Question

1.Calculate the emf for a cell operating with the reaction below at 25^oC in which [Cr⁺³] = 1.0 X 10^-3M and [Ni⁺²] = 1.5M.

3Ni⁺² (aq) + 2Cr (s) g>>>3Ni (s) + 2Cr⁺³ (aq)

Answer 1

Lets find Eo 1st

from data table:

Eo(Cr3+/Cr(s)) = -0.74 V

Eo(Ni2+/Ni(s)) = -0.25 V

As per given reaction/cell notation,

cathode is (Ni2+/Ni(s))

anode is (Cr3+/Cr(s))

Eocell = Eocathode - Eoanode

= (-0.25) - (-0.74)

= 0.49 V

Number of electron being transferred in balanced reaction is 6

So, n = 6

we have below equation to be used:

E = Eo - (2.303*RT/nF) log {[Cr3+]^2/[Ni2+]^3}

Here:

2.303*R*T/F

= 2.303*8.314*298.0/96500

= 0.0591

So, above expression becomes:

E = Eo - (0.0591/n) log {[Cr3+]^2/[Ni2+]^3}

E = 0.49 - (0.0591/6) log (0.001^2/1.5^3)

E = 0.49-(-6.433*10^-2)

E = 0.5543 V

Answer: 0.554 V