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Redox Reactions

Question
CBSEENCH11006783

Balance the following equations in basic medium by ion-electron method and oxidation number method:

straight P subscript 4 left parenthesis straight s right parenthesis space plus space OH to the power of minus left parenthesis aq right parenthesis space rightwards arrow space space PH subscript 3 left parenthesis straight g right parenthesis space plus space HPO subscript 2 superscript minus left parenthesis aq right parenthesis

Solution

(a) Ion-electron method:
1. Write the oxidation and reduction half-reactions by observing the changes in oxidation numbers.
   Oxidation half-reaction.
            0               +1
           straight P subscript 4 left parenthesis straight s right parenthesis space space rightwards arrow space space space space space straight H subscript 2 PO subscript 2 superscript minus left parenthesis aq right parenthesis
Reduction half-reaction.
            0                -3
           straight P subscript 4 left parenthesis straight s right parenthesis space space space rightwards arrow space space space PH subscript 3 left parenthesis straight g right parenthesis
2. Balancing the oxidation half reaction.
 (i)Balancing P atoms by multiply straight H subscript 2 PO subscript 2 superscript minus by 4
               straight P subscript 4 left parenthesis straight s right parenthesis space space rightwards arrow space space space space 4 straight H subscript 2 PO subscript 2 superscript minus left parenthesis aq right parenthesis
(ii) Add 4 electrons towards RHS to balance the charges. 
                 straight P subscript 4 left parenthesis straight s right parenthesis space space space rightwards arrow space space space 4 straight H subscript 2 PO subscript 2 superscript minus left parenthesis aq right parenthesis
(iii) Add 8OHtowards LHS to balance the charges
              straight P subscript 4 left parenthesis straight s right parenthesis space plus space 8 OH to the power of minus space left parenthesis aq right parenthesis space rightwards arrow space space 4 straight H subscript 2 PO subscript 2 superscript minus left parenthesis aq right parenthesis space plus space 4 straight e to the power of minus
         (Balanced oxidation half reaction)
3. Balancing the reduction half reaction
   (i) Balance P atoms by multiplying PH3 by 4
               0              -3
           straight P subscript 4 left parenthesis straight s right parenthesis space rightwards arrow space space 4 PH subscript 3 left parenthesis straight g right parenthesis
(ii) Add 12 electrons towards LHS  to balance the charge on P
             straight P subscript 4 left parenthesis straight s right parenthesis space plus space 12 straight e to the power of minus space space rightwards arrow space space space 4 PH subscript 3 left parenthesis straight g right parenthesis
(iii) Balance oxygen and hydrogen atoms by adding 12H2O towards LHS and 12OH- towards 
RHS
straight P subscript 4 left parenthesis straight s right parenthesis space plus space 12 straight H subscript 2 straight O left parenthesis straight l right parenthesis space plus 12 straight e to the power of minus space space space rightwards arrow space space 4 PH subscript 3 left parenthesis straight g right parenthesis space plus space 12 OH to the power of minus left parenthesis aq right parenthesis
                                        [Balanced reduction half reaction]
4. Multiply balanced oxidation half-reaction by 3 and add it to the balanced reduction half-reaction, we have
3 straight P subscript 4 left parenthesis straight s right parenthesis space plus 24 OH to the power of minus left parenthesis aq right parenthesis space plus space straight P subscript 4 left parenthesis straight s right parenthesis space plus 12 space straight H subscript 2 straight O left parenthesis straight l right parenthesis space space rightwards arrow space space 12 straight H subscript 2 PO subscript 4 superscript 2 minus end superscript space plus space 4 PH subscript 3 left parenthesis straight g right parenthesis space plus 12 OH to the power of minus left parenthesis aq right parenthesis
or space 4 straight P subscript 4 left parenthesis straight s right parenthesis space plus 12 OH to the power of minus left parenthesis aq right parenthesis space plus 12 straight H subscript 2 straight O left parenthesis straight l right parenthesis space rightwards arrow space space 12 straight H subscript 2 PO subscript 4 superscript 2 minus end superscript space plus space 4 PH subscript 3
or space straight P subscript 4 left parenthesis straight s right parenthesis space plus 3 OH to the power of minus left parenthesis aq right parenthesis space plus 3 straight H subscript 2 straight O left parenthesis straight l right parenthesis space space rightwards arrow space space space 3 straight H subscript 2 PO subscript 4 superscript 2 minus end superscript space plus space PH subscript 3 

This is the balanced redox equation,
(b) Oxidation number method
(i) The skeleton equation along with oxidation number of each atom is
0                                -3 +  1    +1  +1   -2
straight P subscript 4 left parenthesis straight s right parenthesis space plus OH to the power of minus left parenthesis aq right parenthesis space space rightwards arrow space space space PH subscript 3 left parenthesis straight g right parenthesis space plus space straight H subscript 2 space space straight P space space space straight O subscript 2 superscript minus
(ii) The oxidation number of P increases by 1 per atom while that of P decreases by 3 per atom.

P4 acts both as an oxidising as well as reducing agent
Total space increase space in space straight O. straight N. space of space straight P subscript 4 space space in space straight H subscript 2 PO subscript 2 superscript minus space equals space 1 space cross times space 4 space equals space 4
Total space decrease space in space straight O. straight N. space of space straight P subscript 4 space in space PH subscript 3 space equals space 3 space cross times space 4 space equals space 12
left parenthesis iii right parenthesis space Equalise space the space increase divided by decrease space in space straight O. straight N. space by space multiplying space PH subscript 3 space by space 1 space and space straight H subscript 2 PO subscript 2 superscript minus space by space 3
       straight P subscript 4 left parenthesis straight s right parenthesis space plus space OH to the power of minus left parenthesis aq right parenthesis space space rightwards arrow space space space PH subscript 3 left parenthesis straight g right parenthesis space plus space 3 straight H subscript 2 PO subscript 2 superscript minus left parenthesis aq right parenthesis
(iv) Balance O atoms by multiplying OHby 6
straight P subscript 4 left parenthesis straight s right parenthesis space plus space 6 OH to the power of minus left parenthesis aq right parenthesis space space rightwards arrow space space space PH subscript 3 left parenthesis straight g right parenthesis space plus space 3 straight H subscript 2 PO subscript 2 superscript minus left parenthesis aq right parenthesis
(v) Balance H atoms by adding three H2O towards L.H.S. and three OH- towards R.H.S.
straight P subscript 4 left parenthesis straight s right parenthesis space plus space 6 OH to the power of minus left parenthesis aq right parenthesis space plus space 3 straight H subscript 2 straight O left parenthesis straight l right parenthesis space space rightwards arrow space space PH subscript 3 left parenthesis straight g right parenthesis space plus space 3 straight H subscript 2 PO subscript 2 superscript minus left parenthesis aq right parenthesis space plus space 3 OH to the power of minus left parenthesis aq right parenthesis

This is the balanced equation.