Chemistry - Electrochemistry Question with Solution | TestHub

(a)$\mathrm{Cd}\left(\mathrm{s}\right)+2\mathrm{Ni}{\left(\mathrm{OH}\right)}_3\left(\mathrm{s}\right)\to \mathrm{CdO}\left(\mathrm{s}\right)+2\mathrm{Ni}{\left(\mathrm{OH}\right)}_2\left(\mathrm{s}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$ 
Discharge of secondary Battery

(b)$\mathrm{Zn}\left(\mathrm{Hg}\right)+\mathrm{HgO}\left(\mathrm{s}\right)\to \mathrm{ZnO}\left(\mathrm{s}\right)+\mathrm{Hg}\left(\mathrm{l}\right)$ 
(Primary Battery Mercury cell)

(c) $2{\mathrm{PbSO}}_4\left(\mathrm{s}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\to \mathrm{Pb}\left(\mathrm{s}\right)+{\mathrm{PbO}}_2\left(\mathrm{s}\right)+2{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)$ Charging of secondary Battery

(d) $2{\mathrm{H}}_2\left(\mathrm{g}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)\to 2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)-$ Fuel cell

Mercury cell, suitable for low current devices like hearing aids, watches, etc. consists of zinc - mercury amalgam as anode and a paste of $\mathrm{HgO}$ and carbon as the cathode. The electrolyte is a paste of $\mathrm{KOH}$ and $\mathrm{ZnO}$. The electrode reactions for the cell are given below:

Anode: $\mathrm{Zn}\left(\mathrm{Hg}\right)+2{\mathrm{OH}}^{-}\to \mathrm{ZnO}\left(\mathrm{s}\right)+{\mathrm{H}}_2\mathrm{O}+2{\mathrm{e}}^{-}$

Cathode: $\mathrm{HgO}+{\mathrm{H}}_2\mathrm{O}+2{\mathrm{e}}^{-}\to \mathrm{Hg}\left(\mathrm{l}\right)+2{\mathrm{OH}}^{-}$

Another important secondary cell is the nickel-cadmium cell which has longer life than the lead storage cell but more expensive to manufacture.

The overall reaction during discharge is:

$\mathrm{Cd}\left(\mathrm{s}\right)+2\mathrm{Ni}{\left(\mathrm{OH}\right)}_3\left(\mathrm{s}\right)\to \mathrm{CdO}\left(\mathrm{s}\right)+2\mathrm{Ni}{\left(\mathrm{OH}\right)}_2\left(\mathrm{s}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

Fuel cell using ${\mathrm{H}}_2$ and ${\mathrm{O}}_2$ produces electricity.

Cathode: ${\mathrm{O}}_2\left(\mathrm{g}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)+4{\mathrm{e}}^{-}\to 4{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)$

Anode: $2{\mathrm{H}}_2\left(\mathrm{g}\right)+4{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)\to 4{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)+4{\mathrm{e}}^{-}$

Overall reaction being:

$2{\mathrm{H}}_2\left(\mathrm{g}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)\to 2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

(ii) Cell reaction of lead storage battery during recharge -

The reverse of oxidation at anode during discharge is the reduction reaction at cathode during recharging.

${\mathrm{Pb}}_{\left(\mathrm{aq}\right)}^{2+}+2{\mathrm{e}}^{-}\to {\mathrm{Pb}}_{\left(\mathrm{s}\right)}$

${\mathrm{PbSO}}_{4\left(\mathrm{s}\right)}\to {\mathrm{Pb}}_{\left(\mathrm{aq}\right)}^{2+}+{\mathrm{SO}}_{4\left(\mathrm{aq}\right)}^{2-}$

Final Reaction: ${\mathrm{PbSO}}_{4\left(\mathrm{s}\right)}+2{\mathrm{e}}^{-}\to {\mathrm{Pb}}_{\left(\mathrm{s}\right)}+{\mathrm{SO}}_{4\left(\mathrm{aq}\right)}^{2-}$