Answer

A lead storage battery consists of a lead anode, a grid of lead packed with lead oxide $\left(\mathrm{PbO_2}\right)$ as the cathode, and a $38 \%$ solution of sulphuric acid $\left(\mathrm{H_2} \mathrm{SO_4}\right)$ as an electrolyte.

When the battery is in use, the following cell reactions take place:

At anode: $\mathrm{Pb_{(s)}}+\mathrm{SO_{4(a q)}}^{2-} \longrightarrow \mathrm{PbSO_{4(s)}}+2 \mathrm{e}^{-}$

At cathode: $\mathrm{PbO_{2(s)}}+\mathrm{SO_{4(a q)}}^{2-}+4 \mathrm{H_{(a q)}}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{PbSO_{4(s)}}+2 \mathrm{H_2} \mathrm{O_(f)}$

The overall cell reaction is given by,

$$ \mathrm{Pb_{(s)}}+\mathrm{PbO_{2(s)}}+2 \mathrm{H_2} \mathrm{SO_{4(a q)}} \longrightarrow 2 \mathrm{PbSO_{4(s)}}+2 \mathrm{H_2} \mathrm{O_{(l)}} $$

When a battery is charged, the reverse of all these reactions takes place.

Hence, on charging,

$$ \mathrm{PbSO_{4(s)}} \text { present at the anode and cathode is converted into } \mathrm{Pb_{(s)}} \mathrm{PbO_{2(s)}} $$

respectively.

Answer

Methane and methanol can be used as fuels in fuel cells.

Answer

In the process of corrosion, due to the presence of air and moisture, oxidation takes place at a particular spot of an object made of iron. That spot behaves as the anode. The reaction at the anode is given by,

$$ \mathrm{Fe_{(s)}} \longrightarrow \mathrm{Fe}^{2+}{ _{(a q)}}+2 \mathrm{e}^{-} $$

Electrons released at the anodic spot move through the metallic object and go to another spot of the object.

There, in the presence of $\mathrm{H}^{+}$ions, the electrons reduce oxygen. This spot behaves as the cathode. These $\mathrm{H}^{+}$ions come either from $\mathrm{H_2} \mathrm{CO_3}$, which are formed due to the dissolution of carbon dioxide from air into water or from the dissolution of other acidic oxides from the atmosphere in water.

The reaction corresponding at the cathode is given by,

$\mathrm{O_{2(\mathrm{~g})}}+4 \mathrm{H_{(a q)}^{+}}+4 \mathrm{e}^{-} \longrightarrow 2 \mathrm{H_2} \mathrm{O_{(l)}}$

The overall reaction is:

$ 2 \mathrm{Fe_{(s)}}+\mathrm{O_{2(g)}}+4 \mathrm{H_{(a q)}^{+}} \longrightarrow 2 \mathrm{Fe_{(a q)}^{2+}} +2 \mathrm{H_2} \mathrm{O_{(l)}}$

Also, ferrous ions are further oxidized by atmospheric oxygen to ferric ions. These ferric ions combine with moisture, present in the surroundings, to form hydrated ferric oxide $\left(\mathrm{Fe_2} \mathrm{O_3}, x \mathrm{H_2} \mathrm{O}\right)_{\text {i.e., rust. }}$

Hence, the rusting of iron is envisaged as the setting up of an electrochemical cell.