Examples of Cell Potential:

In both the ways the two half-cell reactions are to be written essentially in order to determine n, the number of electrons included in the cell reaction. To further understand this, consider following examples

Example 1.1: A cell is set up as follows:

Zn |Zn²⁺ (a = 5 ×10^-3)|| Cu ²⁺ (a = 2 ×10^-2 ) Cu

The standard potentials are: E^o_Cu²⁺ _/Cu = 0.337 V and E^o _Zn²⁺_/Zn = -0.763 V

a)  Calculate the cell potential.

b) Indicate the polarity of the electrodes and the direction of the spontaneous reaction.

a) The cell diagram specifies that the overall cell process involves the oxidation of Zn to Zn²⁺ and the reduction of Cu²⁺ to Cu metal.

An electrode reactions written as reductions are:

Zn²⁺ + 2e ↔    Zn   .... (i) ; E^o = - 0.763 V

Cu²⁺ + 2e ↔   Cu   .... (ii); E^o = + 0.337 V

Individual electrode potentials could be calculate by Nernst equation as:

E_Cu²⁺_/Cu = Eº_Cu²⁺_/Cu +(0.0591/2) log a_Cu²⁺                                                    ... (1.39)

= 0.337 + (0.0591/2) log (2 ×10^-2)

= 0.337 - 0.050

= 0.287 V

E_Zn²⁺ / _Zn = Eº_Zn²⁺_{/ Zn}+( 0.0591/2) log a_Zn²⁺                                               ... (1.40)

 E_Zn²⁺ / _Zn = -0.763+( 0.0591/2) log(5 ×10^-3 )

= - 0.763 - 0.068

= - 0.831 V

It is apparent that the copper half-cell has a higher reduction potential, hence it will act as cathode. Obviousl y the zinc half-cell works as the anode. So that the interlinked half-cells will result in the reactions

Cu²⁺ + 2e ↔ Cu, and

Zn - 2e ↔ Zn²⁺

and an overall cell reaction will be

Zn + Cu²⁺  ↔   Zn²⁺ + Cu                                                           ... (1.41)

The voltage among the electrodes is built as:

E_cell = E_right - E_left = E_cu²⁺_/Cu - E _Zn²⁺_/Zn

E_cell  = ECu²⁺ / Cu -   E_zn²⁺ _{/ zn}

or  E_cell = 0.287 - (- 0.831) =1.118 V

Alternatively, the cell potential can be calculated from the expression

E_cell = Eº_cell + (0.0591/2) log (^aCu²⁺/^aZn²⁺)

 =(Eº_Cu^2+/_Cu - Eº_Zn^2+/_Zn)+(0.0591/2)log 2×10^-2/5×10^-3

= {0.337 - (- 0.763)} + (0.0591/2) log 4

= 1.100 + 0.018

= 1.118 V

b) The cell reaction, as written within Eq. 1.38, tends to happen spontaneously as having positive potential.

From Eqs. 1.38 And 1.39 it is clear that the copper electrode will be the positive terminal and the zinc electrode will work as the negative terminal of the cell.