Question

Balance the following reactions:

1. $Cu+H_2{SO}_4-\to CuS{O}_4+{SO}_2\uparrow +H_{2}O$

2. $Fe++H_{2}O-\to {Fe}_3{O}_4+H_2\uparrow$

3. ${CH}_4+O_2\to C{O}_2+H_{2}O$

Answer 1

(1) $Cu+H_{2}S{O}_4\rightleftharpoons CuS{O}_4+S{O}_2+H_{2}O$

$Cu\to C{u}^{2+}+2e^{-}$

$S{O}_4^{2-}\to S{O}_2+2H_{2}O+4H^{+}+2e^{-}$

$\therefore Cu+H_{2}S{O}_4+2H^{+}\rightleftharpoons C{u}^{2+}+S{O}_2+2H_{2}O$

Adding a $S{O}_4^{2-}$ on both sides

$\Rightarrow Cu+2H_{2}S{O}_4\rightleftharpoons CuS{O}_4+S{O}_2+2H_{2}O$

(2) $Fe+H_{2}O\to F{e}_3{O}_4+H_2$

$H_{2}O\to H_2+H_{2}O+2H^{+}+2e^{-}$

$SFe\to F{e}_3{O}_4+8H^{+}+8e^{-}+4H_{2}O$

$3Fe+4H_{2}O+4H_{2}O+8H^{+}\to 4H_2+4H_{2}O+F{e}_3{O}_4+8H^{+}$

$\Rightarrow 3Fe+4H_{2}O\to F{e}_3{O}_4+4H_2$

$C{H}_4+2O_2\rightleftharpoons C{O}_2+2H_{2}O$