Question

For the redox reaction,
$xF{e}^{2+}+yC{r}_2{O}_7^{2-}+z{H}^{+}\to F{e}^{3+}+C{r}^{3+}+H_{2}O$
x, y and z are

• A. (x)-(3),(y)-(1),(z)-(14)
• B. (x)-(6),(y)-(1),(z)-(7)
• C. (x)-(6),(y)-(2),(z)-(14)
• D. (x)-(6),(y)-(1),(z)-(14)

Answer 1

Answer: (D)

To balance a redox reaction, balance effected atoms
$(I)F{e}^{2+}\to F{e}^{3+}$ (Fe-balanced)
(II) Take oxidation number of two Cr-atom $[C{r}_2{O}_7^{2-}\to 2C{r}_6^{3+}]$ (Cr-balanced) Take total (oxidation number) of each atom (oxidised reduced) cross-multiple by change in oxidation number and add
$I\times 6+II\times I$
$6F{e}^{2+}+C{r}_2{O}_7^{2-}\to 6F{e}^{3+}+2C{r}^{3+}$ Balance O, by $H_{2}O\to .......+7H_{2}O$
Balance H, by $H+.....+14H+$
Thus, $x=6$
$y=1$
$z=14$