Question

A voltaic cell employs the following redox reaction: 2Fe3+(aq)+3Mg(s)→2Fe(s)+3Mg2+(aq) Calculate the cell potential at 25 ∘C under each of the following conditions.

A) [Fe3+]= 1.9×10−3 M ; [Mg2+]= 2.80 M

B) [Fe3+]= 2.80 M ; [Mg2+]= 1.9×10−3 M

Answer 1

Ans:

standard reduction potentials for Fe and Mg
Fe⁺³ + 3 e^- --> Fe .....Eo = - 0.04 volts
Mg⁺² + 2 e^- --> Mg ...... Eo = -2.37 volts

Eo for 2Fe³⁺(aq)+3Mg(s)--> 2Fe(s)+3Mg²⁺(aq) .... which has 6 moles of electrons transfered
Eo = +2.37 +(- 0.04) = +2.33 volts

.
Calculate the cell potential at A. [Fe³⁺]=1.9×10⁻³ M; [Mg2+]=2.80 M

nernst:
E = Eo - (0.0592 /n) (log Q)

Her n = 6
E = 2.33 - (0.0592 /6) (log [products] / [reactants])

E = 2.33 - (0.009867) (log [Mg⁺²]³ / [Fe⁺³]²)

E = 2.33 - (0.009867) (log [2.8]³ / [1.9*10^-3]²)

E = 2.33 - (0.009867) [log (21.952 / (3.61*10^-6)]

E = 2.33 - (0.009867) (log (6.08*10⁶)

E = 2.33 - (0.009867) (6.7839)

E = 2.33 - 0.06693

E = 2.263 Volts


B. [Fe³⁺]=2.80 M; [Mg²⁺]=1.9*10^-3 M


nernst:
E = Eo - (0.0592 /n) (log Q)
E = 2.33 - (0.0592 /6) (log [products] / [reactants])

E = 2.33 - (0.009867) (log [Mg⁺²]³ / [Fe⁺³]²)

E = 2.33 - (0.009867) (log [1.9*10^-3]³ / [2.8]²)

E = 2.33 - (0.009867) [log (6.859*10^-9 / 7.84)]

E = 2.33 - (0.009867) [log (0.874*10^-9)]

E = 2.33 - (0.009867) (-9.058)

E = 2.33 - (-0.08937)

E = 2.33 + 0.08937

Eo = 2.419 volts