Question

Consider an electrochemical cell, where [Cr²⁺] = 0.15 M and [Al³⁺] = 0.0040 M, based on the following reaction:​

3 Cr²⁺(aq)  +  2 Al(s)  → 3 Cr(s)  +  2 Al³⁺(aq)

The standard reduction potentials are as follows:​

Cr²⁺(aq)  +  2 e^-  → Cr(s) E° = -0.91 V
Al³⁺(aq)  +  3 e^-  → Al(s) E° = -1.66 V

What is the cell potential at 25 °C?

Select one:

a. 0.71 V

b. 0.73 V

c. 0.75 V

d. 0.77 V

​Can you please hsow working :)

Answer 1

3 Cr2+(aq)  +  2 Al(s)  → 3 Cr(s)  +  2 Al3+(aq)

n = 6 ( number of electrons transfered in this reaction)

E cell = Eocell - (RT/nF) lnQ

at 25oC E cell = Eocell - (0.059/n) log Q

Cr2+(aq)  +  2 e-  → Cr(s) E° = -0.91 V
Al3+(aq)  +  3 e-  → Al(s) E° = -1.66 V

Al(s)  → Al3+(aq)  +  3 e- E° = +1.66 V

Eocell = -0.91 V  +1.66 V => 0.75 V

Ecell = 0.75 V - (0.059 /6 * log ( 0.00402 / 0.153)

Ecell = 0.75 + 0.02

Ecell = 0.77 V (option d)