E°(cell) = E°(cathode) - E°(anode)
E°(cell) = (-0.440) - (-0.400) = -0.04 V
Electrons transferred in reaction (n) = 2
Reaction quotient (Q) = [Cd2+] / [Fe2+] = 0.01/0.75 = 0.01333
Using nernst equation at 298 K;
E(cell) = E°(cell) - (0.0592 * log Q) / n
E = (-0.04) - (0.0592 * log (0.01333)) / 2
E = 0.0155 V ...Answer
Since E is positive , reaction is spontaneous.
Be sure to answer all parts. Consider the following reaction: Cd(s) + Fe2+(aq) → Cd2+(aq) +...
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