JEE Main 2014 — Redox Reactions Question with Solution

$[F{e}^{2+}+C_2{O}_4^{2-}\to F{e}^{3+}+2C{O}_2+3e^{-}]\times 2(oxidation)$
$C{r}_2{O}_7^{2-}+14H^{+}+6e^{-}\to 2C{r}^{3+}+7H_{2}O(reduction)$
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${\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}+2{\mathrm{Fe}}_{ }^{2+}+2{\mathrm{C}}_2{\mathrm{O}}_4^{2-}+14{\mathrm{H}}_{}^{+}\stackrel{}{\to }2{\mathrm{Cr}}^{3+}+2{\mathrm{Fe}}^{3+}+4{\mathrm{CO}}_2+7{\mathrm{H}}_2\mathrm{O}$

Hence, total 6 electrons are involved in the net redox reaction to happen.