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Solving for the Cell Potential

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    A method used by many professors and LSC specialists to calculate \(E^o_{cell}\) :

    1. Write out the reduction half reaction and the oxidation half reaction, stacking one over the other.
    2. Balance the half reactions and multiply each by any necessary numbers so that the number of electrons lost is equal to the number of electrons gained.
    3. Add the two half reactions and their corresponding voltages, the standard reduction voltage for the reduction half reaction and the voltage for the oxidation half reaction (the reverse sign of the standard reduction potential for the species in the oxidation half reaction).
    4. The superscript zeros on the \(E_o\) voltage symbols indicate standard conditions, i.e., all solutions at 1.0 M and all gases at 1.0 atm and all at 298 K.

    Remember: When comparing two reduction half reactions, the larger half reaction value is the more easily reduced.

    Example: The logical Approach

    From the Standard Reduction Table, the following values are obtained:

    Species Pair Reduction Half Reaction \(E^o\) Gibbs Energy Difference
    \[F_2(g)/F^- \] \[F_2(g) + 2e^- \rightarrow 2F^-(aq)\] \[2.87\, \text{V}\]

    \[ \Delta G^o= -nFE^o_{cell} \] Spontaneous (as drawn)

    \[Li(s)/Li^+\] \[2Li^+(aq) + 2e^- \rightarrow 2Li(s)\] \[-3.04\, \text{V}\]

    \[ \Delta G^o= -nFE^o_{cell} \] Not Spontaneous (as drawn)

    This is the lower voltage, so it will be the oxidation half reaction. Reverse it and change its sign. Thus,

    \( F_2(g) + 2e^- \rightarrow 2F^-(aq) \;\;\;\; E^o = 2.87 \, V\) Eoreduction

    \( 2Li(s) \rightarrow 2Li^+(aq) + 2e^- \;\;\;\; E^o = +3.04 V\) + Eoox(sign of standard reduction voltage reversed)

    \( F_2(g) + 2Li(s) \rightarrow 2F^-(aq) + 2Li^+(aq) E^o_{cell} = 5.91 V\) \( E^o_{cell}\)

    The overall formula for the above process would be

    \[E^o_{cell} = E^o_{red} + E^o_{ox}\]

    Example: The alternative Memorization Approach

    To solve for the standard cell voltage of an electrochemical cell, the book uses:

    1. Eocell = Eo(cathode) - Eo(anode) (Note the equivalent formulas to this equation below:

    Eocell = Eo(cathode) - Eo(anode) =

    Eocell = Eo(right) - Eo(left) =

    Eocell = Eo(reduction half cell) - Eo(oxidation half cell)

    For all of the formulas above, standard REDUCTION values are plugged into the proper position without your changing the sign. The negative in the formula will change the sign of the Eo(anode) or Eo(left).

    So remember, the text's formulas (any formulae in number 1 above) requires that you use standard REDUCTION values for BOTH half cells in a cell; the negative sign in the formula reverses the reduction value of the cell species in the oxidation half reaction so that they really have an oxidation half cell voltage value. Thus,

    \[E^o_{cell} = E^o_{cathode} - E^o_{anode}\]


    \(E^o_{cell} = 2.87 \, \text{V} - (-3.04\,\text{V}) = 5.91\, \text{V}\]

    For the equation Eocell= Eored+ Eoox, you change the sign of the standard reduction value from the table and it becomes Eoox , then you use this value in the equation.)

    Contributors and Attributions

    Solving for the Cell Potential is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.

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