Using the nernst equation calculate the cell voltage for: fe(s) + cd2+(aq) â fe2+(aq) + cd(s) when the [fe2+] = 0.10 m and [cd2+] = 1.4 m. potentially useful information: fe2+ + 2eâ â fe(s); îµ0 = -0.44 v cd2+ + 2eâ â cd(s); îµ0 = -0.40 v

The Nernst equation is:

E = E° - RTlnK/nF
where
E° is the standard potential voltage
R is the universal gas constant = 8.314 J/mol·K
K is the reaction quotient
n is the number of moles electrons transferred
F is Faraday's constant = 96,500 C/mol e⁻

Let's determine K first. The overall reaction is:

Fe(s) + Cd²⁺(aq) --> Fe²⁺(aq) + Cd(s)
Accounting for aqueous phases only, 
K = [products]/[reactants] = [Fe²⁺]/[Cd²⁺] = 0.10 M/1.4 M = 1/14

From the reactions written, you can see that 2 electrons were transferred. So, n = 2.

Lastly, the value for E⁰ is the sum of individual E⁰ of the reactions.
E⁰ = E⁰cathode - E⁰anode
Cathode reaction: Fe(s) --> 2e + Fe²⁺ (E⁰cathode = 0.44 V)
Anode reaction: Cd²⁺ + 2e --> Cd(s) (E⁰anode = -0.4 V)
Thus,
E⁰ = 0.44 - -0.4 = 0.84 V

Substituting the values (assume T at room temperature = 298 K),
E = 0.84 - (8.314)(298 K)(ln 1/14)/(2)(96,500)
E = 0.87 V

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