# Calculate the cell potential at 25 C under each of the following conditions.?

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+]=2.4*10^-3 M; [Mg2+]=3.20 M

B. [Fe3+]=3.20 M; [Mg2+]=2.3*10^-3 M

Relevance

my text has these standard reduction potentials

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

would be +2.37 & - 0.04 = +2.33 volts

since texts rarely agree, use your reduction potentials if they differ from what my text provides

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Calculate the cell potential at A. [Fe3+]=2.4*10^-3 M; [Mg2+]=3.20 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 [3.2]^3 / [2.4*10^-3]^2)

E = 2.33 - (0.009867) [log (32.768 / 5.76 e-6)]

E = 2.33 - (0.009867) (log 5.69 e6)

E = 2.33 - (0.009867) (6.755)

E = 2.33 - 0.067

E = 2.26 Volts

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Calculate the cell potential at B. [Fe3+]=3.20 M; [Mg2+]=2.3*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 [2.3*10^-3]^3 / [3.2]^2)

E = 2.33 - (0.009867) [log (1.217 e-8 / 10.24)]

E = 2.33 - (0.009867) [log (1.19 e-9)]

E = 2.33 - (0.009867) (-8.925)

E = 2.33 - (-0.088)

E = 2.33 + 0.088

Eo = 2.42 volts

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the only other thing to point out...

is

did you mean to say:

A. [Fe3+]=2.4*10^-3 M

B. [Mg2+]=2.3*10^-3 M

or did you mean to have both be the same concentration upon switching

• 7 years ago

pls can i quickly gt d ans to the question...... what is the formula of a bimetallic galvanic cell emf?? thanks