Question

I know that the answer is C. 8, but can you explain step by step how to get that answer?

Answer 1

8

Step-by-step explanation:

Oxidation:

`Fe^(2+) -->Fe^(3+)+e^(-)`

Reduction:

`Cr_(2)O_(7)^(2-)+6e^(-) -->2Cr^(3+)`

We have to equalise the number of moles of electrons gained and lost in a redox reaction in order to get a balanced reaction.

Hence we have to multiply the oxidation reaction throughout by 6.

and adding the two half-reactions we obtain:

`6Fe^(2+)+Cr_(2)O_(7)^(2-) -->6Fe^(3+)+2Cr^(3+)`

Still the total charge and number of oxygen is not balanced.

Since the reaction takes place in acidic conditions, we will add required number of H+ to the appropriate side to balance the charge and add half the amount of H2O to balance the hydrogen atoms.

We add 14 H+ on LHS and 7H2O on RHS to obtain:

`6Fe^(2+)+Cr_(2)O_(7)^(2-)+14H^(+) -->6Fe^(3+)+2Cr^(3+)+7H_(2)O`

Sum of coefficients of product cations = 6+2 = 8