Question

(A) What type of battery is lead storage battery? Write the anode and the cathode reactions and the overall reaction occurring in a lead storage battery when current is drawn from it.
(B) In the button cell, widely used in watches, the following reaction takes place.
\[Z{n_{(s)}} + A{g_2}{O_{(s)}} + {H_2}{O_{(l)}} \to Z{n^{2 + }}_{(aq)} + 2A{g_{(s)}} + 2O{H^ - }_{(aq)}\]
Determine \[{E^0}\] and \[\Delta {G^0}\] for the reaction (Given that \[{E^0}_{(A{g^ + }/Ag)} = + 0.80V\] and \[{E^0}_{(Z{n^{2 + }}/Zn)} = - 0.76V\] )

Answer 1

Hint:
For Question (A):
Batteries are classified into two types depending upon their ability to get recharged. As the name suggests, Lead batteries involve change in oxidation states of Pb-atom as it undergoes oxidation and reduction.

For Question (B):
\[{E^0}\] is known as the Standard electric potential of the cell which can be derived by following formula.\[{E^0}_{cell} = {E^0}_{cathode} - {E^0}_{anode}\]
\[\Delta {G^0}\] is Standard Gibbs energy of the cell which can be found if the number of electrons involved in the reaction, and \[{E^0}\] value is known.

Complete step-by-step answer:
For Question (A):
- Lead storage battery is a secondary type of battery. Now let’s see the construction of a lead storage battery. It is obvious that Lead storage batteries should have use of lead in its structure.
- So, Anode of Lead storage battery is made up of Pb metal and cathode is made of solid \[Pb{O_2}\]. Let’s see the anode and cathode reactions.
Anode: \[P{b_{(s)}} + S{O_4}{^{2 - }_{(aq)}} \to PbS{O_4}_{(s)} + 2{e^ - }\]
Cathode: \[Pb{O_2}_{(s)} + S{O_4}{^{2 - }_{(aq)}} + 4{H^ + }_{(aq)} + 2{e^ - } \to PbS{O_4}_{(s)} + 2{H_2}{O_{(l)}}\]
We also need to give the overall reaction for this battery.
Now, to write the overall reaction we will just add reactants and products of both anode and cathode reactions and cancel out the same components if present on both sides in the same amount.
Overall reaction : \[P{b_{(s)}} + Pb{O_2}_{(s)} + 2{H_2}S{O_4}_{(aq)} \to 2PbS{O_4}_{(s)} + 2{H_2}{O_{(l)}}\]

For Question (B):
We are given the reaction that occurs in the button cell. Now, with the below given formula, we can find the potential of the cell.
So, Applying the formula of Standard potential of cell,
\[{E^0}_{cell} = {E^0}_{cathode} - {E^0}_{anode}\]
Here, in a given cell, \[Z{n_{(s)}}\] is anode and \[A{g_2}{O_{(s)}}\] is cathode. So, formula can be written as,
\[{E^0}_{cell} = {E^0}_{A{g^ + }/Ag} - {E^0}_{Zn/Z{n^{2 + }}}\]………………(1)
Putting the value of potentials available in the question into equation (1),
\[{E^0}_{cell} = 0.80V - ( - 0.76V)\]
\[{E^0}_{cell} = 0.80V + 0.76V\]
\[{E^0}_{cell} = 1.56V\]
Now, we can find the Gibbs energy of the cell if the number of electrons involved in the process and standard potential of the cell is known. The formula is given below.
\[\Delta {G^0} = - nF{E^0}_{cell}\]…………………(2)
We will have to write the anode and cathode reaction now to clarify the reaction.
Anode Reaction: \[Z{n_{(s)}} \to Z{n^{2 + }}_{(aq)} + 2{e^ - }\]
Cathode Reaction: \[A{g_2}{O_{(s)}} + {H_2}{O_{(l)}} \to 2A{g^ + }_{(aq)} + 2O{H^ - }_{(aq)}\]
As we can see in the reaction, two electrons are involved in the process. So n=2.
Faraday constant \[F = 96500Cmo{l^{ - 1}}\]
We found in question that \[{E^0}_{cell} = 1.56V\]
Now, putting all these values in equation (2),
\[\Delta {G^0} = - 2 \times 96500 \times 1.56\]
\[\Delta {G^0} = - 301080Jmo{l^{ - 1}}\]
converting it into \[kJmo{l^{ - 1}}\] unit,
\[\Delta {G^0} = - 301.08kJmo{l^{ - 1}}\]
So, we found in this question that \[{E^0}_{cell} = 1.56V\] and \[\Delta {G^0} = - 301.08kJmo{l^{ - 1}}\].

Additional Information:
- To charge Lead storage battery current if drawn into it in a reversed direction. So anode and cathode reactions are also reversed. In which \[PbS{O_4}_{(s)}\] is converted to \[P{b_{(s)}}\] at anode and \[PbS{O_4}_{(s)}\] is converted to \[Pb{O_2}_{(s)}\] at cathode.
- We can also find Gibbs energy by the other formula given below.
\[\Delta {G^0} = - RT\ln K\]

Note:
For Question (A):
- Keep in mind that at anode only oxidation reactions occur and at cathode only reduction reactions occur. Do not get confused with that. While writing the overall reaction, make sure that any reactant or product is not being missed.

For Question (B):
- Make sure that you find n value properly in \[\Delta {G^0} = - nF{E^0}_{cell}\] , as it may lead to error.
- Always write the value of any property in units. Be careful while converting one unit to another as they can also be a source of error.