Half-Cell Potentials

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If a cell is constructed from two half-cells, and a voltmeter included in the conducting wire, then the observed potential difference is a measure of the difference in reducing power of the two couples. In this way, the reducing power of a particular couple can be specified as the potential of its half-cell relative to another half-cell. This can only ever be a relative measure as it is not possible to measure the potential of a half-cell in isolation.

By choosing a particular half-cell (couple) as a reference standard, the reducing power of any other half-cell can be specified as a particular potential. For this purpose the reference standard chosen is the standard hydrogen half-cell . This consists of the couple H⁺ (aq)/H₂(g) (with platinised platinum as the solid conductor), the pressure of H₂(g) being 101 kPa exactly, and the concentration of H⁺(aq) being 1 M exactly. The potential of this half-cell is then defined as zero at all temperatures. If any other half-cell is connected to this reference half-cell, the observed potential difference between them is then a measure of the electrode potential, E , of the half-cell under test. If the components of this half-cell also have concentrations of 1 M exactly for a solute and pressure equal to 101 kPa exactly for a gas, then the potential difference is defined as the standard electrode-potential , E^o. This is also sometimes called the reduction potential.

The potential difference of the cell composed of the Pt / H⁺ / H₂ and Cu²⁺ / Cu couples (all components in their standard states) is 0.34 volt and the copper half-cell is positive with respect to the hydrogen half-cell. Hence the value of E^o for the Cu²⁺ / Cu couple is +0.34 volt. The reaction that occurs is

H2 + Cu2+

Cu + 2H+

It is seen that the Cu²⁺ / Cu couple undergoes reduction while the Pt / H⁺ / H 2 couple undergoes oxidation.

The potential difference of the cell composed of the Pt / H⁺ / H₂ and Zn²⁺ / Zn couples (all components in their standard states) is 0.76 volt and the zinc half-cell is negative with respect to the hydrogen half-cell. Hence the value of E^o for the Zn²⁺ /Zn couple is -0.76 volt. The reaction that occurs is

Zn + 2H2+

H2 + Zn2+

It is seen that the Zn²⁺ / Zn couple undergoes oxidation while the Pt / H⁺ / H₂ couple undergoes reduction.

1. Balance the numbers of all the atoms other than O and H.
2. Balance O by adding H₂O to either side.
3. Balance H by adding H⁺ to either side.
4. Balance the charges by adding electrons to either side.
5. If basic conditions are specified: eliminate any H⁺ present by adding enough OH^- to each side of the reaction equation to turn H⁺ into H₂O.

Example (i)

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Construct the ion/electron half-equation for the reduction of Cr₂O₇^2- to Cr³⁺ in the presence of H⁺.

1. Balance Cr (ie. elements other than O or H)
   
   The reaction is Cr(VI) to Cr(III)
   

   Cr2O72-

   Cr3+

   The Cr atoms are balanced by doubling the Cr 3+
   

   Cr2O72-

   2Cr3+

2. Balance O
   
   To balance for oxygen, 7 atoms must be added to the right-hand side of the equation. Since an ion/electron half-equation is for the oxidation or reduction of only one species (and hence for a change in oxidation number of only one  atom, Cr in this case), the O atoms must have the same oxidation number as those on the left-hand side, namely -II.
   
   It is convenient to insert these as H₂O molecules.
   
   

   Cr2O72-

   2Cr3+ + 7H2O

3. Balance H
   
   To balance for hydrogen, 14 atoms must be added to the  left-hand side. As with oxygen, the H atoms added must have the same oxidation number as those on the right-hand side, namely +I. It is convenient to insert them as H⁺.
   
   

   14H+ + Cr2O72-

   2Cr3+ + 7H2O

4. Balance the charge
   
   There is a total of 12 positive charges on the left side of the equation and 6 positive charges on the right. In order to balance charges, we add electrons (which have a charge of -1). Thus, if we add 6 electrons to the left side, the charges become balanced (both +6).
   
   Note, the purpose of balancing   charges is not to make both sides of the equation zero!
   

   14 H+ + Cr2O72- + 6e-

   2Cr3+ + 7H2O

Note that acidic conditions were specified, so the balanced half-equation for the reduction of Cr(VI) to Cr(III) is

14 H+ + Cr2O72- + 6e-

2Cr3+ + 7H2O

Example (ii)

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Construct the ion/electron half-equation for the oxidation of Ag to Ag₂S through reaction with HS^- in the presence of OH^-.

1. Balance Ag and S (ie. elements other than O or H)
   
   The reaction is Ag(0) to Ag(I). Sulfur remains at -II.

   Ag + HS-

   Ag2S

   The Ag atoms are balanced by doubling the Ag

   2Ag + HS-

   Ag2S

   The S atoms are already balanced.
   
   
2. Balance O
   
   There is no oxygen present in this example to balance.

   2Ag + HS-

   Ag2S

3. Balance H
   
   To balance for hydrogen, add sufficient H⁺ . In our example, this requires the addition of one H⁺ to the right-side of the equation.

   2Ag + HS-

   Ag2S + H+

4. Balance the charge
   
   There is a total of 1 negative charge on the left side of the equation and 1 positive charge on the right. In order to balance charges, we need to add 2 electrons to the right side.
   

   2Ag + HS-

   Ag2S + H+ + 2e-

   Note that basic conditions were specified, so we need to complete Step 5 as well so as to remove any acid present.
   
5. Basic conditions
   
   There is 1 H⁺ present in our current equation. To eliminate this we need to add sufficient OH^- to both sides of the equation so that the H⁺ will be neutralised.

   2Ag + HS- + OH-

   Ag2S + H+ + OH- + 2e-

               
   This can be simplified to:

   2Ag + HS- + OH-

   Ag2S + H2O + 2e-

   If water molecules had existed on the left-side of the equation, we could further simplify by cancelling out water molecules.

The final balanced reaction is thus:

2Ag + HS- + OH-

Ag2S + H2O + 2e-

Alternative Method

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Note that there is an alternative method for handling basic conditions that is given in some textbooks. This involves a different set of steps as outlined below:

1. Balance the numbers of all the atoms other than O and H.
2. Balance O by adding H₂O to either side.
3. Balance H by adding H₂O to the side deficient in H and an equal amount of OH^- to the other side.
4. Balance the charges by adding electrons to either side.
   

However, it is suggested that you use the steps given at the top of the page as these are simpler and easier to remember as they do not involve a change in procedure between acidic and basic conditions.