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.

Part B

[Fe3+]= 1.2×10⁻³ M ; [Mg2+]= 3.25 M

Express your answer in units of volts.

Part C

[Fe3+]= 3.25 M ; [Mg2+]= 1.2×10⁻³ M

Express your answer in units of volts.

Answer 1

Standard reduction potentials are:

Fe+3 + 3e- = Fe(s) E₀=-0.037

Mg(s) = Mg+2 E₀=+2.356
--------------------------------------------------------------------
2Fe+3 + 3Mg = 2Fe(s) + 3Mg+2 E^o = -0.037+2.356= +2.319

Nernst Equation:

E = E^o –( 0.0592/n *log {[Mg+2]³/ [Fe+3]²}) ;

Total 6 no. of electrons are transferred hence n= 6

Part B:

[Fe3+]= 1.2×10⁻³ M ; [Mg2+]= 3.25 M

E = +2.319 - 0.0592/6 * log [(3.25)³/ (1.2 x 10^-3)²]

E = +2.319 - 0.0098 * 7.377

E = +2.319 – 0.0722 = +2.2467 V

Part C:

[Fe3+]= 3.25 M ; [Mg2+]= 1.2×10⁻³ M

E = +2.319 - 0.0592/6 * log [(1.2 x 10^-3)³/ (3.25)²]
E = +2.319 - 0.0098 * (-9.786)
E = = +2.319 + 0.0959 = +2.4150 V