Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: Win,odd Dhamnekar on August 28, 2022, 11:13:00 AM
-
Be sure to answer all parts.
Consider the following reaction:
Cd(s) + Fe2+(aq) :rarrow: Cd2+ + Fe(s)
E°(Fe2+/Fe)= -0.4400 V, E°(Cd2+/Cd)= -0.4000 V.
Compute the emf for this reaction at 25°C if [Fe]2+= 0.75 M, [Cd]2+= 0.010 M.
Will the reaction occur spontaneously at these conditions?
My answer: E°Fe(cathode)(+)- E°Cd(anode)(-)= [itex]\frac{0.059}{2}\log{\frac{[0.01]}{[0.75]}} = -0.06V[/itex]
-0.4400 V-(-0.06)=-0.3847 V
ΔG°= -2× 96485.3399 J/(V-mol) × -0.3847 V= 7 × 101 kJ/mol
As ΔG° is + ve, this reaction is non-spontaneous.
Are my above answers correct? Are my unit conversions and computation of emf of this reaction correct?
-
No and no. What formula are you trying to use? Are you understanding the difference between E and E°?
-
I think my answers are wrong.
My corrected answers:
For the half reaction(oxidation) Cd(s) :rarrow: Cd2+(aq) + 2 e-, E° = + 0.4000 V
The nernst expression is E = E° - 0.059/n log ([reduction]b/ [ox]a)
[itex]E= -0.4000 V + \frac{0.059}{2}\log{\left(0.010\right)}= -0.46 V[/itex]
For the half reaction(reduction) Fe2+(aq) + 2 e- :rarrow: Fe(s), E° = -0.4400 V
[itex] E= -0.4400 V + \frac{0.059}{2}\log{\left(\frac{1}{[0.75]}\right)} = -0.444 V [/itex]
E for the complete reaction Ereduction(cathode)+- Eox(anode)-= -0.444 V -(-0.46 V)= 0.02 V
As E° for the whole reaction is +ve, we can say this reaction is spontaneous.
Are these above answers correct?
-
Terminology may not be uniform among all textbooks and teachers. I prefer to use spontaneous and non-spontaneous for ΔG and to use favorable and unfavorable for ΔG°. My suggestion to you is to follow the the usage in your textbook.