Question

Balance each of the following oxidation-reduction by using the oxidation states method.

1. ${\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{6}}\left(g\right)\mathbf{+}{\mathbf{O}}_{\mathbf{2}}\left(g\right)\mathbf{\to }{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(g\right)\mathbf{+}{\mathbf{CO}}_{\mathbf{2}}\left(g\right)$
2. $\mathbf{Mg}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\mathbf{HCl}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{\to }{\mathbf{Mg}}^{\mathbf{2}\mathbf{+}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\mathbf{Cl}}^{\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$
3. $\mathbf{Cu}\left(s\right)\mathbf{+}{\mathbf{Ag}}^{\mathbf{+}}\left(\mathrm{aq}\right)\mathbf{\to }{\mathbf{Cu}}^{\mathbf{2}\mathbf{+}}\left(\mathrm{aq}\right)\mathbf{+}\mathbf{Ag}\left(s\right)$
4. $\mathbf{Zn}\left(s\right)\mathbf{+}{\mathbf{H}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}\left(\mathrm{aq}\right)\mathbf{\to }{\mathbf{ZnSO}}_{\mathbf{4}}\left(\mathrm{aq}\right)\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\left(g\right)$

Short answer

1. $2{\mathrm{C}}_2{\mathrm{H}}_6\left(\mathrm{g}\right)+7{\mathrm{O}}_2\left(\mathrm{g}\right)\to 6{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)+4{\mathrm{CO}}_2\left(\mathrm{g}\right)$
   
2. $\mathrm{Mg}\left(\mathrm{s}\right)+2\mathrm{HCl}\left(\mathrm{aq}\right)\to {\mathrm{Mg}}^{2+}\left(\mathrm{aq}\right)+2{\mathrm{Cl}}^{-}\left(\mathrm{aq}\right)+2{\mathrm{H}}_2\left(\mathrm{g}\right)$
   
3. $\mathrm{Cu}\left(\mathrm{s}\right)+2{\mathrm{Ag}}^{+}\left(\mathrm{aq}\right)\to {\mathrm{Cu}}^{2+}\left(\mathrm{aq}\right)+2\mathrm{Ag}\left(\mathrm{s}\right)$
   
4. $\mathrm{Zn}\left(\mathrm{s}\right)+{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)\to {\mathrm{ZnSO}}_4\left(\mathrm{aq}\right)+{\mathrm{H}}_2\left(\mathrm{g}\right)$
   

Step-by-step solution

Subpart a)

In this oxidation-reduction, the oxygen molecule is neutral. H-atom has +1 charge. O- atom has -2 charge. Labelling the reaction with charges:

$$\begin{gathered} 2{\mathrm{C}}_2{\mathrm{H}}_6\left(\mathrm{g}\right)+7{\mathrm{O}}_2\left(\mathrm{g}\right)\to 4{\mathrm{CO}}_2\left(\mathrm{g}\right)+6{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right) \\ \stackrel{\uparrow }{+1}\stackrel{\uparrow }{0}\stackrel{\uparrow }{+4}\stackrel{\uparrow }{+1}\stackrel{\uparrow }{-2} \end{gathered}$$

Hence, to balance out the reaction, we have to use $2{\mathrm{CO}}_2$ and $3{\mathrm{H}}_2\mathrm{O}$

${\mathrm{C}}_2{\mathrm{H}}_6\left(\mathrm{g}\right)+\frac{7}{2}{\mathrm{O}}_2\left(\mathrm{g}\right)\to 2{\mathrm{CO}}_2\left(\mathrm{g}\right)+3{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)$

So, it can be written as,

$2{\mathrm{C}}_2{\mathrm{H}}_6\left(\mathrm{g}\right)+7{\mathrm{O}}_2\left(\mathrm{g}\right)\to 4{\mathrm{CO}}_2\left(\mathrm{g}\right)+6{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)$

Subpart b)

In this oxidation-reduction, the magnesium atom is neutral. H-atom in HCl has +1 charge. Cl-atom has -1 charge. Hydrogen molecule is neutral.

Hence, the oxidation state of magnesium changes from 0 to +2 and that of hydrogen changes from $+1$to $0$. So, to balance out the reaction, we have to use 2hydrogen.

$\mathrm{Mg}+2\mathrm{HCl}\to {\mathrm{Mg}}^{2+}+2{\mathrm{Cl}}^{-}+2{\mathrm{H}}_2$

Subpart c)

In this oxidation-reduction, the reactant copper atom is neutral. The product silver atom is neutral. The reactant silver atom has $+1$ charge. And, the product copper atom has $+2$charge.

$\mathrm{Cu}+2{\mathrm{Ag}}^{+}\to {\mathrm{Cu}}^{2+}+2\mathrm{Ag}$

Subpart d)

In this oxidation-reduction, the zinc atom is neutral. The${\mathrm{SO}}_4^{2-}$ attached with hydrogen atom is neutral.

$\mathrm{Zn}\left(\mathrm{s}\right)+{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)\to {\mathrm{ZnSO}}_4\left(\mathrm{aq}\right)+{\mathrm{H}}_2\left(\mathrm{g}\right)$

In the reaction, the reactant side has 1-zinc atom, 2-hydrogen, 1-sulfur and 4 oxygen atoms, which is the same as on the product side.