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Balance the following equation by oxidation number method:
$Fe to the power of 2 plus end exponent space plus space Cr subscript 2 straight O subscript 7 superscript 2 minus end superscript space plus space straight H to the power of plus space space rightwards arrow space space Fe to the power of 3 plus end exponent space plus space 2 Cr to the power of 3 plus end exponent space plus space 3 straight H subscript 2 straight O$

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132.

Balance the following equation by oxidation number method:
$left enclose CrO subscript 4 right parenthesis to the power of 2 minus end exponent space plus space space left parenthesis SO subscript 3 right parenthesis to the power of 2 minus end exponent space space space rightwards arrow space space end enclose open vertical bar Cr left parenthesis OH right parenthesis subscript 4 close vertical bar to the power of 1 minus end exponent space plus space left parenthesis SO subscript 4 right parenthesis to the power of 2 minus end exponent$
(in alkaline medium)

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133.
Balance the following redox reaction by oxidation number method:

$left parenthesis Cr left parenthesis OH right parenthesis subscript 4 superscript minus space plus space straight H subscript 2 straight O subscript 2 space space rightwards arrow space space space space CrO subscript 4 superscript 2 minus end superscript space plus space straight H subscript 2 straight O space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space left parenthesis Basic space medium right parenthesis$

(i) The skeleton equation along with oxidation number of each atom is
(ii) Noting the change in O.N. of above atoms,
Increase in Cr = + 3 to +6 i.e.
Decrease per atom of O = 1 (from O in H2O2 to O in H2O)
Total number of O atom involved = 2
Total decrease for 2 atoms of O = 2 x 1 = 2
(iii) Equalise the increase/decrease in O.N. by multiplying H2O2 by 3 and

(iv)

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134.

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

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135. 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.

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136.

With 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.

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137. Write the various steps for balancing redox equation by ion-electron method (Half reaction method).

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138.

Balance the equation by half reaction method:

$straight I subscript 2 space plus space OH to the power of minus space space space space space rightwards arrow space space space space IO subscript 3 superscript minus space plus space straight I to the power of minus space plus space straight H subscript 2 straight O$

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139.

Balance the following equation by half reaction method in acidic medium:
$MnO subscript 4 superscript minus space plus space Br to the power of minus space space space rightwards arrow space space space Mn to the power of 2 plus end exponent space plus space Br squared$