Cd(s) + Fe2+(aq) → Cd2+(aq) + Fe(s) Eo   (Fe2+ / Fe) = −0.4400 V, Eo   (Cd2+ / Cd) = −0.4000 V Calculate the emf for this reaction at 298 K if [Fe2+] = 0.70 M and [Cd2+] = 0.010 M.

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Cd(s) + Fe2+(aq

Cd2+(aq) + Fe(s)

E

 o

(Fe2+ / Fe) = −0.4400 V, E

 o

(Cd2+ / Cd) = −0.4000 V

Calculate the emf for this reaction at 298 K if [Fe2+] = 0.70 M and [Cd2+] = 0.010 M.

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Step 1

The Nernst equation relates the cell EMF to standard cell potential. At 298 K, the nernst equation has the form:

Step 2

The standard cell potential can be determined from the standard potentials of the cathode and anode. Cathode is where reduction takes place and anode is where oxidation takes place. In the given reaction, Cd undergoes oxidation, thus serves as anode and Fe serves as cathode. Thus, standard cell potential can be calculated as:

Step 3

For the two half reactions involved, two elect...

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