Q.
For the redox reaction,
$xFe^{2+} + yCr_{2}O^{2-}_{7}+ zH^{+} \to Fe^{3+} + Cr^{3+} + H_{2}O$
x, y and z are
Redox Reactions
Solution:
To balance a redox reaction, balance effected atoms
$(I) Fe^{2+} \to Fe^{3+}$ (Fe-balanced)
(II) Take oxidation number of two Cr-atom $ [Cr_{2} O^{2-}_{7} \to 2Cr^{3+}_{6}]$ (Cr-balanced)
Take total (oxidation number) of each atom (oxidised reduced) cross-multiple by change in oxidation number and add
$I\times6+II \times I$
$6Fe^{2+} + Cr_{2}O^{2-}_{7} \to 6Fe^{3+} + 2Cr^{3+}$
Balance O, by $H_{2}O \to ....... + 7H_{2}O$
Balance H, by $H+ .....+ 14H+$
Thus, $x = 6$
$y= 1$
$z = 14$
Balance H, by $H+ .....+ 14H+$
Thus, $x = 6$
$y= 1$
$z = 14$