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Calculating Cell EMF (voltage)

 

Some Standard Potentials

   
Weakest
Oxidant

Oxidants

 

Reductants

E0
(volts)

Strongest
Reductant

Key Concepts

  • The overall electrochemical cell reaction can be written as 2 half-equations:
        1 equation for the reduction reaction and 1 equation for the oxidation reaction

  • The number of electrons gained in the reduction half reaction must equal the number of electrons lost in the oxidation half reaction

  • Eo is a type of energy per electron so it remains unchanged even if we double the numbers of reactants and products in the reaction

  • If an equation is reversed (so that the reactants become the products), the sign of Eo is also reversed

  • The cell's emf (electromotive force, often referred to as the cell voltage), is calculated by adding together the Eo values for each half reaction:
        Eocell = Eoreduction + Eooxidation

  • The reaction is spontaneous in the direction as written if
        Eocell > 0
        (Eocell positive)

  • The reaction is spontaneous in the reverse direction to that written if
        Eocell < 0
        (Eocell negative)

  • A galvanic cell (voltaic cell) produces electricity so the overall cell reaction must have a positive Eocell value
        (Eocell > 0)

  • In practice, cell emf depends on temperature and concentration of reactants and products.
        If the concentration of reactants increases relative to products, the cell reaction becomes more spontaneous and the emf increases.
        As the cell operates, the reactants are used up as more product is formed causing the emf to decrease.

  • An electrolytic cell requires an input of electricity so the overall cell reaction must have a negative Eocell value
        (Eocell < 0)

  • If an electrolytic cell is operating in aqueous solution, then the water is also being reduced at the cathode or formed at the anode, so these equations must also be incorporated into the calculation of Eocell
| K++e K(s) -2.92 /\
| Ba2++2e Ba(s) -2.90 |
| Ca2++2e Ca(s) -2.87 |
| Na++e Na(s) -2.71 |
| Mg2++2e Mg(s) -2.36 |
| Al3++3e Al(s) -1.66 |
| Mn2++2e Mn(s) -1.18 |
| H2O+e ½H2(g)+OH- -0.83 |
| Zn2++2e Zn(s) -0.76 |
| Fe2++2e Fe(s) -0.41 |
| Ni2++2e Ni(s) -0.23 |
| Sn2++2e Sn(s) -0.14 |
| Pb2++2e Pb(s) -0.13 |
| H++e ½H2(g) 0.00 |
| SO42-+4H++2e SO2(aq)+2H2O 0.17 |
| Cu2++2e Cu(s) 0.35 |
| ½O2(g)+H2O+2e 2OH- 0.40 |
| Cu++e Cu(s) 0.52 |
| ½I2(s)+e I- 0.54 |
| ½I2(aq)+e I- 0.62 |
| Fe3++e Fe2+ 0.77 |
| Ag++e Ag(s) 0.80 |
| ½Br2(l)+e Br- 1.07 |
| ½Br2(aq)+e Br- 1.09 |
| ½O2(g)+2H++2e H2O 1.23 |
| ½Cl2(g)+e Cl- 1.36 |
\/ ½Cl2(aq)+e Cl- 1.40 |
Strongest
Oxidant
MnO4-+8H++5e Mn2++4H2O 1.51 Weakest
Reductant
½F2(g)+e F- 2.87

Calculating Cell EMF Examples

  1. Calculate the emf (voltage) for the following reaction:
    Zn(s) + Fe2+ -----> Zn2+ + Fe(s)

    Write the 2 half reactions:
    Zn(s) -----> Zn2+ + 2e
    Fe2+ +2e -----> Fe(s)

    look up the standard electrode potentials in the table above
    Zn2+ + 2e -----> Zn     Eo = -0.76V
    This equation needs to be reversed, so the sign of Eo will also be reversed.
    Zn(s) -----> Zn2+ + 2e     Eo = +0.76V
    Fe2+ +2e -----> Fe(s) Eo = -0.41V

    Add the two equations together:
    Zn(s) -----> Zn2+ + 2e Eo = +0.76V
    Fe2+ + 2e -----> Fe(s) Eo = -0.41V

    Zn(s) + Fe2+ -----> Zn2+ + Fe(s) Eocell = +0.76 + (-0.41) = +0.35V
    Eocell > 0 (Eocell positive) so the reaction is spontaneous as written

  2. Calculate the cell emf (voltage) for the following reaction:
    Br2(aq) + 2Fe2+ -----> 2Br- + 2Fe3+

    Write the two half equations:
    Br2(aq) + 2e -----> 2Br-
    2Fe2+ -----> 2Fe3+ + 2e

    Look up the standard electrode potentials in the table above
    ½Br2(aq) + e -----> Br-     Eo = +1.09V
    This equation needs to be multiplied by 2, however, the value of Eo remains the same
    Br2(aq) + 2e -----> 2Br-     Eo = +1.09V
    Fe3+ + e -----> Fe2+     Eo = +0.77V
    This equation needs to be reversed, the sign of Eo will also be reversed
    Fe2+ -----> Fe3+ + e     Eo = +0.77V
    This equation needs to be multiplied by 2, however, the value of Eo remains the same
    2Fe2+ -----> 2Fe3+ + 2e     Eo = +0.77V

    Add the two equations together:
    Br2(l) + 2e -----> 2Br- Eo = +1.09V
    2Fe2+ -----> 2Fe3+ + 2e Eo = -0.77V

    Br2(aq) + 2Fe2+ -----> 2Br- + 2Fe3+ Eocell = +1.09 + (-0.77) = +0.32V
    Eocell is positive (Eocell > 0) so the reaction is spontaneous in the direction as written

  3. Calculate the cell emf for the following reaction:
    Ni2+ + 2Cl-(aq) -----> Ni(s) + Cl2(g)

    Write the two half equations:
    Ni2+ + 2e -----> Ni(s)
    2Cl-(aq) -----> Cl2(g) + 2e

    Look up the standard electrode potentials in the table above:
    Ni2+ + 2e -----> Ni(s)     Eo = -0.23V
    ½Cl2(g) + e -----> Cl-     Eo = +1.36V
    This equation needs to be reversed, the sign of Eo will also be reversed:
    Cl- -----> ½Cl2(g) + e     Eo = -1.36V
    This equation needs to be multiplied by 2, however, the value of Eo will remain the same:
    2Cl- -----> Cl2(g) + 2e     Eo = -1.36V

    Add the two equations together:
    Ni2+ + 2e -----> Ni(s) Eo = -0.23V
    2Cl- -----> Cl2(g) + 2e Eo = -1.36V

    Ni2+ + 2Cl-(aq) -----> Ni(s) + Cl2(g) Eocell = -0.23 + (-1.36) = -1.59V
    Eocell is negative, (Eocell < 0), so the reaction is NOT spontaneous in the direction as written.
    The reaction would be spontaneous if reversed:
    Ni(s) + Cl2(g) -----> Ni2+ + 2Cl-     Eocell = +1.59V

Calculating Galvanic Cell (Voltaic Cell) EMF (voltage)

The Daniell Cell is a typical Galvanic Cell: Zn|Zn2+||Cu2+|Cu

Write the two half equations:
Zn(s) -----> Zn2+ + 2e
Cu2+ + 2e -----> Cu(s)

Look up the standard potentials in the table above:
Zn2+ + 2e -----> Zn(s)     Eo = -0.76V
This equation needs to be reversed, the sign of Eo will also be reversed:
Zn(s) -----> Zn2+ + 2e     Eo = +0.76V
Cu2+ + 2e -----> Cu(s)     Eo = +0.35V

Add the two equations together:
Zn(s) -----> Zn2+ + 2e Eo = +0.76V
Cu2+ + 2e -----> Cu(s) Eo = +0.35V

Zn(s) + Cu2+ -----> Zn2+ + Cu(s) Eocell = +0.76 + (+0.35) = +1.11V
Eocell must be positive (Eo > 0) in a Galvanic Cell

Calculating EMF (voltage) for Electrolytic Cells containing Aqueous Solutions

A solution of potassium bromide (KBr) can be electrolysed.

Consider the possible reduction reactions:
K+ + e -----> K(s)     Eo = -2.92V
2H2O(l) + 2e -----> H2 + 2OH-(aq)     Eo = -0.83V
The strongest oxidant (weakest reductant) is water (H2O). So H2(g) will be produced at the cathode.

Consider the possible oxidation reactions:
Br2(l) + 2e -----> 2Br-(aq)     Eo = +1.07V
½O2(g) + 2H+(aq) +2e -----> H2O(l)     Eo = +1.23V
The strongest reductant (weakest oxidant) is Br-. So, Br2(l) will be produced at the anode.

Add the equations together:
2H2O(l) + 2e -----> H2(g) + 2OH- Eo = -0.83V
2Br- -----> Br2(l) + 2e Eo = -1.07V

2H2O(l) + 2Br- -----> H2(g) + 2OH- + Br2(l) Eocell = -0.83 + (-1.07) = -1.90V
Eocell for an electrolytic cell must be negative (Eocell < 0).
A minimum of 1.90V must be applied to the cell in order for the electrolytic reaction to occur.

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