Below regarding an electrochemical cell in an automotive lead-acid battery. the cell's anode is made of lead and the cathode is made of lead(iv) oxide. both are submerged in 4.30 m sulfuric acid. the half-reactions are: pbo_2(s) + 3h^+ (aq) + hso_4^- (aq) + 2e^- rightarrow pbso_4(s) + 2h_2o(l) e degree = 1.685 v pbso_4(s) + h^+(aq) + 2e^- rightarrow pb(s) + hso_4^- (aq) e degree = -0.356 v (a) calculate the value of e degree. (b) determine the initial value of e_cell. assume that the first ionization of h_2so_4 is complete and that [h^+] almostequalto [hso_4^-]. (c) find e_cell when the h^+ concentration has dropped by 76.00%. again, assume [h^+] almostequalto [hso_4^-].

a. 2.041 V

b. 2.116 V

c. 2.043 V

Explanation:

a.

In an electrochemical cell;

reduction occurs at the cathode & oxidation at the anode.

Thus, from the question; reduction at cathode=

→

Oxidation at anode;

 →  

The net process in the Cell is obtained by the algebraic sum of the two half-cell reactions:

Overall reaction:

 →  

= 1.685V  -   (- 0.356V)

= 2.041V

b.      

To determine the initial value of E-cell :

In the electrochemical cell; both the anode and the cathode were submerged into 4.30M of Sulfuric acid; therefore:

→    

Assuming:

  such that (;)     = 4.30M

let  : a=   & b=  

=  

 -   ×  log  

=2.041V -    ×  log  

= 2.041V  -  0.02955  ×  log  

= 2.041V  -  0.02955  ×  log  

= 2.041V  -  0.02955  ×  (-2.5339)

= 2.041V   +  0.075V

=2.116V

c.

If concentration of is dropped by 76.00%

 =  4.30M

= 4.30M  -  4.30M  

= 1.032M

 =  2.041V -    ×  log  

=2.041V  -  0.02955  ×  log  

=2.041V  -  0.02955  ×  log  (0.88161954735)

=2.041V  -  0.02955  (- 0.0547)

=2.041V  +   (0.0020)

= 2.043V

the energy to form plasma comes from heat or electricity in the atmosphere!

sunny~ ☺