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

if two electrons have the same n, the same ℓ, and the same ml, then they are in the same orbital. the pauli exclusion principle states that no two identical fermions (a fancy word for electrons and other subatomic particles like electrons) may occupy the same quantum state in an atom simultaneously.