Question

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 ∘ =

Answer 1

Answer:

b. Ag ( s )

d. Ni ( s )

$E^0_{total}$  = 2.97 V

Explanation:

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

In the first cell:

$Ag^+ + e^- \to Ag \ \ \ \ E^0 = 0.80 \ V \\ \\ \\ Al^{3+} + 3e^- \to Al \ \ \ \ E^0 = - 1.66 \ V$

In the second cell:

$Zn^{2+} + 2e^- \to Zn \ \ \ E^0 = -0.76 \ V \\ \\ \\ Ni^{2+} + 2e^- \to Ni \ \ \ E^0 =-0.25 \ V$

Since the cells are connected in series :

$E_{total } > 0$

Thus; Metal plated in the first  cell = Ag

Metal plated in the second cell = Ni

The total potential $E^0_{total}$ = $E^0 {cell \ 1} - E^0 _{cell \ 2}$

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

= 2.97 V