# 23.6: Calculating Standard Cell Potentials

When exposed to moisture, steel will begin to rust fairly quickly. This creates a significant problem for items like nails that are exposed to the atmosphere. The nails can be protected by coated them with zinc metal, making a galvanized nail. The zinc is more likely to oxidize than the iron in the steel, so it prevents rust from developing on the nail.

Figure 23.6.1: Galvanized nails. Image used with permission (Public domain; Raysonho).

### Calculating Standard Cell Potentials

In order to function, any electrochemical cell must consist of two half-cells.The table below can be used to determine the reactions that will occur and the standard cell potential for any combination of two half-cells without actually constructing the cell. The half-cell with the higher reduction potential according to the table will undergo reduction within the cell. The half-cell with the lower reduction potential will undergo oxidation within the cell. If those specifications are followed, the overall cell potential will be a positive value. The cell potential must be positive in order for redox reaction of the cell to be spontaneous in the reverse direction.

Half-Reaction $$E^0 \: \left( \text{V} \right)$$
Table 23.6.1: Standard Reduction Potentials at $$25^\text{o} \text{C}$$
$$\ce{F_2} + 2 \ce{e^-} \rightarrow \ce{F^-}$$ +2.87
$$\ce{PbO_2} + 4 \ce{H^+} + \ce{SO_4^{2-}} + 2 \ce{e^-} \rightarrow \ce{PbSO_4} + 2 \ce{H_2O}$$ +1.70
$$\ce{MnO_4^-} + 8 \ce{H^+} + 5 \ce{e^-} \rightarrow \ce{Mn^{2+}} + 4 \ce{H_2O}$$ +1.51
$$\ce{Au^{3+}} + 3 \ce{e^-} \rightarrow \ce{Au}$$ +1.50
$$\ce{Cl_2} + 2 \ce{e^-} \rightarrow 2 \ce{Cl^-}$$ +1.36
$$\ce{Cr_2O_7^{2-}} + 14 \ce{H^+} + 6 \ce{e^-} \rightarrow 2 \ce{Cr^{3+}} + 7 \ce{H_2O}$$ +1.33
$$\ce{O_2} + 4 \ce{H^+} + 4 \ce{e^-} \rightarrow 2 \ce{H_2O}$$ +1.23
$$\ce{Br_2} + 2 \ce{e^-} \rightarrow 2 \ce{Br^-}$$ +1.07
$$\ce{NO_3^-} + 4 \ce{H^+} + 3 \ce{e^-} \rightarrow \ce{NO} + 2 \ce{H_2O}$$ +0.96
$$2 \ce{Hg^{2+}} + 2 \ce{e^-} \rightarrow \ce{Hg_2^{2+}}$$ +0.92
$$\ce{Hg^{2+}} + 2 \ce{e^-} \rightarrow \ce{Hg}$$ +0.85
$$\ce{Ag^+} + \ce{e^-} \rightarrow \ce{Ag}$$ +0.80
$$\ce{Fe^{3+}} + \ce{e^-} \rightarrow \ce{Fe^{2+}}$$ +0.77
$$\ce{I_2} + 2 \ce{e^-} \rightarrow 2 \ce{I^-}$$ +0.53
$$\ce{Cu^+} + \ce{e^-} \rightarrow \ce{Cu}$$ +0.52
$$\ce{O_2} + 2 \ce{H_2O} + 4 \ce{e^-} \rightarrow 4 \ce{OH^-}$$ +0.40
$$\ce{Cu^{2+}} + 2 \ce{e^-} \rightarrow \ce{Cu}$$ +0.34
$$\ce{Sn^{4+}} + 2 \ce{e^-} \rightarrow \ce{Sn^{2+}}$$ +0.13
$$2 \ce{H^+} + 2 \ce{e^-} \rightarrow \ce{H_2}$$ 0.00
$$\ce{Pb^{2+}} + 2 \ce{e^-} \rightarrow \ce{Pb}$$ -0.13
$$\ce{Sn^{2+}} + 2 \ce{e^-} \rightarrow \ce{Sn}$$ -0.14
$$\ce{Ni^{2+}} + 2 \ce{e^-} \rightarrow \ce{Ni}$$ -0.25
$$\ce{Co^{2+}} + 2 \ce{e^-} \rightarrow \ce{Co}$$ -0.28
$$\ce{PbSO_4} + 2 \ce{e^-} \rightarrow \ce{Pb} + \ce{SO_4^{2-}}$$ -0.31
$$\ce{Cd^{2+}} + 2 \ce{e^-} \rightarrow \ce{Cd}$$ -0.40
$$\ce{Fe^{2+}} + 2 \ce{e^-} \rightarrow \ce{Fe}$$ -0.44
$$\ce{Cr^{3+}} + 3 \ce{e^-} \rightarrow \ce{Cr}$$ -0.74
$$\ce{Zn^{2+}} + 2 \ce{e^-} \rightarrow \ce{Zn}$$ -0.76
$$2 \ce{H_2O} + 2 \ce{e^-} \rightarrow \ce{H_2} + 2 \ce{OH^-}$$ -0.83
$$\ce{Mn^{2+}} + 2 \ce{e^-} \rightarrow \ce{Mn}$$ -1.18
$$\ce{Al^{3+}} + 3 \ce{e^-} \rightarrow \ce{Al}$$ -1.66
$$\ce{Be^{2+}} + 2 \ce{e^-} \rightarrow \ce{Be}$$ -1.70
$$\ce{Mg^{2+}} + 2 \ce{e^-} \rightarrow \ce{Mg}$$ -2.37
$$\ce{Na^+} + \ce{2^-} \rightarrow \ce{Na}$$ -2.71
$$\ce{Ca^{2+}} + 2 \ce{e^-} \rightarrow \ce{Ca}$$ -2.87
$$\ce{Sr^{2+}} + 2 \ce{e^-} \rightarrow \ce{Sr}$$ -2.89
$$\ce{Ba^{2+}} + 2 \ce{e^-} \rightarrow \ce{Ba}$$ -2.90
$$\ce{Rb^+} + \ce{e^-} \rightarrow \ce{Rb}$$ -2.92
$$\ce{K^+} + \ce{e^-} \rightarrow \ce{K}$$ -2.92
$$\ce{Cs^+} + \ce{e^-} \rightarrow \ce{Cs}$$ -2.92
$$\ce{Li^+} + \ce{e^-} \rightarrow \ce{Li}$$ -3.05

Example 23.6.1

Calculate the standard cell potential of a voltaic cell that uses the $$\ce{Ag}$$/$$\ce{Ag^+}$$ and $$\ce{Sn}$$/$$\ce{Sn^{2+}}$$ half-cell reactions. Write the balanced equation for the overall cell reaction that occurs. Identify the anode and the cathode.

Solution:

Step 1: List the known values and plan the problem.

Known

• $$E^0_\ce{Ag} = +0.80 \: \text{V}$$
• $$E^0_\ce{Sn} = -0.14 \: \text{V}$$

Unknown

• $$E^0_\text{cell} = ? \: \text{V}$$

The silver half-cell will undergo reduction because its standard reduction potential is higher. The tin half-cell will undergo oxidation. The overall cell potential can be calculated by using the equation $$E^0_\text{cell} = E^0_\text{red} - E^0_\text{oxid}$$.

Step 2: Solve.

\begin{align} \text{oxidation} \: \left( \text{anode} \right) &: \ce{Sn} \left( s \right) \rightarrow \ce{Sn^{2+}} \left( aq \right) + 2 \ce{e^-} \\ \text{reduction} \: \left( \text{cathode} \right) &: \ce{Ag^+} \left( aq \right) + \ce{e^-} \rightarrow \ce{Ag} \left( s \right) \end{align}

Before adding the two reactions together, the number of electrons lost in the oxidation must equal the number of electrons gained in the reduction. The silver half-cell reaction must be multiplied by two. After doing that and adding to the tin half-cell reaction, the overall equation is obtained.

$\text{overall equation} \: \: \: \: \: \: \ce{Sn} \left( s \right) + 2 \ce{Ag^+} \left( aq \right) \rightarrow \ce{Sn^{2+}} \left( aq \right) + 2 \ce{Ag} \left( s \right)$

The cell potential is calculated.

$E^0_\text{cell} = E^0_\text{red} - E^0_\text{oxid} = +0.80 - \left( -0.14 \: \text{V} \right) = +0.94 \: \text{V}$

The standard cell potential is positive, so the reaction is spontaneous as written. Tin is oxidized at the anode, while silver ion is reduced at the cathode. Note that the voltage for the silver ion reduction is not doubled even though the reduction half-reaction had to be doubled to balance the overall redox equation.

#### Oxidizing and Reducing Agents

A substance which is capable of being reduced very easily is a strong oxidizing agent. Conversely, a substance which is capable of being oxidized very easily is a strong reducing agent. According to the standard cell potential table, fluorine $$\left( \ce{F_2} \right)$$ is the strongest oxidizing agent. It will oxidize any substance below on the table. For example, fluorine will oxidize gold metal according to the following reaction.

$3 \ce{F_2} \left( g \right) + 2 \ce{Au} \left( s \right) \rightarrow 6 \ce{F^-} \left( aq \right) + 2 \ce{Au^{3+}} \left( aq \right)$

Lithium metal $$\left( \ce{Li} \right)$$ is the strongest reducing agent. It is capable of reducing any substance above on the table. For example, lithium will reduce water according to this reaction.

$2 \ce{Li} \left( s \right) + 2 \ce{H_2O} \left( l \right) \rightarrow 2 \ce{Li^+} \left( aq \right) 2 \ce{OH^-} \left( aq \right) + \ce{H_2} \left( g \right)$

Using the table above will allow you to predict whether reactions will occur or not. For example, nickel metal is capable of reducing copper (II) ions, but is not capable of reducing zinc ions. This is because nickel $$\left( \ce{Ni} \right)$$ is below $$\ce{Cu^{2+}}$$, but is above $$\ce{Zn^{2+}}$$ in the table.

### Summary

• Standard cell potential calculations are described.
• Guidelines for making predictions of reaction possibilities using standard cell potentials are given.

### Contributors

• CK-12 Foundation by Sharon Bewick, Richard Parsons, Therese Forsythe, Shonna Robinson, and Jean Dupon.