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Redox Reactions

Question
CBSEENCH11006769

Balance the following equation:
Cr left parenthesis OH right parenthesis subscript 3 space plus space IO subscript 3 superscript minus space space space rightwards arrow space space space CrO subscript 4 superscript 2 minus end superscript space plus space straight I to the power of minus
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space left parenthesis basic space medium right parenthesis space space space space space space space space


Solution

(i) The skeleton equation along with O.N. of each atom isstack Cr with 3 plus on top space left parenthesis space straight O with negative 2 on top space straight H with plus 1 on top right parenthesis subscript 3 with Cr equals space plus 3 below space plus stack left square bracket straight I with plus 5 on top stack straight O subscript 3 superscript minus with negative 2 on top right square bracket space with straight I equals plus 5 below rightwards arrow space left square bracket stack Cr with plus 6 on top stack straight O subscript 4 with negative 2 on top with Cr equals plus 6 below right square bracket space plus stack space stack straight I to the power of minus with negative 1 on top with straight I equals negative 1 below

(ii) Noting the change in O.N. of the above atoms,
   Increase     Cr = +3   to +6  i.e. 3
  Decrease      1 = +5 to  -1  i.e. 6
(iii) Balancing the total increase in O.N. with a total decrease in O.N. by multiplying Cr(OH)3 by 2.
      2 Cr left parenthesis OH right parenthesis subscript 3 plus space space IO subscript 3 superscript minus space space space rightwards arrow space space space 2 CrO subscript 4 superscript 2 minus end superscript space plus space straight I to the power of minus
(iv) Balance Cr atoms by multiplying CrO subscript 4 superscript 2 minus end superscript by 2.
                  2 Cr left parenthesis OH right parenthesis subscript 3 space plus space IO subscript 3 superscript minus space space space rightwards arrow space space 2 CrO subscript 4 superscript 2 minus end superscript space plus space straight I to the power of minus
(v) Balance O atoms by adding OHion on the R.H.S.
          2 Cr left parenthesis OH right parenthesis subscript 3 space plus space IO subscript 3 superscript minus space space space rightwards arrow space space space 2 CrO subscript 4 superscript 2 minus end superscript space plus space straight I to the power of minus space plus space OH to the power of minus
(vi) Balance H atoms by adding H2O and OH- ions as,
              2 Cr left parenthesis OH right parenthesis subscript 3 space plus space IO subscript 3 superscript minus space plus space 5 OH to the power of minus space space rightwards arrow space space space 2 CrO subscript 4 superscript 2 minus end superscript space plus space straight I to the power of minus space plus space 5 straight H subscript 2 straight O space plus space OH to the power of minus
2 Cr left parenthesis OH right parenthesis subscript 3 space plus space IO subscript 3 superscript minus space plus space 4 OH to the power of minus space space rightwards arrow space space space 2 CrO subscript 4 superscript 2 minus end superscript space plus space straight I to the power of minus space plus space 5 straight H subscript 2 straight O