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Consider two cells, the first with Al and Ag electrodes and the second with Zn and Ni electrodes, each in 1.00 M solutions of their ions. If connected as voltaic cells in series, which two metals are plated, and what is the total potential, E ∘ ? Which two metals are plated?


a. Al ( s )

b. Ag ( s )

c. Zn ( s )

d. Ni ( s )


E ∘ =

Respuesta :

ajeigbeibraheem

Answer:

b. Ag ( s )

d. Ni ( s )

[tex]E^0_{total}[/tex]  = 2.97 V

Explanation:

The reduction potentials for the given four (4) electrodes are:

In the first cell:

[tex]Ag^+ + e^- \to Ag \ \ \ \ E^0 = 0.80 \ V \\ \\ \\ Al^{3+} + 3e^- \to Al \ \ \ \ E^0 = - 1.66 \ V[/tex]

In the second cell:

[tex]Zn^{2+} + 2e^- \to Zn \ \ \ E^0 = -0.76 \ V \\ \\ \\ Ni^{2+} + 2e^- \to Ni \ \ \ E^0 =-0.25 \ V[/tex]

Since the cells are connected in series :

[tex]E_{total } > 0[/tex]

Thus; Metal plated in the first  cell = Ag

Metal plated in the second cell = Ni

The total potential [tex]E^0_{total}[/tex] = [tex]E^0 {cell \ 1} - E^0 _{cell \ 2}[/tex]

= (0.80 + 1.66 ) V - (0.25 +0.76 ) V

= 2.97 V

pstnonsonjoku

In the first cell, silver is plated and in the second cell, nickel is plated.

In a voltaic cell, there is the production of electrical energy via a chemical reaction. We have two cells;

  • In the first cell, we have Al and Ag electrodes
  • In the second cell we have Zn and Ni electrodes

In the first cell, silver is plated and the Ecell is 2.46 V while in the second cell, nickel is plated with Ecell of 0.51 V.

Learn more about voltaic cell: https://brainly.com/question/3930479

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