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Z n 2 + + 2 e - Z n ( E 0 = - 0 . 76 V )

C u 2 + + 2 e - C u ( E 0 = + 0 . 34 V )

We know from demonstrations, and also by looking at the sign of the electrode potential, that when these two half cells are combined, zinc will be the oxidation half-reaction and copper will be the reduction half-reaction. A voltmeter connected to this cell will show that the zinc electrode is more negative than the copper electrode. The reading on the meter will show the potential difference between the two half cells. This is known as the electromotive force (emf) of the cell.

Electromotive Force (emf)

The emf of a cell is defined as the maximum potential difference between two electrodes or half cells in a voltaic cell. emf is the electrical driving force of the cell reaction. In other words, the higher the emf, the stronger the reaction.

Standard emf (E c e l l 0 )

Standard emf is the emf of a voltaic cell operating under standard conditions (i.e. 100 kPa, concentration = 1 mol.dm - 3 and temperature = 298 K). The symbol 0 denotes standard conditions.

When we want to represent this cell, it is shown as follows:

Z n | Z n 2 + ( 1 m o l . d m - 3 ) | | C u 2 + ( 1 m o l . d m - 3 ) | C u

The anode half cell (where oxidation takes place) is always written on the left . The cathode half cell (where reduction takes place) is always written on the right .

It is important to note that the potential difference across a cell is related to the extent to which the spontaneous cell reaction has reached equilibrium. In other words, as the reaction proceeds and the concentration of reactants decreases and the concentration of products increases, the reaction approaches equilibrium. When equilibrium is reached, the emf of the cell is zero and the cell is said to be 'flat'. There is no longer a potential difference between the two half cells, and therefore no more current will flow.

Uses of standard electrode potential

Standard electrode potentials have a number of different uses.

Calculating the emf of an electrochemical cell

To calculate the emf of a cell, you can use any one of the following equations:

E ( c e l l ) 0 = E 0 (right) - E 0 (left) ('right' refers to the electrode that is written on the right in standard cell notation. 'Left' refers to the half-reaction written on the left in this notation)

E ( c e l l ) 0 = E 0 (reduction half reaction) - E 0 (oxidation half reaction)

E ( c e l l ) 0 = E 0 (oxidising agent) - E 0 (reducing agent)

E ( c e l l ) 0 = E 0 (cathode) - E 0 (anode)

So, for the Zn-Cu cell,

E ( c e l l ) 0 = 0.34 - (-0.76)

= 0.34 + 0.76

= 1.1 V

The following reaction takes place:

C u ( s ) + A g + ( a q ) C u 2 + ( a q ) + A g ( s )

  1. Represent the cell using standard notation.
  2. Calculate the cell potential (emf) of the electrochemical cell.
  1. C u 2 + + 2 e - C u (E o V = 0.16V)

    A g + + e - A g (E o V = 0.80V)

  2. Both half-reactions have positive electrode potentials, but the silver half-reaction has a higher positive value. In other words, silver does not form ions easily, and this must be the reduction half-reaction. Copper is the oxidation half-reaction. Copper is oxidised, therefore this is the anode reaction. Silver is reduced and so this is the cathode reaction.

  3. C u | C u 2 + ( 1 m o l . d m - 3 ) | | A g + ( 1 m o l . d m - 3 ) | A g
  4. E ( c e l l ) 0 = E 0 (cathode) - E 0 (anode)

    = +0.80 - (+0.34)

    = +0.46 V

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Source:  OpenStax, Siyavula textbooks: grade 12 physical science. OpenStax CNX. Aug 03, 2011 Download for free at http://cnx.org/content/col11244/1.2
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