Question

What is the standard cell potential, E°_cell, for the spontaneous voltaic cell made by coupling a Ca⁺²|Ca (E°_red=-2.756V) half-cell and a Al⁺³|Al (E°_red=-1.662V) half-cell? Report your answer in volts, V.

Please show where you get the numbers, and write clearly please :)

Answer 1

Half reaction:

Ca⁺² + 2 e^- Ca E_red° = -2.756 V ----(1)

Al⁺³ + 3e^- Al E_red° = -1.662 V -----(1)

Mutiply (1) by 3 and (2) by 2 to make overall reaction

3 Ca⁺² + 6e^- 3 Ca E_red° = -2.756 V ---(3)

2 Al⁺³ + 6 e^- 2 Al E_red​​​​​​° = -1.662 V ---(4)

NOTE : E° will not be multiplied by 2 or 3.

Gibbs free energy (∆G°) = - n FE°

For the reaction to be spontaneous, ∆G° must be negative and for ∆G° to be negative, E° must be positive.

Therefore, (3) reaction will be reversed, as its E° is more negative than (4).

3 Ca 3 Ca⁺² + 6 e^- E_oxid° = 2.756 V

2 Al⁺³ + 6 e^- 2 Al E_red° = -1.662 V

Oveall reaction

3 Ca + 2 Al⁺³ 3 Ca⁺² + 2 Al

So, standard cell potential (E_cell°) = 2.756 -1.662

= 1.094 V