Question

Write the final balanced reaction for the following redox reaction:
$C{r}_2{O}_7^{2-}+F{e}^{2+}\to C{r}^{3+}+F{e}^{3+}$

• A. $10H^{+}+C{r}_2{O}_7^{2-}+5F{e}^{2+}\to 2C{r}^{3+}+5H_{2}O+5F{e}^{3+}$
• B. $14H^{+}+C{r}_2{O}_7^{2-}+4F{e}^{2+}\to 2C{r}^{3+}+7H_{2}O+4F{e}^{3+}$
• C. $12H^{+}+C{r}_2{O}_7^{2-}+6F{e}^{2+}\to 2C{r}^{3+}+6H_{2}O+6F{e}^{3+}$
• D. $14H^{+}+C{r}_2{O}_7^{2-}+6F{e}^{2+}\to 2C{r}^{3+}+7H_{2}O+6F{e}^{3+}$

Answer 1

The correct option is D $14H^{+}+C{r}_2{O}_7^{2-}+6F{e}^{2+}\to 2C{r}^{3+}+7H_{2}O+6F{e}^{3+}$

Oxidation half-reaction Reduction half-reaction

$[F{e}^{+2}⟶F{e}^{+3}+e^{-}]\times 6$ $[14H^{+}+C{r}_2{O}_7^{-2}+6e^{-}⟶2C{r}^{+3}+7H_{2}O]\times 1$

$⟶$ Multiply whole respective equation with the numbers which makes electrons balance, then add both equations.

Finally we get,

$14H^{+}+C{r}_2{O}_7^{-2}+6F{e}^{+2}⟶2C{r}^{+3}+7H_{2}O+6F{e}^{+3}$