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     E^o = -0.76V
   This equation needs to be reversed, so the sign of E^o will also be reversed.
   Zn(s) -----> Zn²⁺ + 2e     E^o = +0.76V
   Fe²⁺ +2e -----> Fe(s) E^o = -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^-     E^o = +1.09V
   This equation needs to be multiplied by 2, however, the value of E^o remains the same
   Br₂(aq) + 2e -----> 2Br^-     E^o = +1.09V
   Fe³⁺ + e -----> Fe²⁺     E^o = +0.77V
   This equation needs to be reversed, the sign of E^o will also be reversed
   Fe²⁺ -----> Fe³⁺ + e     E^o = +0.77V
   This equation needs to be multiplied by 2, however, the value of E^o remains the same
   2Fe²⁺ -----> 2Fe³⁺ + 2e     E^o = +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^-     E^o = +1.36V
   This equation needs to be reversed, the sign of E^o will also be reversed:
   Cl^- -----> ½Cl₂(g) + e     E^o = -1.36V
   This equation needs to be multiplied by 2, however, the value of E^o will remain the same:
   2Cl^- -----> Cl₂(g) + 2e     E^o = -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)

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)     E^o = -0.76V
This equation needs to be reversed, the sign of E^o will also be reversed:
Zn(s) -----> Zn²⁺ + 2e     E^o = +0.76V
Cu²⁺ + 2e -----> Cu(s)     E^o = +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)     E^o = -2.92V
2H₂O(l) + 2e -----> H₂ + 2OH^-(aq)     E^o = -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)     E^o = +1.07V
½O₂(g) + 2H⁺(aq) +2e -----> H₂O(l)     E^o = +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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