Fe2+ + H+ + Cr2o72- = Cr3+ + Fe 3+ +H2O. Balance the given reaction byOxidation number method.

Balance the given reaction by Half reaction method. MnO4- + H+ +Br- =Mn2+ +Br2+H2O

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1. For the oxidation no. method, we follow the following steps,

For the equation; Fe+2 + H+ + Cr2O72-  -->Cr+3 + Fe+3 + H2O

  1. Oxidation no. of each species is calculated for LHS,  as:

For Cr2O72- , the oxidation no. of Cr = 2 (x) + -2 x 7 = -2, or  x = +6

 2). Determination of oxidation and reduction parts,

 Here, we see that iron, is oxidized and Cr is reduced as: Fe from (+2 to +3 ) and Cr from (+6 to +3)

Fe+2 -> Fe+3 + 1e-

2Cr+6 + 6e- -> 2Cr+3

3). Balancing no. of electrons lost = no. of electrons gained

6Fe+2 -> 6Fe+3 + 6e-

2Cr+6 + 6e- -> 2Cr+3

4.) Now the half reactions are added as:

6Fe+2 + 2Cr+6 -->6Fe+3 + 2Cr+3

5).Now, the main reaction is written as:

6Fe+2 + Cr2O72-  --> 6Fe+3 + 2Cr+3

6.) Charged is balanced on LHS and RHS as

Charge on LHS = +12 -2 = +10

Charge on RHS = +18 + 6 = +24

So, we need to add +10 charge on left side to balance the reaction charge and so we add 10 H+ on left side as:

6Fe+2 + Cr2O72- + 14H+ -->6Fe+3 + 2Cr+3

And, at the right side, the no. of oxygen atoms is balanced by adding 7 water molecule as;

 6Fe+2 + Cr2O72- + 14H+ -->6Fe+3 + 2Cr+3 + 7H2O (the balanced equation)

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