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When lead metal is added to a beaker of HCl(aq), a gas is produced. Knowing that lead is oxidized and that hydrogen is reduced, write the balanced net equation for the reaction. (Use the lowest possible coefficients. Include states-of-matter under the given conditions in your answer.) chemPadHelp How many electrons are transferred in the balanced equation? What quantity of useful work can be obtained when Pb is added directly to the beaker of HCl?

Respuesta :

Answer:

Two electrons are transferred

Useful work would be 25090 J

Explanation:

When lead metal is added to a beaker of HCl, the lead is oxidized and hydrogen reduced.

Lead will lose two electrons and each atom of hydrogen will accept one electron

The oxidation reaction is:

[tex]Pb(s)--->Pb^{+2}(aq)+2e[/tex]   [tex]E_{0}=-0.13 V[/tex]

The reduction reaction is:

[tex]H^{+}+e--->\frac{1}{2}H_{2}(g)[/tex] [tex]E_{0}= 0V[/tex]

Overall equation is:

[tex]Pb(s)+2H^{+} ---> Pb^{+2}+H_{2}(g)[/tex]

Useful work is ΔG

ΔG = -nFE°

E° = [tex]E^{0}_{cathode}-E^{0}_{anode}[/tex]

E°=0-(-0.13) = 0.13 V

F = 96500 C

n = 2

ΔG = -2X96500X(0.13)= -25090 J

pstnonsonjoku

-25.1 KJ of wok is derived from the cell.

The net ionic equation for the reaction between lead metal and HCl is obtained as follows;

  • Molecular reaction equation;

Pb(s) + 2HCl(aq) -------> PbCl2(aq) + H2(g)

  • Complete ionic equation;

Pb(s) + 2H^+(aq) + 2Cl^-(aq) -------> Pb^2+(aq) +  2Cl^-(aq) + H2(g)

  • Net ionic equation;

Pb(s) +  2H^+(aq)  -------> Pb^2+(aq) + H2(g)

We can see that two electrons were transferred in the process.

E°cell = E°cathode - E°anode

E°cathode = 0 V

E°anode = -0.13 V

E°cell =  0V - (-0.13 V)

E°cell = 0.13 V

But ΔG = -nFE°cell

n =Number of electrons transferred = 2

F = Faraday's constant = 96500 C

E°cell = Cell potential

ΔG = -(2 × 96500  ×  0.13)

ΔG = -25.1 KJ

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