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1 day ago

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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^-].

1 answer:

castortr0y [2.9K]1 day ago

5 0

Answer:

a. 2.041 V

b. 2.116 V

c. 2.043 V

Explanation:

a.

In an electrochemical cell; reductions take place at the cathode, while oxidations occur at the anode.

Therefore, based on the question; reduction at the cathode=

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

Oxidation at the anode;

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

The total reaction in the cell is achieved by summing the two half-reaction equations algebraically:

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 find the initial E-cell value:

In this electrochemical cell, both electrode materials were placed into 4.30M Sulfuric acid; thus:

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

Assuming:

$H^{+} = HSO_{4}^{-}$ where (;) $H_{2} SO_{4}$ = 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 the concentration of $H^{+}$ decreases by 76.00%

$H^{+} = HSO_{4}^{-}$ = 4.30M

= 4.30M - 4.30M $\frac{76}{100}$

= 1.032M

= 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

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