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Redox Reactions — Chemistry STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceChemistryRedox Reactions5 Marks
Question
Balance the following equations in basic medium by ion-electron method and oxidation number methods and identify the oxidising agent and the reducing agent.$\text{P}_4(\text{s})+\text{OH}^{-}(\text{aq})\rightarrow\text{PH}_3(\text{g})+\text{HPO}_2^-(\text{aq})$
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Answer



$P_4$ acts both as an oxidising as well as a reducing agent.
Oxidation number method:
Total decrease in O.N. of $P_4$ in $PH_3 = 3 \times 4 = 12$
Total increase in O.N. of $P_4$ in $\text{H}_2\text{PO}_2^{-}$ = 1 × 4 = 4
Therefore, to balance increases decreases in O.N. multiply $PH_3$ by 1 and $\text{H}_2\text{PO}_2^{-}$ by 3, we have,
$\text{P}_4(\text{s})+\text{OH}^{-}(\text{aq})\rightarrow\text{PH}_3(\text{g})+3\text{H}_2\text{PO}_2^{-}(\text{aq})$
To balance O atoms, multiply $OH^-$ by 6, we have,
$\text{P}_4(\text{s})+6\text{OH}^{-}(\text{aq})\rightarrow\text{PH}_3(\text{g})+3\text{H}_2\text{PO}_2^{-}(\text{aq})$
To balance H atoms, add $3H_2O$ to L.H.S. and $3OH^-$​​​​​​​ to the R.H.S., we have,
$\text{P}_4(\text{s})+6\text{OH}^{-}(\text{aq})+3\text{H}_2\text{O(l)}\rightarrow\\\text{PH}_3(\text{g})+3\text{H}_2\text{PO}_2^{-}(\text{aq})+3\text{OH}^-(\text{aq})$
or $\text{P}_4(\text{s})+3\text{OH}^{-}(\text{aq})+3\text{H}_2\text{O(l)}\rightarrow\\\text{PH}_3(\text{g})+3\text{H}_2\text{PO}_2^{-}(\text{aq})\ ....(\text{i})$
Thus, eq (i) represents the correct balanced equation.
Ion electron method. The two half reactions are:
Oxidation half reaction:
$\text{P}_4(\text{s})\rightarrow\text{H}_2\text{PO}^-_2(\text{aq})\ .....(\text{ii})$
Balancing P atoms, we have,
$\ \ \ \ \stackrel{{0}}{\ \ \ \ \ \ \ \hbox{P}_4(\text{s})}\rightarrow\stackrel{{+1}}{\ \ \ \ \ \ \ 4\hbox{H}_2\text{PO}^-_2(\text{aq})}$
Balance O.N. by adding electrons,
$\text{P}_4(\text{s})\rightarrow4\text{H}_2\text{PO}^-_2(\text{aq})+4\text{e}^-$
Balance charge by adding 8 $OH^-$ ions,
$\text{P}_4(\text{s})+8\text{OH}^-(\text{aq})\rightarrow4\text{H}_2\text{PO}^-_2(\text{aq})+4\text{e}^-\ .....(\text{iii})$
O and H get automatically balanced. Thus, eq. (iii) represents the balanced oxidation half reaction.
Reduction half reaction:
$\ \ \ \ \stackrel{{0}}{\ \ \ \ \ \ \ \hbox{P}_4(\text{s})}\rightarrow\stackrel{{-3}}{\ \ \ \ \ \ \ \hbox{PH}_3(\text{g})}\ .....(\text{iv})$
Balancing P atoms, we have,
$\text{P}_4(\text{s})\rightarrow4\text{PH}_3(\text{g})$
Balance O.N. by adding electrons,
$\text{P}_4(\text{s})+12\text{e}^-\rightarrow4\text{PH}_3(\text{g})$
Balance charge by adding $12OH^-$​​​​​​​ ions,
$\text{P}_4(\text{s})+12\text{e}^-\rightarrow4\text{PH}_3(\text{g})+12\text{OH}^-(\text{aq})$
Balance O atoms, by adding $12H_2O$ to L.H.S. of above equation.
$\text{P}_4(\text{s})+12\text{H}_2\text{O(l)}+12\text{e}^{-}\rightarrow4\text{PH}_3(\text{g})+12\text{OH}^{-}(\text{aq})\ .....(\text{v})$
To cancel out electrons, multiply eq. (iii) by 3 and add it to eq. (v), we have,
$4\text{P}_4(\text{s})+24\text{OH}^{-}(\text{aq})+12\text{H}_2\text{O(l)}\rightarrow\\4\text{PH}_3(\text{aq})+12\text{H}_2\text{PO}_2^-(\text{aq})+12\text{H}_2\text{O(l)}+12\text{OH}^{-}(\text{aq})\ ......(\text{vi})$
Or $\text{P}_4(\text{g})+3\text{OH}^{-}(\text{aq})+3\text{H}_2\text{O(l)}\rightarrow\text{PH}_3(\text{aq})+3\text{H}_2\text{PO}_2^-(\text{aq})$
Thus, eq. (vi) represents the correct balanced equation.

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