Question

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

Answer 1

Answer:

8

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