Answer:
8
Explanation:
Oxidation:
![Fe^{2+} -->Fe^{3+}+e^{-}](https://tex.z-dn.net/?f=Fe%5E%7B2%2B%7D%20--%3EFe%5E%7B3%2B%7D%2Be%5E%7B-%7D)
Reduction:
![Cr_{2}O_{7}^{2-}+6e^{-} -->2Cr^{3+}](https://tex.z-dn.net/?f=Cr_%7B2%7DO_%7B7%7D%5E%7B2-%7D%2B6e%5E%7B-%7D%20--%3E2Cr%5E%7B3%2B%7D)
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+}](https://tex.z-dn.net/?f=6Fe%5E%7B2%2B%7D%2BCr_%7B2%7DO_%7B7%7D%5E%7B2-%7D%20--%3E6Fe%5E%7B3%2B%7D%2B2Cr%5E%7B3%2B%7D)
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](https://tex.z-dn.net/?f=6Fe%5E%7B2%2B%7D%2BCr_%7B2%7DO_%7B7%7D%5E%7B2-%7D%2B14H%5E%7B%2B%7D%20--%3E6Fe%5E%7B3%2B%7D%2B2Cr%5E%7B3%2B%7D%2B7H_%7B2%7DO)
Sum of coefficients of product cations = 6+2 = 8