Question

um what...?
Cadmium and zinc are used in a cell to produce cadmium metal. Using the standard reduction potentials shown, calculate the cell potential and identify the type of cell.
Cd2+(aq) + 2e– Cd(s) Eº = –0.40 V
Zn2+(aq) + 2e– Zn(s) Eº = –0.76 V

Answer 1

@jdoe0001 @thomaster @Australopithecus

Answer 2

@aaronq

Answer 3

This sentence "Cadmium and zinc are used in a cell to produce cadmium metal." implies that \(Cd^{2+}\) will oxidize \(Zn_{(s)}\) to produce \(Cd_{(s)}\). You have to add up the individual potentials. The net equation will look like this:

\(\sf Cd^{2+}+Zn_{(s)}→Zn^{2+}+Cd_{(s)}\)

So the individual equations must be:

\(\sf Cd^{2+} + 2e^– → Cd_{(s)} ~~~ \xi^o = –0.40 V\) Cathode

\( \sf Zn(s) →2e^– + Zn^{2+} ~~~ \xi^o = \color{red}+0.76 V\) Anode


\(ξ^o_{cell}=ξ^o_{cathode}+ξ^o_{anode}\)

\(ξ^o_{cell}=–0.40 V+0.76 V=0.36V\)

Answer 4

The answer choices are also foreign to me though

A. +0.36 V; galvanic
B. –0.36 V; electrolytic
C. –1.16 V; electrolytic
D. +1.16 V; galvanic

Answer 5

@aaronq

Answer 6

it's A. The potential is +0.36 V, and this is a galvanic cell because it flows spontaneously.

If the cell potential was less than 0 V, then it would be electrolytic, which are non-spontaneous. This means that it would need an external source of power to drive the cell.

The spontaneity of a cell can be assessed with: \(\Delta G^o=-nF\xi^o \), where F is faraday's constant and n represents the moles of electrons.