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Redox Reactions

Question
CBSEENCH11006770

Write the net ionic equation for the reaction of potassium dichromate (VI), K2Cr2O7 with sodium sulphite, Na2SO3, in an acid solution to give chromium (III) ion and the sulphate ion.

Solution

(i) The skeleton ionic equation is
        space space Cr subscript 2 straight O subscript 7 superscript 2 minus end superscript left parenthesis aq right parenthesis space plus space SO subscript 3 superscript 2 minus end superscript left parenthesis aq right parenthesis space space space rightwards arrow space space space Cr to the power of 3 plus end exponent left parenthesis aq right parenthesis space plus space SO subscript 4 superscript 2 minus end superscript left parenthesis aq right parenthesis space space space
(ii) Write the O.N. of each atom and identify the atoms which undergo a change in O.N.

           

(iii) Calculate the total increase and decrease in O.N.
Since there are two Cr atoms on L.H.S. and only one on R.H.S., therefore multiply Cr3+on R.H.S. of (1) by 2 and thus a total decrease in O.N. of Cr is 2 x 3 = 6.
(iv) Equalise the increase/decrease in O.N. by multiplying SO subscript 3 superscript 2 minus end superscript space by space 3 space.
Cr subscript 2 straight O subscript 7 superscript 2 minus end superscript space left parenthesis aq right parenthesis space plus space 3 SO subscript 3 superscript 2 minus end superscript left parenthesis aq right parenthesis space space rightwards arrow space space space 2 Cr to the power of 3 plus end exponent left parenthesis aq right parenthesis space plus space SO subscript 4 superscript 2 minus end superscript left parenthesis aq right parenthesis
(v) Balance S atoms by multiplying SO subscript 4 superscript 2 minus end superscript by 3.
              Cr subscript 2 straight O subscript 7 superscript 2 minus end superscript left parenthesis aq right parenthesis space plus 3 SO subscript 3 superscript 2 minus end superscript left parenthesis aq right parenthesis space space rightwards arrow space space 2 Cr to the power of 3 plus end exponent left parenthesis aq right parenthesis space plus space 3 SO subscript 4 superscript 2 minus end superscript left parenthesis aq right parenthesis
(vi) Balance O atoms by adding 4H2O molecules towards the R.H.S.
Cr subscript 2 straight O subscript 7 superscript 2 minus end superscript left parenthesis aq right parenthesis space plus space 3 SO subscript 3 superscript 2 minus end superscript left parenthesis aq right parenthesis space space space rightwards arrow space space space 2 Cr to the power of 3 plus end exponent left parenthesis aq right parenthesis space plus space 3 SO subscript 4 superscript 2 minus end superscript left parenthesis aq right parenthesis space plus space 4 straight H subscript 2 straight O left parenthesis straight l right parenthesis
(vii) Balance H atoms by adding 8H ions on the L.H.S., since the reaction, occurs in the acidic medium.
     Cr subscript 2 straight O subscript 7 superscript 2 minus end superscript left parenthesis aq right parenthesis space plus space 3 SO subscript 3 superscript 2 minus end superscript space left parenthesis aq right parenthesis space plus space 8 straight H to the power of plus left parenthesis aq right parenthesis space space rightwards arrow space space space 2 Cr to the power of 3 plus end exponent left parenthesis aq right parenthesis space plus space 3 SO subscript 4 superscript 2 minus end superscript left parenthesis aq right parenthesis space plus space 4 straight H subscript 2 straight O left parenthesis straight l right parenthesis