Elements and ions differ in the ease with which they are reduced or oxidized. Reduction half-reactions for several ions and elements are listed in Table 14.3. After each equation is a number called the reduction potential (symbol, Eo), which compares the tendency for the listed reaction to occur with the tendency of aqueous hydrogen ion to be reduced:
2 H+ + 2 e- H2 E0 = 0.0 V
The reduction potential for the reduction of hydrogen ion to hydrogen has been arbitrarily assigned a value of 0.0 volts (V). A substance whose reduction potential is less than that of hydrogen ion (has a negative value) is reduced less easily than is hydrogen ion. A substance whose reduction potential is positive (is greater than that of a hydrogen ion) is more easily reduced than is hydrogen.
The reduction potentials listed were measured at 25°C and 1 atm pressure, using solutions in which one mole of reactant was dissolved in 1 L solution. At different concentrations, temperatures, or pressures, the values of the reduction potentials are slightly different.
Polyatomic ions (MnO4-, Cr2O72-, NO3-, and so on) are often involved in electron-transfer reactions. Reduction potentials for these ions are also shown in Table 14.3. Why do we show the whole ion instead of just the element? The element is not alone in solution but is part of the polyatomic ion. For example, in the permanganate ion, MnO4-, manganese has an oxidation number of +7. Manganese in this state is not an ion but part of a covalently bonded polyatomic ion. The half-reactions show only substances that actually exist in solution.
|Oxidized form||Reduced form||E0 (volts)|
|Li+ + e-||Li||-3.05|
|K+ + e-||K||-2.93|
|Mg2+ + 2 e-||Mg||-2.38|
|Zn2+ + 2 e-||Zn||-0.76|
|Fe2+ + 2 e -||Fe||-0.44|
|2 H+ + 2 e-||H2||0.00|
|SO22- + 4 H+ + 2 e-||H2SO3 + H2O||0.20|
|Cu2+ + 2 e-||Cu||0.34|
|I2 + 2 e-||2 I-||0.54|
|Fe3+ + e-||Fe2+||0.77|
|Hg22+ + 2 e-||2 Hg||0.79|
|Ag+ + e-||Ag||0.80|
|NO3- + 4 H+ + 3 e-||NO + 2 H2O||0.96|
|Br2 + 2 e-||2 Br-||1.09|
|O2 + 4 H+ + 4 e-||2 H2O||1.23|
|Cr2O72- + 14 H+ + 6 e-||2 Cr3+ + 7 H2O||1.33|
|Cl2 + 2 e-||2 Cl-||1.36|
|MnO4- + 8 H+ + 5 e-||Mn2+ + 4 H2O||1.49|
A. The Use of Reduction Potential Tables in Writing Redox Equations
We have seen that an oxidation-reduction equation is the sum of an oxidation half-reaction and a reduction half-reaction. Reduction half-reactions can be obtained from a table of reduction potentials, but what about oxidation half-reactions? These equations can be obtained in a balanced form by reversing the reduction equations of the table. Let us see how equations such as those in Table 14.3 can be used in balancing redox equations.
When copper reacts with dilute nitric acid, copper(II) ion and nitrogen oxide (NO) are formed:
Cu + HNO3 Cu2+ + NO
Write the balanced ionic equation for this reaction.
Copper has changed oxidation number from 0 to +2; it has been oxidized. We can find the equation for the reduction of copper ion in the table and reverse it to represent oxidation.
Cu Cu2+ + 2 e-
The nitrogen of nitric acid was reduced in this reaction. We can find the equation for this reduction of nitric acid to nitrogen oxide in the table.
NO3- + 4 H+ + 3 e- NO + 2 H2O
We can balance the loss and gain of electrons by multiplying the oxidation half-reaction by 3 and the reduction half-reaction by 2. Adding the resulting equations gives us the equation
3 Cu + 2 NO3- + 8 H+ + 6 e- 4 H2O + 2 NO + 3 Cu2+ + 6 e-
After canceling electrons, we have the balanced equation for this reaction.
B. The Spontaneity of Redox
A table of reduction potentials allows prediction of the spontaneity of a redox reaction. We saw earlier that a spontaneous reaction has a negative free energy change (see Section 13.1A). In a spontaneous reaction, the sum of the potentials of the half-reactions is positive. The potential of the reduction half-reaction is obtained from a table of reduction potentials such as Table 14.3; the potential of the oxidation half-reaction is equal in magnitude to the reduction potential of the reversed reduction half-reaction but is opposite in sign. For example, we can use the reduction half-reaction
Mg2+ + 2 e- MgO E0 = -2.38 V
to write the oxidation half-reaction
MgO Mg2++ 2 e- E0 = +2.38 V
The sum of the potentials of a redox reaction can be calculated as follows. For the redox reaction
Fe + Cu2+ Fe2+ + Cu
the half-reactions are:
oxidation: Fe Fe2+ + 2 e-
reduction: Cu2+ + 2 e- Cu
The E0 of the reduction half-reaction is obtained from Table 14.3:
Cu2+ + 2 e- Cu E0 = +0.34 V
Table 14.3 also gives the potential for the reaction
Fe2+ + 2 e- Fe E0 = -0.44 V
If the equation is reversed, we must also reverse the sign of the E0 for the oxidation half-reaction
Fe Fe2+ + 2 e- E0 = +0.44 V
The sum of the potentials for the overall reaction is:
E0 = - 0.34 V - 0.44 V = + 0.78 V
The reaction is spontaneous and will occur without the net input of energy.
Consider another situation. Suppose a student proposed preparation of hydrogen gas by adding copper metal to hydrochloric acid. Is this proposal sensible? The equation for the proposed reaction is:
Cu + 2 HCl Cu2+ + 2 Cl- + H2(g)
The half-reactions for this equation and their potentials are:
oxidation: Cu Cu2+ + 2 e- E0= -0.34 V
reduction: 2 H+ + 2 e- H2(g) E0 = 0.0 V
The potential for the reduction half-reaction was obtained from Table 14.3. The potential for the oxidation half-reaction was obtained by reversing the equation and sign of the corresponding reduction half-reaction in Table 14.3. The sum of the potentials is negative.
E0 = -0.34 V + 0.0 V = -0.34 V
The reaction is not spontaneous. The student will not be able to prepare hydrogen gas by this reaction. In summary, a redox reaction is spontaneous if the sum of the potentials of its half-reactions is positive. If the sum of the potentials is negative, the reaction will not occur without the net input of energy.
Will acidified potassium dichromate solution oxidize chloride ion? The equation for the proposed reaction is:
2 Cl- + 14 H+ + Cr2O72- 2 Cr3+ + Cl2 + 7 H2O
The half reactions for the equation are:
oxidation: 2Cl- Cl2 + 2 e- Eo = - 1.36 V
reduction: Cr2O72- + 14 H+ 2 Cr3+ + 7 H2O Eo = + 1.33 V
The potential of the reduction half-reaction was obtained from Table 14.3. The potential for the oxidation of chloride ion was obtained by reversing the sign of the potential for the half-reaction
Cl2 + 2 e- 2 Cl-
The sum of these potentials is negative.
Eo = -1.36 V + 1.33 V V= - 0.03 V
Acidified potassium dichromate will not, under standard conditions, oxidize chloride ion.
C. Activity Series
The activity of an element determines how easily the free element becomes an ion. For metals to become ions, they must be oxidized:
Li Li+ + e-
This equation is the reverse of the equation shown in Table 14.3. A metal is said to be very active (easily oxidized) if the reduction potential of its ion is near the top of the table (has a very large negative value). For example, lithium has a reduction potential of - 3.05 V. Lithium is a very active metal and is very easily oxidized to lithium ion.
An activity series of the metals lists the metals in order of decreasing ease of oxidation. In terms of reduction potentials, this listing would be in order of increasing reduction potentials. Table 14.4 shows an activity series for metals. This activity series can be used to predict displacement reactions (see Section 8.2C). A metal will displace any metal below it in the series. For example, copper is above silver in the activity series. Copper will displace silver, as shown in the equation
Cu + 2 Ag+ Cu2+ + 2 Ag
The activity of a nonmetal also depends on the relative ease with which it forms ions. For a nonmetal, the formation of ions involves reduction. The reduction potential of chlorine measures the tendency of the following reaction to occur:
Cl2 + 2 e- 2 Cl-
The most active nonmetal has the lowest reduction potential. Table 14.5 is an activity series for the halogens. Any nonmetal in the series will displace a nonmetal to its left in the series.
Thus, Table 14.5 predicts that chlorine will displace bromine as shown in the following equation:
2 NaBr + Cl2 2 NaCl + Br2
Use Tables 14.3, 14.4 and 14.5 to predict the spontaneity of the following reactions. If the reaction goes as written, balance the equation. If the reaction will not go as written but requires the inut of energy, cross out the products and write "no reaction".
a. Ni + Zn2+ Ni2+ + Zn
b. I2 + 2 Br- Br2 + 2 I-
c. Fe + 2 Ag+ Fe2+ + 2 Ag
a. No reaction. Nickel is below zinc in the activity series of metals (Table 14.4). It will not displace zinc.
b. No reaction. Iodine is to the left of bromine in the nonmetal activity series (Table 14.5). Iodine will not displace bromine.
c. Fe + 2 Ag+ Fe2+ + 2 Ag. Iron is above silver in the activity series (Table 14.4). It will displace silver.