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=
[tex]PbO_{2(s)} + 3H_{(aq)}^{+} +HSO_{4}_{(aq)} +2e^{-}[/tex] → [tex]PbSO_{4(s)} + 2H_{2} O_{(l)} E^{0}_{(cathode)} = 1.685V[/tex]
Oxidation at anode;
[tex]Pb_{s} + HSO^{-}_{4}[/tex] → [tex]PbSO_{4(s)}+H^{+}+2e^{-} E^{0}_{anode} =0.356V[/tex]
The net process in the Cell is obtained by the algebraic sum of the two half-cell reactions:
Overall reaction:
[tex]PbO_{2(s)}+Pb_{s}+2HSO^{-}_{4(aq)}+ 2 H^{+}_{(aq)}[/tex] → [tex]PbSO_{4(s)} + 2H_{2} O_{(l)}[/tex]
[tex]E^{0} _{cell} = E^{0} _{cathode}- E^{0} _{anode}[/tex]
= 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:
[tex]H_{2} SO_{4}[/tex] → [tex]H^{+} + HSO_{4}^{-}[/tex]
Assuming:
[tex]H^{+} = HSO_{4}^{-}[/tex] such that (;) [tex]H_{2} SO_{4}[/tex] = 4.30M
let : a= [tex]H^{+}[/tex] & b= [tex]HSO_{4}^{-}[/tex]
[tex]E^{i} _{cell}[/tex] =
[tex]E^{0} _{cell}[/tex] - [tex]\frac{0.0591}{2}[/tex] × log [tex]\frac{1}{(a)^{2} (b)^{2} }[/tex]
=2.041V - [tex]\frac{0.0591}{2}[/tex] × log [tex]\frac{1}{(4.30)^{2} (4.30)^{2} }[/tex]
= 2.041V - 0.02955 × log [tex]\frac{1}{341.8801}[/tex]
= 2.041V - 0.02955 × log [tex]{0.0029250026}[/tex]
= 2.041V - 0.02955 × (-2.5339)
= 2.041V + 0.075V
=2.116V
c.
If concentration of [tex]H^{+}[/tex] is dropped by 76.00%
[tex]H^{+} = HSO_{4}^{-}[/tex] = 4.30M
= 4.30M - 4.30M [tex]\frac{76}{100}[/tex]
= 1.032M
[tex]E_{cell}[/tex] = 2.041V - [tex]\frac{0.0591}{2}[/tex] × log [tex]\frac{1}{(1.032)^{2} (1.032)^{2} }[/tex]
=2.041V - 0.02955 × log [tex]\frac{1}{1.13427612058}[/tex]
=2.041V - 0.02955 × log (0.88161954735)
=2.041V - 0.02955 (- 0.0547)
=2.041V + (0.0020)
= 2.043V
