Stoichiometry
6.0 Redox Reactions
6.0 Redox Reactions
Those reactions which involve the exchange of electrons are called redox reactions. For the calculation of n-factor of oxidising agent or reducing agent, the method depends upon the change in oxidation state of the species considered. We will discuss them one by one.
(i) When only one atom undergoing either reduction or oxidation e.g.
\[\mathop {\mathop {MnO_4^ - }\limits^{ + 7} }\limits_{n = 5} \xrightarrow{{{H^ + }}}\mathop {M{n^{2 + }}}\limits^{ + 2} \]
In such a case, we consider the change in oxidation state of atom undergoing oxidation or reduction change per molecule as the n-factor of the species.
$n-factor$ = $|\left( { + 2} \right)1{\text{ }}-{\text{ }}\left( { + 7} \right)1|{\text{ }} = {\text{ }}5$
\[n - factor{\text{ }} = {\text{ }}\left| {\left( { + 3} \right){\text{ }} \times {\text{ }}1{\text{ }}-{\text{ }}\left( { + 2} \right){\text{ }} \times {\text{ }}1} \right|{\text{ }} = {\text{ }}1\]
\[n - factor{\text{ }} = {\text{ }}\left| {\left( { + 4} \right){\text{ }} \times {\text{ }}2{\text{ }}-{\text{ }}\left( { + 3} \right){\text{ }} \times {\text{ }}2} \right|{\text{ }} = {\text{ }}2\]
\[n - factor{\text{ }} = {\text{ }}\left| {\left( { + 3} \right){\text{ }} \times {\text{ }}2{\text{ }}--{\text{ }}\left( { + 6} \right) \times {\text{ }}2} \right|{\text{ }} = {\text{ }}6\]
\[\mathop {\mathop {F{e^{2 + }}}\limits^{ + 2} }\limits_{n = 1} \xrightarrow{{}}\mathop {F{e^{3 + }}}\limits^{ + 3} \]
\[\mathop {\mathop {{C_2}O_4^{ - 2}}\limits^{ + 3 \times 2} }\limits_{n = 2} \xrightarrow{{}}\mathop {2C{O_2}}\limits^{ + 4 \times 2} \]
\[\mathop {\mathop {C{r_2}O_7^{2 - }}\limits^{ + 6 \times 2} }\limits_{n = 6} \xrightarrow{{}}\mathop {2C{r^{3 + }}}\limits^{ + 3 \times 2} \]
(ii) Salts which reacts in such a way that only one atom undergoes change in oxidation state but appears in two products with the same oxidation state:
In such a case the method of calculation of n-factor remains the same i.e., we will calculate the change in oxidation state of the atom per mole of that substance (reactant).
\[\mathop {C{r_2}O_7^{2 - }}\limits^{ + 6 \times 2} \xrightarrow{{}}\mathop {C{r^{3 + }}}\limits^{ + 3} + \mathop {C{r^{3 + }}}\limits^{3 + } \]
In this example, oxidation state of $Cr$ changes from $+6$ to $+3$ in both the products. So
n-factor $=$ $\left| {\left( { + 6} \right){\text{ }} \times {\text{ }}2{\text{ }}-{\text{ }}\left( { + 3} \right){\text{ }} \times {\text{ }}2} \right|{\text{ }} = {\text{ }}6$
(iii) Salts which react in such a way that only one atom undergoes change in oxidation state but goes in two products with different oxidation state as a result of either only oxidation or only reduction.
\[\mathop {3MnO_4^ - }\limits^{ + 7} \xrightarrow{{}}\mathop {2M{n^{2 + }}}\limits^{ + 2} + \mathop {M{n^{ + 6}}}\limits^{ + 6} \]
In such a case, it is impossible to calculate the $n-$factor until and unless one knows that how much of $MnO_4^ - $ is changing to $M{n^{2 + }}$ and how much to $M{n^{6 + }}$ and if one knows the balanced equation then there is no need of calculation of $n-$factor.
Nevertheless in such case the $n-$factor can be calculated by deducing the total change in oxidation state divided by total number of atom undergoing reduction/oxidation change. So, for the calculation of $n-$factor in the above example, out of three moles of $MnO_4^ - $, two moles are being converted to $M{n^{2 + }}$ and one mole changes to $M{n^{6 + }}$. So total decrease in oxidation state of $Mn$.
\[{{\text{ = }}\left| {{\text{ }}\left[ {2{\text{ }} \times {\text{ }}\left( { + 2} \right){\text{ }}-{\text{ }}2{\text{ }} \times {\text{ }}\left( { + 7} \right)} \right]{\text{ }}} \right|{\text{ }} + {\text{ }}\left| {{\text{ }}\left[ {1{\text{ }} \times {\text{ }}\left( { + 6} \right){\text{ }}-{\text{ }}1{\text{ }} \times {\text{ }}\left( { + 7} \right)} \right]} \right|}\]\[{ = \left| {{\text{ }}4{\text{ }}-{\text{ }}14{\text{ }}} \right|{\text{ }} + {\text{ }}\left| {{\text{ }}6{\text{ }}-{\text{ }}7{\text{ }}} \right|{\text{ }} = {\text{ }}11}\]
So, $n-$factor = $\frac{{11}}{3}$
(iv) Salts which react in such a way that only one atom undergoes change in oxidation state in two product, in one product with changed oxidation state and in other product with same oxidation state as that of reactant.
In such case also one cannot calculate the $n-$factor without knowing the balanced chemical equation because one must know how much of atom has changed its oxidation state. For example.
\[{K_2}C{r_2}{O_7} + {\text{ }}14HCl \to 2KCl{\text{ }} + {\text{ }}2CrC{l_3} + {\text{ }}3C{l_2} + {\text{ }}7{H_2}O\]
Let us calculate the $n-$factor of $HCl$. Out of $14$ moles of $Cl– (in HCl)$ only $6$ moles of $Cl–$ are changing its oxidation state from $–1$ to $0$ in the product $C{l_2}$ and the oxidation state of remaining 8 $Cl–$ ions remains same in $KCl$ and $CrC{l_3}$. So, total no. of moles of electrons lost by $14$ moles of $HCl$ is $6$. So each mole of $HCl$ takes up $\frac{6}{{14}}$ i.e., $\frac{3}{{7}}$ moles of electrons and hence $n-$ factor of $HCl$ is $\frac{3}{{7}}$.
(v) Salts which react in such a way that two or more atoms in the salt undergoes change in oxidation states as a result of either oxidation or reduction. Let us consider the following example,
\[Fe{C_2}{O_4} \to F{e^{3 + }} + {\text{ }}2C{O_2}\]
In this case, the oxidation of both $F{e^{2 + }}$ and ${C^{3 + }}$ are changing from + 2 and +3 to +3 and +4 respectively. In such a case we will calculate the n-factor of the salt as the total increase or decrease in oxidation state per mole of the salt. As one can see that one mole of $Fe{C_2}{O_4}$ contains one mole of $F{e^{2 + }}$ and one mole of ${C_2}{O_4}^{2-}$– (i.e. 2 carbon atoms per mole of ${C_2}{O_4}^{2-}$
Total change in oxidation state
$$ = {\text{ }}\left| {{\text{ }}1{\text{ }} \times {\text{ }}\left( { + 2} \right){\text{ }}-{\text{ }}1{\text{ }} \times {\text{ }}\left( { + 3} \right){\text{ }}} \right|{\text{ }} + {\text{ }}\left| {{\text{ }}2{\text{ }} \times {\text{ }}\left( { + 3} \right){\text{ }}-{\text{ }}2{\text{ }} \times {\text{ }}\left( { + 4} \right){\text{ }}} \right|$$$$ = {\text{ }}1{\text{ }} + {\text{ }}2{\text{ }} = {\text{ }}3$$
So, $n-$factor of $Fe{C_2}{O_4}$ is $3$.
(vi) Salts which react in such a way that two atoms undergoing change in oxidation state but one undergoing oxidation and other reduction reaction. In such a case one has to calculate the change in oxidation state of either the atom being oxidized or the atom being reduced. For example.
\[\mathop {(N}\limits^{ - 3 \times 2} {H_4}{)_2}{\mathop {Cr}\limits^{ + 6 \times 2} _2}{O_7}\xrightarrow{{}}\mathop {{N_2}}\limits^{0 \times 2} + \mathop {C{r_2}}\limits^{ + 3 \times 2} {O_3} + 4{H_2}O\]
In this reaction, the oxidation state of N is increasing by 6 units and that of Cr is decreasing by 6 unit. So, we can consider either oxidation or reduction product for the calculation of n-factor and it will be the same.
$n-$factor of ${\left( {N{H_4}} \right)_2}C{r_2}{O_7}$ considering oxidation
$ = \left| {\left( { - 3} \right) \times 2-\left( 0 \right) \times 2} \right| = 6$
$n-$factor of ${\left( {N{H_4}} \right)_2}C{r_2}{O_7}$ considering reduction
$ = \left| {\left( { + 6} \right) \times 2-\left( { + 3} \right) \times 2} \right| = 6$
(vii) Species which undergoes disproportionation reaction: Those reaction in which oxidant and reductant are the same species or the same element from the species is getting oxidized as well as reduced.
When the number of moles of atoms being oxidized is equal to the number of moles of atoms being reduced.
The $n-$factor can be calculated by knowing the balanced chemical equation and considering any of the change taking place. Say for example,
\[2{H_2}{O_2} \to 2{H_2}O{\text{ }} + {\text{ }}{O_2}\]
Out of $2$ moles of ${H_2}{O_2}$ consumed in the reaction, one mole of ${H_2}{O_2}$ is being oxidized as \[({H_2}{O_2} \to {O_2})\] and one mole of ${H_2}{O_2}$ is being reduced as, \[({H_2}{O_2} \to 2{H_2}O).\] First consider the oxidation reaction
\[\mathop {{H_2}{O_2}}\limits^{( - 1) \times 2} \xrightarrow{{}}\mathop {{O_2}}\limits^{2 \times 0} \]
n-factor = $\left| {2{\text{ }} \times {\text{ }}0{\text{ }}-{\text{ }}\left( {-1} \right){\text{ }} \times {\text{ }}2} \right|{\text{ }} = {\text{ }}2$
Again, considering reduction reaction
\[\mathop {{H_2}{O_2}}\limits^{( - 1) \times 2} \xrightarrow{{}}\mathop {2{H_2}O}\limits^{( - 2) \times 2} \]
n-factor = $\left| {\left( { - 2} \right){\text{ }} \times {\text{ }}2{\text{ }}-{\text{ }}\left( { - 1} \right){\text{ }} \times {\text{ }}2} \right|{\text{ }} = {\text{ }}2$
So, $n-$factor of ${H_2}{O_2}$ either considering oxidation or reduction reaction is same i.e. $2$.
