Textbook Chapters | Study Guides | MCAT 2015 Topics | Online Presentations | About | Privacy Policy


Biology Study Guide Topics

Endocrine System | Lymphatic System | Blood | Circulatory System | Skull Bones | Human Skull and Brain | Tissue Types | The Cell | DNA | Anatomy Models | Electron Transport Chain | History of Microbiology | Human Anatomy | Punnett Squares | What is Mitosis | What is Life | Macromolecules | Cellular Respiration | DNA Replication | Enzymes | Pathogenic Bacteria | Natural Selection | Punnett Squares | Transcription and Translation | Exam Notes | Viruses | Osmosis | Protists | Genetic Code | Mendelian Genetics | Meiosis | Sensory Processing | Amino Acids |



Online Presentations

Bones of the Human Skull | Tissue Types | Selective and Differential Media

Classroom Activities

Recombinant DNA Cut And Tape Classroom Activity


Standard Reduction Potentials

By the end of this section, you will be able to:

The cell potential in [link] (+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 [link] 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)+2eH2(g,1 atm)E°=0 V2H+(aq, 1M)+2eH2(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). The voltage is defined as zero for all temperatures.

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.
The figure shows a beaker just over half full of a blue liquid. A glass tube is partially submerged in the liquid. Bubbles, which are labeled “H subscript 2 ( g )” are rising from the dark grey square, labeled “P t electrode” at the bottom of the tube. A curved arrow points up to the right, indicating the direction of the bubbles. A black wire which is labeled “P t wire” extends from the dark grey square up the interior of the tube through a small port at the top. A second small port extends out the top of the tube to the left. An arrow points to the port opening from the left. The base of this arrow is labeled “H subscript 2 ( g ) at 1 a t m.” A light grey arrow points to a diagram in a circle at the right that illustrates the surface of the P t electrode in a magnified view. P t atoms are illustrated as a uniform cluster of grey spheres which are labeled “P t electrode atoms.” On the grey atom surface, the label “e superscript negative” is shown 4 times in a nearly even vertical distribution to show electrons on the P t surface. A curved arrow extends from a white sphere labeled “H superscript plus” at the right of the P t atoms to the uppermost electron shown. Just below, a straight arrow extends from the P t surface to the right to a pair of linked white spheres which are labeled “H subscript 2.” A curved arrow extends from a second white sphere labeled “H superscript plus” at the right of the P t atoms to the second electron shown. A curved arrow extends from the third electron on the P t surface to the right to a white sphere labeled “H superscript plus.” Just below, an arrow points left from a pair of linked white spheres which are labeled “H subscript 2” to the P t surface. A curved arrow extends from the fourth electron on the P t surface to the right to a white sphere labeled “H superscript plus.” Beneath this atomic view is the label “Half-reaction at P t surface: 2 H superscript plus ( a q, 1 M ) plus 2 e superscript negative right pointing arrow H subscript 2 ( g, 1 a t m ).”

A galvanic cell consisting of a SHE and Cu2+/Cu half-cell can be used to determine the standard reduction potential for Cu2+ ([link]). 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) + 2eCathode (reduction):Cu2+(aq)+2eCu(s)¯Overall:Cu2+(aq)+H2(g)2H+(aq)+Cu(s)Anode (oxidation):H2(g)2H+(aq) + 2eCathode (reduction):Cu2+(aq)+2eCu(s)¯Overall:Cu2+(aq)+H2(g)2H+(aq)+Cu(s)

The standard reduction potential 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.

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°
A galvanic cell can be used to determine the standard reduction potential of Cu2+.
This figure contains a diagram of an electrochemical cell. Two beakers are shown. Each is just over half full. The beaker on the left contains a clear, colorless solution and is labeled below as “1 M H superscript plus.” The beaker on the right contains a blue solution and is labeled below as “1 M C u superscript 2 plus.” A glass tube in the shape of an inverted U connects the two beakers at the center of the diagram. The tube contents are colorless. The ends of the tubes are beneath the surface of the solutions in the beakers and a small grey plug is present at each end of the tube. The beaker on the left has a glass tube partially submersed in the liquid. Bubbles are rising from the grey square, labeled “Standard hydrogen electrode” at the bottom of the tube. A curved arrow points up to the right, indicating the direction of the bubbles. A black wire extends from the grey square up the interior of the tube through a small port at the top to a rectangle with a digital readout of “positive 0.337 V” which is labeled “Voltmeter.” A second small port extends out the top of the tube to the left. An arrow points to the port opening from the left. The base of this arrow is labeled “H subscript 2 ( g ).” The beaker on the right has an orange-brown strip that is labeled “C u strip” at the top. A wire extends from the top of this strip to the voltmeter. An arrow points toward the voltmeter from the left which is labeled “e superscript negative flow.” Similarly, an arrow points away from the voltmeter to the right. A curved arrow extends from the standard hydrogen electrode in the beaker on the left into the surrounding solution. The tip of this arrow is labeled “H plus.” An arrow points downward from the label “e superscript negative” on the C u strip in the beaker on the right. A second curved arrow extends from another “e superscript negative” label into the solution below toward the label “C u superscript 2 plus” in the solution. A third “e superscript negative” label positioned at the lower left edge of the C u strip has a curved arrow extending from it to the “C u superscript 2 plus” label in the solution. A curved arrow extends from this “C u superscript 2 plus” label to a “C u” label at the lower edge of the C u strip.

Using the SHE as a reference, other standard reduction potentials can be determined. Consider the cell shown in [link], 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)+2ecathode (reduction):2Ag+(aq)+2e2Ag(s)¯overall:2Ag+(aq)+H2(g)2H+(aq)+2Ag(s)anode (oxidation):H2(g)2H+(aq)+2ecathode (reduction):2Ag+(aq)+2e2Ag(s)¯overall:2Ag+(aq)+H2(g)2H+(aq)+2Ag(s)

The standard reduction potential 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 [link],

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)+2ecathode (reduction):2Ag+(aq)+2e2Ag(s)¯overall:Cu(s)+2Ag+(aq)Cu2+(aq)+2Ag(s)anode (oxidation):Cu(s)Cu2+(aq)+2ecathode (reduction):2Ag+(aq)+2e2Ag(s)¯overall:Cu(s)+2Ag+(aq)Cu2+(aq)+2Ag(s)
Ecell°=Ecathode°Eanode°=EAg+/Ag°ECu2+/Cu°=0.80 V0.34 V=0.46 VEcell°=Ecathode°Eanode°=EAg+/Ag°ECu2+/Cu°=0.80 V0.34 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 [link]. A more complete list is provided in Appendix L.

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.
This figure contains a diagram of an electrochemical cell. Two beakers are shown. Each is just over half full. The beaker on the left contains a clear, colorless solution which is labeled “H N O subscript 3 ( a q ).” The beaker on the right contains a clear, colorless solution which is labeled “A g N O subscript 3 ( a q ).” A glass tube in the shape of an inverted U connects the two beakers at the center of the diagram and is labeled “Salt bridge.” The tube contents are colorless. The ends of the tubes are beneath the surface of the solutions in the beakers and a small grey plug is present at each end of the tube. The label “2 N a superscript plus” appears on the upper right portion of the tube. A curved arrow extends from this label down and to the right. The label “2 N O subscript 3 superscript negative” appears on the upper left portion of the tube. A curved arrow extends from this label down and to the left. The beaker on the left has a glass tube partially submerged in the liquid. Bubbles are rising from the grey square, labeled “SHE anode” at the bottom of the tube. A curved arrow points up to the right. The labels “2 H superscript plus” and “2 N O subscript 3 superscript negative” appear on the liquid in the beaker. A black wire extends from the grey square up the interior of the tube through a small port at the top to a rectangle with a digital readout of “positive 0.80 V” which is labeled “Voltmeter.” A second small port extends out the top of the tube to the left. An arrow points to the port opening from the left. The base of this arrow is labeled “H subscript 2 ( g ).” The beaker on the right has a silver strip that is labeled “A g cathode.” A wire extends from the top of this strip to the voltmeter. An arrow points toward the voltmeter from the left which is labeled “e superscript negative flow.” Similarly, an arrow points away from the voltmeter to the right. The solution in the beaker on the right has the labels “N O subscript 3 superscript negative” and “A g superscript plus” on the solution. A curved arrow extends from the A g superscript plus label to the A g cathode. Below the left beaker at the bottom of the diagram is the label “Oxidation half-reaction: H subscript 2 ( g ) right pointing arrow 2 H superscript plus ( a q ) plus 2 e superscript negative.” Below the right beaker at the bottom of the diagram is the label “Reduction half-reaction: 2 A g superscript plus ( a q ) right pointing arrow 2 A g ( s ).”
Selected Standard Reduction Potentials at 25 °C
Half-Reaction E° (V)
F2(g)+2e2F(aq)F2(g)+2e2F(aq) +2.866
PbO2(s)+SO42−(aq)+4H+(aq)+2ePbSO4(s)+2H2O(l)PbO2(s)+SO42−(aq)+4H+(aq)+2ePbSO4(s)+2H2O(l) +1.69
MnO4(aq)+8H+(aq)+5eMn2+(aq)+4H2O(l)MnO4(aq)+8H+(aq)+5eMn2+(aq)+4H2O(l) +1.507
Au3+(aq)+3eAu(s)Au3+(aq)+3eAu(s) +1.498
Cl2(g)+2e2Cl(aq)Cl2(g)+2e2Cl(aq) +1.35827
O2(g)+4H+(aq)+4e2H2O(l)O2(g)+4H+(aq)+4e2H2O(l) +1.229
Pt2+(aq)+2ePt(s)Pt2+(aq)+2ePt(s) +1.20
Br2(aq)+2e2Br(aq)Br2(aq)+2e2Br(aq) +1.0873
Ag+(aq)+eAg(s)Ag+(aq)+eAg(s) +0.7996
Hg22+(aq)+2e2Hg(l)Hg22+(aq)+2e2Hg(l) +0.7973
Fe3+(aq)+eFe2+(aq)Fe3+(aq)+eFe2+(aq) +0.771
MnO4(aq)+2H2O(l)+3eMnO2(s)+4OH(aq)MnO4(aq)+2H2O(l)+3eMnO2(s)+4OH(aq) +0.558
I2(s)+2e2I(aq)I2(s)+2e2I(aq) +0.5355
NiO2(s)+2H2O(l)+2eNi(OH)2(s)+2OH(aq)NiO2(s)+2H2O(l)+2eNi(OH)2(s)+2OH(aq) +0.49
Cu2+(aq)+2eCu(s)Cu2+(aq)+2eCu(s) +0.337
Hg2Cl2(s)+2e2Hg(l)+2Cl(aq)Hg2Cl2(s)+2e2Hg(l)+2Cl(aq) +0.26808
AgCl(s)+2eAg(s)+Cl(aq)AgCl(s)+2eAg(s)+Cl(aq) +0.22233
Sn4+(aq)+2eSn2+(aq)Sn4+(aq)+2eSn2+(aq) +0.151
2H+(aq)+2eH2(g)2H+(aq)+2eH2(g) 0.00
Pb2+(aq)+2ePb(s)Pb2+(aq)+2ePb(s) −0.126
Sn2+(aq)+2eSn(s)Sn2+(aq)+2eSn(s) −0.1262
Ni2+(aq)+2eNi(s)Ni2+(aq)+2eNi(s) −0.257
Co2+(aq)+2eCo(s)Co2+(aq)+2eCo(s) −0.28
PbSO4(s)+2ePb(s)+SO42−(aq)PbSO4(s)+2ePb(s)+SO42−(aq) −0.3505
Cd2+(aq)+2eCd(s)Cd2+(aq)+2eCd(s) −0.4030
Fe2+(aq)+2eFe(s)Fe2+(aq)+2eFe(s) −0.447
Cr3+(aq)+3eCr(s)Cr3+(aq)+3eCr(s) −0.744
Mn2+(aq)+2eMn(s)Mn2+(aq)+2eMn(s) −1.185
Zn(OH)2(s)+2eZn(s)+2OH(aq)Zn(OH)2(s)+2eZn(s)+2OH(aq) −1.245
Zn2+(aq)+2eZn(s)Zn2+(aq)+2eZn(s) −0.7618
Al3+(aq)+3eAl(s)Al3+(aq)+3eAl(s) −1.662
Mg2(aq)+2eMg(s)Mg2(aq)+2eMg(s) −2.372
Na+(aq)+eNa(s)Na+(aq)+eNa(s) −2.71
Ca2+(aq)+2eCa(s)Ca2+(aq)+2eCa(s) −2.868
Ba2+(aq)+2eBa(s)Ba2+(aq)+2eBa(s) −2.912
K+(aq)+eK(s)K+(aq)+eK(s) −2.931
Li+(aq)+eLi(s)Li+(aq)+eLi(s) −3.04

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

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 [link], the reactions involved in the galvanic cell, both written as reductions, are

Au3+(aq)+3eAu(s)EAu3+/Au°=+1.498 VAu3+(aq)+3eAu(s)EAu3+/Au°=+1.498 V
Ni2+(aq)+2eNi(s)ENi2+/Ni°=−0.257 VNi2+(aq)+2eNi(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)+2eEanode°=ENi2+/Ni°=−0.257 VCathode (reduction): Au3+(aq)+3eAu(s)Ecathode°=EAu3+/Au°=+1.498 VAnode (oxidation):Ni(s)Ni2+(aq)+2eEanode°=ENi2+/Ni°=−0.257 VCathode (reduction): Au3+(aq)+3eAu(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.

Check Your Learning 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.

Answer:

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

Key Concepts and Summary

Assigning the potential of the standard hydrogen electrode (SHE) as zero volts allows the determination of standard reduction potentials, , for half-reactions in electrochemical cells. As the name implies, standard reduction potentials use standard states (1 bar or 1 atm for gases; 1 M for solutes, often at 298.15 K) and are written as reductions (where electrons appear on the left side of the equation). The reduction reactions are reversible, so standard cell potentials can be calculated by subtracting the standard reduction potential for the reaction at the anode from the standard reduction for the reaction at the cathode. When calculating the standard cell potential, the standard reduction potentials are not scaled by the stoichiometric coefficients in the balanced overall equation.

Key Equations

Chemistry End of Chapter Exercises

For each reaction listed, determine its standard cell potential at 25 °C and whether the reaction is spontaneous at standard conditions.

(a) Mg(s)+Ni2+(aq)Mg2+(aq)+Ni(s)Mg(s)+Ni2+(aq)Mg2+(aq)+Ni(s)

(b) 2Ag+(aq)+Cu(s)Cu2+(aq)+2Ag(s)2Ag+(aq)+Cu(s)Cu2+(aq)+2Ag(s)

(c) Mn(s)+Sn(NO3)2(aq)Mn(NO3)2(aq)+Sn(s)Mn(s)+Sn(NO3)2(aq)Mn(NO3)2(aq)+Sn(s)

(d) 3Fe(NO3)2(aq)+Au(NO3)3(aq)3Fe(NO3)3(aq)+Au(s)3Fe(NO3)2(aq)+Au(NO3)3(aq)3Fe(NO3)3(aq)+Au(s)

(a) +2.115 V (spontaneous); (b) +0.4626 V (spontaneous); (c) +1.0589 V (spontaneous); (d) +0.727 V (spontaneous)

For each reaction listed, determine its standard cell potential at 25 °C and whether the reaction is spontaneous at standard conditions.

(a) Mn(s)+Ni2+(aq)Mn2+(aq)+Ni(s)Mn(s)+Ni2+(aq)Mn2+(aq)+Ni(s)

(b) 3Cu2+(aq)+2Al(s)2Al3+(aq)+2Cu(s)3Cu2+(aq)+2Al(s)2Al3+(aq)+2Cu(s)

(c) Na(s)+LiNO3(aq)NaNO3(aq)+Li(s)Na(s)+LiNO3(aq)NaNO3(aq)+Li(s)

(d) Ca(NO3)2(aq)+Ba(s)Ba(NO3)2(aq)+Ca(s)Ca(NO3)2(aq)+Ba(s)Ba(NO3)2(aq)+Ca(s)

Determine the overall reaction and its standard cell potential at 25 °C for this reaction. Is the reaction spontaneous at standard conditions?

Cu(s)Cu2+(aq)Au3+(aq)Au(s)Cu(s)Cu2+(aq)Au3+(aq)Au(s)

3Cu(s)+2Au3+(aq)3Cu2+(aq)+2Au(s);3Cu(s)+2Au3+(aq)3Cu2+(aq)+2Au(s); +1.16 V; spontaneous

Determine the overall reaction and its standard cell potential at 25 °C for the reaction involving the galvanic cell made from a half-cell consisting of a silver electrode in 1 M silver nitrate solution and a half-cell consisting of a zinc electrode in 1 M zinc nitrate. Is the reaction spontaneous at standard conditions?

Determine the overall reaction and its standard cell potential at 25 °C for the reaction involving the galvanic cell in which cadmium metal is oxidized to 1 M cadmium(II) ion and a half-cell consisting of an aluminum electrode in 1 M aluminum nitrate solution. Is the reaction spontaneous at standard conditions?

3Cd(s)+2Al3+(aq)3Cd2+(aq)+2Al(s);3Cd(s)+2Al3+(aq)3Cd2+(aq)+2Al(s); −1.259 V; nonspontaneous

Determine the overall reaction and its standard cell potential at 25 °C for these reactions. Is the reaction spontaneous at standard conditions? Assume the standard reduction for Br2(l) is the same as for Br2(aq).


Pt(s)H2(g)H+(aq)Br2(aq)Br(aq)Pt(s)Pt(s)H2(g)H+(aq)Br2(aq)Br(aq)Pt(s)

Glossary

standard cell potential (Ecell°)(Ecell°)
the cell potential when all reactants and products are in their standard states (1 bar or 1 atm or gases; 1 M for solutes), usually at 298.15 K; can be calculated by subtracting the standard reduction potential for the half-reaction at the anode from the standard reduction potential for the half-reaction occurring at the cathode
standard hydrogen electrode (SHE)
the electrode consists of hydrogen gas bubbling through hydrochloric acid over an inert platinum electrode whose reduction at standard conditions is assigned a value of 0 V; the reference point for standard reduction potentials
standard reduction potential (E°)
the value of the reduction under standard conditions (1 bar or 1 atm for gases; 1 M for solutes) usually at 298.15 K; tabulated values used to calculate standard cell potentials

This page was adapted from the textbook made available by OpenStax College under a Creative Commons License 4.0 International. Download for free at http://cnx.org/content/col11448/latest/



The pdf study guides on this site are licensed under a Creative Commons 4.0 International License. Please note that this only applies to the pdf format study guides. Some of the figures available under creative commons can be found here: ccimages

© 2009-2016 Anthony D'Onofrio, PhD