jacqueline398

jacqueline398

04.09.2020 •

Chemistry

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The picture explains it.

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Ответ:

lilxlawney

04.06.2020

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

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