Question

Write cell reactions in lead storage batteries during discharge.

Answer 1

Lead-acid batteries (also known as lead storage batteries) can store a lot of charge in it and can also provide high current for short periods of time. The battery uses sponge lead and lead peroxide for the conversion of chemical energy into electrical energy. They are also capable of being recharged and can also have moderate power density and a good response time.

Complete Solution :
In order to understand the discharging of lead storage batteries, we have to understand its design.
In a lead storage battery, the negative and positive plates were made out of lead foil in the form of two spirals, separated with a sheet of cloth and then coiled up. The cells initially have low capacity. A slow process known as forming was required in order to corrode lead foils which creates lead dioxide on plates and roughening them to increase the surface area. The battery consists of six pieces of two-volt cells which are connected in series. Each cell is composed of several negative and positive electrodes made up of pure spongy lead and lead oxide respectively. The electrodes connected in parallel are immersed in a dilute solution of sulfuric acid.

- Now, we are interested in the discharge of lead storage batteries, so let’s talk about it.
During the discharged state of lead storage battery both positive and negative plates become lead(II) sulphate. The electrolyte loses most of its dissolved sulfuric acid and becomes water. The discharge process is powered by the conduction of electrons from the negative plate to the positive plate in the external circuit connected with the battery. The discharge process is driven by the reduction in energy when $2{H^ + }(aq)$ of the acid react with ${O^{2- }}$ ions of $Pb{O_2}$ to form the strong O-H bonds in water( −880 kJ per 18 g of water). This process can also compensate for the unfavourable formation of $P{b^{2 + }}(aq)$ ions or lead sulphate.
Now, let’s have a look on the reactions occurring on the negative and positive plates:
Negative plate reaction:
$Pb(s) + HS{O_4}^ - (aq) \to PbS{O_4}(s) + {H^ + }(aq) + 2{e^ - }$

As electrons accumulate, they create an electric field which attracts hydrogen ions and repels sulphate ions leading to formation of a double-layer near the surface of the solution. Unless charge is allowed to flow out of the electrode, the hydrogen ions screen the charged electrode from the solution which resists further reaction.
Positive plate reaction:
$Pb{O_2}(s) + HS{O_4}^ - (aq) + 3{H^ + }(aq) + 2{e^ - } \to PbS{O_4}(s) + 2{H_2}O(l)$

On combining these two reactions, one can determine the overall reaction:
$Pb(s) + Pb{O_2}(s) + 2HS{O_4}^ - (aq) + 2{H^ + }(aq) \to 2PbS{O_4}(s) + 2{H_2}O(l)$

Note: The net reaction of discharging is just opposite to net charging reaction i.e. the right side of reaction during charging is the left side during discharging and vice versa. But the individual reactions (reactions occurring at the cathode and anode) during charging is not exactly opposite to the individual reaction during discharging.