarrows\ce{Mn}^{2+}+4\ce{H}_{2}\ce{O} \] Left hand side (LHS) total charge \(=+2\)
    Right hand side (RHS) total charge \(=+2\)
    And \[ \ce{H}_{2}\ce{SO}_{3}+\ce{H}_{2}\ce{O}\rightleftarrows\ce{SO}_{4}^{2-}+4\ce{H}^{+}+2e \] LHS total charge \(=0\)
    RHS total charge \(=0\)
    If all three steps have been done correctly, one half-reaction will have electrons on the left hand side of the equation (the reduction half-reaction), while the other half-reaction will have electrons on the right hand side of the equation (the oxidation half-reaction).

  4. The balanced half-reactions can be termed electrochemical equations, since they involve the electrons. To obtain a chemical equation for the reaction, we need to eliminate the electrons from the overall equation.
    Hence the fourth step is the addition of the two half-reactions in such a way that the electrons cancel out. This may require multiplying one equation or both, by an appropriate integer.
    For the two half-reactions above:
    \[\begin{align*} 2 & \left\{ 5e+8\ce{H}^{+}\ce{MnO}_{4}^{-}\rightleftarrows\ce{Mn}^{2+}+4\ce{H}_{2}\ce{O}\right\} \\ 5 & \left\{ \ce{H}_{2}\ce{SO}_{3}+\ce{H}_{2}\ce{O}\rightleftarrows\ce{SO}_{4}^{2-}+4\ce{H}^{+}+2e\right\} \end{align*}\] The first reaction is multiplied by two and the second reaction is multiplied by five, and both are added to give: \[ 5\ce{H}_{2}\ce{SO}_{3}+2\ce{MnO}_{4}^{-}+16\ce{H}^{+}+5\ce{H}_{2}\ce{O}+10e\rightleftarrows2\ce{Mn}^{2+}+5\ce{SO}_{4}^{2-}+20\ce{H}^{+}+8\ce{H}_{2}\ce{O}+10e \] Note the equal number of electrons on each side of the equation cancel out.

  5. The final step is to cancel the electrons and simplify the chemical equation by eliminating any chemical species common to each side of the equation.

LHS RHS Balance
\(16\ce{H}^{+}\) \(20\ce{H}^{+}\) \(20-16=4\ce{H}^{+}\)on RHS
\(5\ce{H}_{2}\ce{O}\) \(8\ce{H}_{2}\ce{O}\) \(8-5=3\ce{H}_{2}\ce{O}\) on RHS

\[ 5\ce{H}_{2}\ce{SO}_{3}+2\ce{MnO}_{4}^{-}\rightleftarrows2\ce{Mn}^{2+}+5\ce{SO}_{4}^{2-}+3\ce{H}_{2}\ce{O}+\ce{4H}^{+} \]