I think my answers are wrong.

My corrected answers:

For the half reaction(oxidation) Cd(s)

Cd

^{2+}(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) Fe

^{2+}(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?