$\text{P}_4(\text{s})+\text{OH}^{-}(\text{aq})\rightarrow\text{PH}_3(\text{g})+\text{HPO}_2^-(\text{aq})$
P4 acts both as an oxidising as well as a reducing agent.
Oxidation number method:
Total decrease in O.N. of P4 in PH3 = 3 × 4 = 12
Total increase in O.N. of P4 in $\text{H}_2\text{PO}_2^{-}$ = 1 × 4 = 4
Therefore, to balance increases decreases in O.N. multiply PH3 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 3H2O 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 12H2O 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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$2\text{A}\rightleftharpoons\text{B}+\text{C}$ is 2 × 10-3. At a given time, the composition of reaction mixture is [A] = [B] = [C] = 3 × 10-4M. In which direction the reaction will proceed?
$\text{CCI}_4(\text{g})\xrightarrow{ \ \ \ \ \ }\text{C}(\text{g})+4\text{CI}(\text{g})$
and calculate bond enthalpy of C – Cl in CCl4(g).$\Delta_\text{vap}\text{H}^\ominus(\text{CCI}_4)=30.5\text{kJ} \ \text{mol}^{-1}.$
$\Delta_\text{f}\text{H}^\ominus(\text{CCI}_4)=-135.5\text{kJ} \ \text{mol}^{-1}.$
$\Delta_\text{a}\text{H}^\ominus(\text{C})=-715.0\text{kJ} \ \text{mol}^{-1},$ where $\Delta_\text{a}\text{H}^\ominus$ is enthalpy of atomisation
$\Delta_\text{a}\text{H}^\ominus(\text{CI}_2)=242\text{kJ} \ \text{mol}^{-1}$