Chemistry

# 17.3Standard Reduction Potentials

Chemistry17.3 Standard Reduction Potentials

### Learning Objectives

By the end of this section, you will be able to:
• Determine standard cell potentials for oxidation-reduction reactions
• Use standard reduction potentials to determine the better oxidizing or reducing agent from among several possible choices

The cell potential in Figure 17.4 (+0.46 V) results from the difference in the electrical potentials for each electrode. While it is impossible to determine the electrical potential of a single electrode, we can assign an electrode the value of zero and then use it as a reference. The electrode chosen as the zero is shown in Figure 17.6 and is called the standard hydrogen electrode (SHE). The SHE consists of 1 atm of hydrogen gas bubbled through a 1 M HCl solution, usually at room temperature. Platinum, which is chemically inert, is used as the electrode. The reduction half-reaction chosen as the reference is

$2H+(aq, 1M)+2e−⇌H2(g,1 atm)E°=0 V2H+(aq, 1M)+2e−⇌H2(g,1 atm)E°=0 V$

E° is the standard reduction potential. The superscript “°” on the E denotes standard conditions (1 bar or 1 atm for gases, 1 M for solutes). This voltage is defined as zero for all temperatures.

Figure 17.6 Hydrogen gas at 1 atm is bubbled through 1 M HCl solution. Platinum, which is inert to the action of the 1 M HCl, is used as the electrode. Electrons on the surface of the electrode combine with H+ in solution to produce hydrogen gas.

A galvanic cell consisting of a SHE and Cu2+/Cu half-cell can be used to determine the standard reduction potential for Cu2+ (Figure 17.7). In cell notation, the reaction is

$Pt(s)│H2(g,1 atm)│H+(aq,1M)║Cu2+(aq,1M)│Cu(s)Pt(s)│H2(g,1 atm)│H+(aq,1M)║Cu2+(aq,1M)│Cu(s)$

Electrons flow from the anode to the cathode. The reactions, which are reversible, are

$Anode (oxidation):H2(g)⟶2H+(aq) + 2e−Cathode (reduction):Cu2+(aq)+2e−⟶Cu(s)¯Overall:Cu2+(aq)+H2(g)⟶2H+(aq)+Cu(s)Anode (oxidation):H2(g)⟶2H+(aq) + 2e−Cathode (reduction):Cu2+(aq)+2e−⟶Cu(s)¯Overall:Cu2+(aq)+H2(g)⟶2H+(aq)+Cu(s)$

The standard cell potential, E°cell, can be determined by subtracting the standard reduction potential for the reaction occurring at the anode from the standard reduction potential for the reaction occurring at the cathode. The minus sign is necessary because oxidation is the reverse of reduction. The minus sign is necessary because oxidation is the reverse of reduction.

$Ecell°=Ecathode°−Eanode°Ecell°=Ecathode°−Eanode°$
$+0.34 V=ECu2+/Cu°−EH+/H2°=ECu2+/Cu°−0=ECu2+/Cu°+0.34 V=ECu2+/Cu°−EH+/H2°=ECu2+/Cu°−0=ECu2+/Cu°$
Figure 17.7 A galvanic cell can be used to determine the standard reduction potential of Cu2+.

Using the SHE as a reference, other standard reduction potentials can be determined. Consider the cell shown in Figure 17.8, where

$Pt(s)│H2(g,1 atm)│H+(aq, 1M)║Ag+(aq, 1M)│Ag(s)Pt(s)│H2(g,1 atm)│H+(aq, 1M)║Ag+(aq, 1M)│Ag(s)$

Electrons flow from left to right, and the reactions are

$anode (oxidation):H2(g)⟶2H+(aq)+2e−cathode (reduction):2Ag+(aq)+2e−⟶2Ag(s)¯overall:2Ag+(aq)+H2(g)⟶2H+(aq)+2Ag(s)anode (oxidation):H2(g)⟶2H+(aq)+2e−cathode (reduction):2Ag+(aq)+2e−⟶2Ag(s)¯overall:2Ag+(aq)+H2(g)⟶2H+(aq)+2Ag(s)$

The standard cell potential, E°cell, can be determined by subtracting the standard reduction potential for the reaction occurring at the anode from the standard reduction potential for the reaction occurring at the cathode. The minus sign is needed because oxidation is the reverse of reduction.

$Ecell°=Ecathode°−Eanode°Ecell°=Ecathode°−Eanode°$
$+0.80 V=EAg+/Ag°−EH+/H2°=EAg+/Ag°−0=EAg+/Ag°+0.80 V=EAg+/Ag°−EH+/H2°=EAg+/Ag°−0=EAg+/Ag°$

It is important to note that the potential is not doubled for the cathode reaction.

The SHE is rather dangerous and rarely used in the laboratory. Its main significance is that it established the zero for standard reduction potentials. Once determined, standard reduction potentials can be used to determine the standard cell potential, $Ecell°,Ecell°,$ for any cell. For example, for the cell shown in Figure 17.4,

$Cu(s)│Cu2+(aq,1M)║Ag+(aq,1M)│Ag(s)Cu(s)│Cu2+(aq,1M)║Ag+(aq,1M)│Ag(s)$
$anode (oxidation):Cu(s)⟶Cu2+(aq)+2e−cathode (reduction):2Ag+(aq)+2e−⟶2Ag(s)¯overall:Cu(s)+2Ag+(aq)⟶Cu2+(aq)+2Ag(s)anode (oxidation):Cu(s)⟶Cu2+(aq)+2e−cathode (reduction):2Ag+(aq)+2e−⟶2Ag(s)¯overall:Cu(s)+2Ag+(aq)⟶Cu2+(aq)+2Ag(s)$
$Ecell°=Ecathode°−Eanode°=EAg+/Ag°−ECu2+/Cu°=0.80 V−0.34 V=0.46 VEcell°=Ecathode°−Eanode°=EAg+/Ag°−ECu2+/Cu°=0.80 V−0.34 V=0.46 V$

Again, note that when calculating $Ecell°,Ecell°,$ standard reduction potentials always remain the same even when a half-reaction is multiplied by a factor. Standard reduction potentials for selected reduction reactions are shown in Table 17.2. A more complete list is provided in Appendix L.

Figure 17.8 A galvanic cell can be used to determine the standard reduction potential of Ag+. The SHE on the left is the anode and assigned a standard reduction potential of zero.
Selected Standard Reduction Potentials at 25 °C
Half-Reaction E° (V)
$F2(g)+2e−⟶2F−(aq)F2(g)+2e−⟶2F−(aq)$ +2.866
$PbO2(s)+SO42−(aq)+4H+(aq)+2e−⟶PbSO4(s)+2H2O(l)PbO2(s)+SO42−(aq)+4H+(aq)+2e−⟶PbSO4(s)+2H2O(l)$ +1.69
$MnO4−(aq)+8H+(aq)+5e−⟶Mn2+(aq)+4H2O(l)MnO4−(aq)+8H+(aq)+5e−⟶Mn2+(aq)+4H2O(l)$ +1.507
$Au3+(aq)+3e−⟶Au(s)Au3+(aq)+3e−⟶Au(s)$ +1.498
$Cl2(g)+2e−⟶2Cl−(aq)Cl2(g)+2e−⟶2Cl−(aq)$ +1.35827
$O2(g)+4H+(aq)+4e−⟶2H2O(l)O2(g)+4H+(aq)+4e−⟶2H2O(l)$ +1.229
$Pt2+(aq)+2e−⟶Pt(s)Pt2+(aq)+2e−⟶Pt(s)$ +1.20
$Br2(aq)+2e−⟶2Br−(aq)Br2(aq)+2e−⟶2Br−(aq)$ +1.0873
$Ag+(aq)+e−⟶Ag(s)Ag+(aq)+e−⟶Ag(s)$ +0.7996
$Hg22+(aq)+2e−⟶2Hg(l)Hg22+(aq)+2e−⟶2Hg(l)$ +0.7973
$Fe3+(aq)+e−⟶Fe2+(aq)Fe3+(aq)+e−⟶Fe2+(aq)$ +0.771
$MnO4−(aq)+2H2O(l)+3e−⟶MnO2(s)+4OH−(aq)MnO4−(aq)+2H2O(l)+3e−⟶MnO2(s)+4OH−(aq)$ +0.558
$I2(s)+2e−⟶2I−(aq)I2(s)+2e−⟶2I−(aq)$ +0.5355
$NiO2(s)+2H2O(l)+2e−⟶Ni(OH)2(s)+2OH−(aq)NiO2(s)+2H2O(l)+2e−⟶Ni(OH)2(s)+2OH−(aq)$ +0.49
$Cu2+(aq)+2e−⟶Cu(s)Cu2+(aq)+2e−⟶Cu(s)$ +0.34
$Hg2Cl2(s)+2e−⟶2Hg(l)+2Cl−(aq)Hg2Cl2(s)+2e−⟶2Hg(l)+2Cl−(aq)$ +0.26808
$AgCl(s)+e−⟶Ag(s)+Cl−(aq)AgCl(s)+e−⟶Ag(s)+Cl−(aq)$ +0.22233
$Sn4+(aq)+2e−⟶Sn2+(aq)Sn4+(aq)+2e−⟶Sn2+(aq)$ +0.151
$2H+(aq)+2e−⟶H2(g)2H+(aq)+2e−⟶H2(g)$ 0.00
$Pb2+(aq)+2e−⟶Pb(s)Pb2+(aq)+2e−⟶Pb(s)$ −0.1262
$Sn2+(aq)+2e−⟶Sn(s)Sn2+(aq)+2e−⟶Sn(s)$ −0.1375
$Ni2+(aq)+2e−⟶Ni(s)Ni2+(aq)+2e−⟶Ni(s)$ −0.257
$Co2+(aq)+2e−⟶Co(s)Co2+(aq)+2e−⟶Co(s)$ −0.28
$PbSO4(s)+2e−⟶Pb(s)+SO42−(aq)PbSO4(s)+2e−⟶Pb(s)+SO42−(aq)$ −0.3505
$Cd2+(aq)+2e−⟶Cd(s)Cd2+(aq)+2e−⟶Cd(s)$ −0.4030
$Fe2+(aq)+2e−⟶Fe(s)Fe2+(aq)+2e−⟶Fe(s)$ −0.447
$Cr3+(aq)+3e−⟶Cr(s)Cr3+(aq)+3e−⟶Cr(s)$ −0.744
$Mn2+(aq)+2e−⟶Mn(s)Mn2+(aq)+2e−⟶Mn(s)$ −1.185
$Zn(OH)2(s)+2e−⟶Zn(s)+2OH−(aq)Zn(OH)2(s)+2e−⟶Zn(s)+2OH−(aq)$ −1.245
$Zn2+(aq)+2e−⟶Zn(s)Zn2+(aq)+2e−⟶Zn(s)$ −0.7618
$Al3+(aq)+3e−⟶Al(s)Al3+(aq)+3e−⟶Al(s)$ −1.662
$Mg2(aq)+2e−⟶Mg(s)Mg2(aq)+2e−⟶Mg(s)$ −2.372
$Na+(aq)+e−⟶Na(s)Na+(aq)+e−⟶Na(s)$ −2.71
$Ca2+(aq)+2e−⟶Ca(s)Ca2+(aq)+2e−⟶Ca(s)$ −2.868
$Ba2+(aq)+2e−⟶Ba(s)Ba2+(aq)+2e−⟶Ba(s)$ −2.912
$K+(aq)+e−⟶K(s)K+(aq)+e−⟶K(s)$ −2.931
$Li+(aq)+e−⟶Li(s)Li+(aq)+e−⟶Li(s)$ −3.04
Table 17.2

Tables like this make it possible to determine the standard cell potential for many oxidation-reduction reactions.

### Example 17.4

#### Cell Potentials from Standard Reduction Potentials

What is the standard cell potential for a galvanic cell that consists of Au3+/Au and Ni2+/Ni half-cells? Identify the oxidizing and reducing agents.

#### Solution

Using Table 17.2, the reactions involved in the galvanic cell, both written as reductions, are
$Au3+(aq)+3e−⟶Au(s)EAu3+/Au°=+1.498 VAu3+(aq)+3e−⟶Au(s)EAu3+/Au°=+1.498 V$
$Ni2+(aq)+2e−⟶Ni(s)ENi2+/Ni°=−0.257 VNi2+(aq)+2e−⟶Ni(s)ENi2+/Ni°=−0.257 V$

Galvanic cells have positive cell potentials, and all the reduction reactions are reversible. The reaction at the anode will be the half-reaction with the smaller or more negative standard reduction potential. Reversing the reaction at the anode (to show the oxidation) but not its standard reduction potential gives:

$Anode (oxidation):Ni(s)⟶Ni2+(aq)+2e−Eanode°=ENi2+/Ni°=−0.257 VCathode (reduction): Au3+(aq)+3e−⟶Au(s)Ecathode°=EAu3+/Au°=+1.498 VAnode (oxidation):Ni(s)⟶Ni2+(aq)+2e−Eanode°=ENi2+/Ni°=−0.257 VCathode (reduction): Au3+(aq)+3e−⟶Au(s)Ecathode°=EAu3+/Au°=+1.498 V$

The least common factor is six, so the overall reaction is

$3Ni(s)+2Au3+(aq)⟶3Ni2+(aq)+2Au(s)3Ni(s)+2Au3+(aq)⟶3Ni2+(aq)+2Au(s)$

The reduction potentials are not scaled by the stoichiometric coefficients when calculating the cell potential, and the unmodified standard reduction potentials must be used.

$Ecell°=Ecathode°−Eanode°=1.498 V−(−0.257 V)=1.755 VEcell°=Ecathode°−Eanode°=1.498 V−(−0.257 V)=1.755 V$

From the half-reactions, Ni is oxidized, so it is the reducing agent, and Au3+ is reduced, so it is the oxidizing agent.

A galvanic cell consists of a Mg electrode in 1 M Mg(NO3)2 solution and a Ag electrode in 1 M AgNO3 solution. Calculate the standard cell potential at 25 °C.

$Mg(s)+2Ag+(aq)⟶Mg2+(aq)+2Ag(s)Ecell°=0.7996 V−(−2.372 V)=3.172 VMg(s)+2Ag+(aq)⟶Mg2+(aq)+2Ag(s)Ecell°=0.7996 V−(−2.372 V)=3.172 V$

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