Question

9. 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. standard conditions

B. [Fe3+]= 1.7×10⁻³ M ; [Mg2+]= 2.30 M

C. [Fe3+]= 2.30 M ; [Mg2+]= 1.7×10⁻³ M

express answers in volts

Answer 1

9)  2Fe3+(aq)+3Mg(s)→2Fe(s)+3Mg2+(aq)

   At cathode (reduction): Fe³⁺ + 3e-  -----------> Fe Eo = -.0.04 V

   At anode (oxidation) : Mg --------> Mg²⁺ + 2e- Eo = +2.37

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A) E^ocell = potential at cathode + potential anode

= - 0.04 V + 2.37 V

= + 2.33 V

E^ocell = + 2.33 V

B) 2Fe³⁺(aq)+3Mg(s)→2Fe(s)+3Mg²⁺(aq)

no of electrons transferred = 6

Given that [Fe³⁺]= 1.7×10⁻³ M ,  [Mg²⁺]= 2.30 M

At 25oC,

  E_cell = E^o_cell - [0.059/n] Iog {[Mg²⁺]³ / [Fe³⁺]²} (solids cannot be taken into consideration)

= 2.33 - [0.059/6] Iog {(2.3)³ / (1.7×10⁻³)²}

= + 2.26 V

Ecell = + 2.26 V

Therefore,

cell potential = + 2.26 V

C) Given that [Fe³⁺]= 2.30 M ; [Mg²⁺]= 1.7×10⁻³ M

At 25oC,

    E_cell = E^o_cell - [0.059/n] Iog {[Mg²⁺]³ / [Fe³⁺]²} (solids cannot be taken into consideration)

= 2.33 - [0.059/6] Iog {(1.7×10⁻³)³ / (2.3)²}

= + 2.42 V

Ecell = + 2.42 V

Therefore,

cell potential = + 2.42 V