[Win,odd Dhamnekar]

Be sure to answer all parts.

Consider the following reaction:

Cd(s) + Fe²⁺(aq) 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? 

[mjc123]

No and no. What formula are you trying to use? Are you understanding the difference between E and E°?

[Win,odd Dhamnekar]

 I think my answers are wrong.
My corrected answers:
 
For the half reaction(oxidation) Cd(s)  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^- 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 E_{reduction(cathode)+}- E_ox(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?

[Babcock_Hall]

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.