Question

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^-].

Answer 1

Answer:

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=

$PbO_{2(s)} + 3H_{(aq)}^{+} +HSO_{4}_{(aq)} +2e^{-}$ → $PbSO_{4(s)} + 2H_{2} O_{(l)} E^{0}_{(cathode)} = 1.685V$

Oxidation at anode;

$Pb_{s} + HSO^{-}_{4}$  →  $PbSO_{4(s)}+H^{+}+2e^{-} E^{0}_{anode} =0.356V$

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

Overall reaction:

$PbO_{2(s)}+Pb_{s}+2HSO^{-}_{4(aq)}+ 2 H^{+}_{(aq)}$  →  $PbSO_{4(s)} + 2H_{2} O_{(l)}$

$E^{0} _{cell} = E^{0} _{cathode}- E^{0} _{anode}$

= 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:

$H_{2} SO_{4}$ →     $H^{+} + HSO_{4}^{-}$

Assuming:

$H^{+} = HSO_{4}^{-}$   such that (;)  $H_{2} SO_{4}$   = 4.30M

let  : a=  $H^{+}$ & b=  $HSO_{4}^{-}$

$E^{i} _{cell}$ =  

$E^{0} _{cell}$  -  $\frac{0.0591}{2}$ ×  log   $\frac{1}{(a)^{2} (b)^{2} }$

=2.041V -  $\frac{0.0591}{2}$  ×  log  $\frac{1}{(4.30)^{2} (4.30)^{2} }$

= 2.041V  -  0.02955  ×  log  $\frac{1}{341.8801}$

= 2.041V  -  0.02955  ×  log  ${0.0029250026}$

= 2.041V  -  0.02955  ×  (-2.5339)

= 2.041V   +  0.075V

=2.116V

c.

If concentration of $H^{+}$ is dropped by 76.00%

$H^{+} = HSO_{4}^{-}$  =  4.30M

= 4.30M  -  4.30M   $\frac{76}{100}$

= 1.032M

$E_{cell}$  =  2.041V -  $\frac{0.0591}{2}$  ×  log  $\frac{1}{(1.032)^{2} (1.032)^{2} }$

=2.041V  -  0.02955  ×  log  $\frac{1}{1.13427612058}$

=2.041V  -  0.02955  ×  log  (0.88161954735)

=2.041V  -  0.02955  (- 0.0547)

=2.041V  +   (0.0020)

= 2.043V