10.3 Standard reduction potentials (Page 2/5)

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Cu (s) │ {Cu}^{2+} (a q, 1 M) ║ {Ag}^{+} (a q, 1 M) │ Ag (s)

\begin{array}{l} \underline{\begin{array}{l} anode (oxidation): Cu (s) ⟶ {Cu}^{2+} (a q) + {2e}^{−} \\ cathode (reduction): 2 {Ag}^{+} (a q) + {2e}^{−} ⟶ 2Ag (s) \end{array}} \\ overall: Cu (s) + {2Ag}^{+} (a q) ⟶ {Cu}^{2+} (a q) + 2Ag (s) \end{array}

E_{cell}^{°} = E_{cathode}^{°} - E_{anode}^{°} = E_{{Ag}^{+} /Ag}^{°} - E_{{Cu}^{2+} /Cu}^{°} = 0.80 V - 0.34 V = 0.4 6 V

Again, note that when calculating $E_{cell}^{°},$ 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 .

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 ).” — 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)
$F_{2} (g) + {2e}^{−} ⟶ {2F}^{−} (a q)$	+2.866
${PbO}_{2} (s) + {SO}_{4}^{2−} (a q) + {4H}^{+} (a q) + {2e}^{−} ⟶ {PbSO}_{4} (s) + {2H}_{2} O (l)$	+1.69
${MnO}_{4}^{−} (a q) + {8H}^{+} (a q) + {5e}^{−} ⟶ {Mn}^{2+} (a q) + {4H}_{2} O (l)$	+1.507
${Au}^{3+} (a q) + {3e}^{−} ⟶ Au (s)$	+1.498
${Cl}_{2} (g) + {2e}^{−} ⟶ {2Cl}^{−} (a q)$	+1.35827
$O_{2} (g) + {4H}^{+} (a q) + {4e}^{−} ⟶ {2H}_{2} O (l)$	+1.229
${Pt}^{2+} (a q) + {2e}^{−} ⟶ Pt (s)$	+1.20
${Br}_{2} (a q) + {2e}^{−} ⟶ {2Br}^{−} (a q)$	+1.0873
${Ag}^{+} (a q) + e^{−} ⟶ Ag (s)$	+0.7996
${Hg}_{2}^{2+} (a q) + {2e}^{−} ⟶ 2Hg (l)$	+0.7973
${Fe}^{3+} (a q) + e^{−} ⟶ {Fe}^{2+} (a q)$	+0.771
${MnO}_{4}^{−} (a q) + {2H}_{2} O (l) + {3e}^{−} ⟶ {MnO}_{2} (s) + {4OH}^{−} (a q)$	+0.558
$I_{2} (s) + {2e}^{−} ⟶ {2I}^{−} (a q)$	+0.5355
${NiO}_{2} (s) + {2H}_{2} O (l) + {2e}^{−} ⟶ {Ni(OH)}_{2} (s) + {2OH}^{−} (a q)$	+0.49
${Cu}^{2+} (a q) + {2e}^{−} ⟶ Cu (s)$	+0.337
${Hg}_{2} {Cl}_{2} (s) + {2e}^{−} ⟶ 2Hg (l) + {2Cl}^{−} (a q)$	+0.26808
$AgCl (s) + {2e}^{−} ⟶ Ag (s) + {Cl}^{−} (a q)$	+0.22233
${Sn}^{4+} (a q) + {2e}^{−} ⟶ {Sn}^{2+} (a q)$	+0.151
${2H}^{+} (a q) + {2e}^{−} ⟶ H_{2} (g)$	0.00
${Pb}^{2+} (a q) + {2e}^{−} ⟶ Pb (s)$	−0.126
${Sn}^{2+} (a q) + {2e}^{−} ⟶ Sn (s)$	−0.1262
${Ni}^{2+} (a q) + {2e}^{−} ⟶ Ni (s)$	−0.257
${Co}^{2+} (a q) + {2e}^{−} ⟶ Co (s)$	−0.28
${PbSO}_{4} (s) + {2e}^{−} ⟶ Pb (s) + {SO}_{4}^{2−} (a q)$	−0.3505
${Cd}^{2+} (a q) + {2e}^{−} ⟶ Cd (s)$	−0.4030
${Fe}^{2+} (a q) + {2e}^{−} ⟶ Fe (s)$	−0.447
${Cr}^{3+} (a q) + {3e}^{−} ⟶ Cr (s)$	−0.744
${Mn}^{2+} (a q) + {2e}^{−} ⟶ Mn (s)$	−1.185
${Zn(OH)}_{2} (s) + {2e}^{−} ⟶ Zn (s) + {2OH}^{−} (a q)$	−1.245
${Zn}^{2+} (a q) + {2e}^{−} ⟶ Zn (s)$	−0.7618
${Al}^{3+} (a q) + {3e}^{−} ⟶ Al (s)$	−1.662
${Mg}^{2} (a q) + {2e}^{−} ⟶ Mg (s)$	−2.372
${Na}^{+} (a q) + e^{−} ⟶ Na (s)$	−2.71
${Ca}^{2+} (a q) + {2e}^{−} ⟶ Ca (s)$	−2.868
${Ba}^{2+} (a q) + {2e}^{−} ⟶ Ba (s)$	−2.912
$K^{+} (a q) + e^{−} ⟶ K (s)$	−2.931
${Li}^{+} (a q) + e^{−} ⟶ Li (s)$	−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 Au ³⁺ /Au and Ni ²⁺ /Ni half-cells? Identify the oxidizing and reducing agents.

Solution

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

{Au}^{3+} (a q) + 3 e^{−} ⟶ Au (s) E_{{Au}^{3+} /Au}^{°} = +1.498 V

{Ni}^{2+} (a q) + 2 e^{−} ⟶ Ni (s) E_{{Ni}^{2+} /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:

\begin{array}{l} Anode (oxidation): Ni (s) ⟶ {Ni}^{2+} (a q) + {2e}^{−} E_{anode}^{°} = E_{{Ni}^{2+} /Ni}^{°} = −0.257 V \\ {Cathode (reduction): Au}^{3+} (a q) + {3e}^{−} ⟶ Au (s) E_{cathode}^{°} = E_{{Au}^{3+} /Au}^{°} = +1.498 V \end{array}

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

3Ni (s) + {2Au}^{3+} (a q) ⟶ {3Ni}^{2+} (a q) + 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.

E_{cell}^{°} = E_{cathode}^{°} - E_{anode}^{°} = 1.498 V - (−0.2 57 V) = 1.7 55 V

From the half-reactions, Ni is oxidized, so it is the reducing agent, and Au ³⁺ is reduced, so it is the oxidizing agent.

Check your learning

A galvanic cell consists of a Mg electrode in 1 M Mg(NO ₃ ) ₂ solution and a Ag electrode in 1 M AgNO ₃ solution. Calculate the standard cell potential at 25 °C.

Answer:

$Mg (s) + 2 {Ag}^{+} (a q) ⟶ {Mg}^{2+} (a q) + 2 Ag (s) E_{cell}^{°} = 0.7 996 V - (−2.3 72 V) = 3.17 2 V$

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Key concepts and summary

Assigning the potential of the standard hydrogen electrode (SHE) as zero volts allows the determination of standard reduction potentials, E° , 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

$E_{cell}^{°} = E_{cathode}^{°} - E_{anode}^{°}$

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) $Mg (s) + {Ni}^{2+} (a q) ⟶ {Mg}^{2+} (a q) + Ni (s)$

(b) $2 {Ag}^{+} (a q) + Cu (s) ⟶ {Cu}^{2+} (a q) + 2Ag (s)$

(d) $3 {Fe(NO}_{3})_{2} (a q) + {Au(NO}_{3})_{3} (a q) ⟶ {3Fe(NO}_{3})_{3} (a q) + Au (s)$

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

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

(a) $Mn (s) + {Ni}^{2+} (a q) ⟶ {Mn}^{2+} (a q) + Ni (s)$

(b) $3 {Cu}^{2+} (a q) + 2Al (s) ⟶ {2Al}^{3+} (a q) + 2Cu (s)$

(d) ${Ca(NO}_{3})_{2} (a q) + Ba (s) ⟶ {Ba(NO}_{3})_{2} (a q) + Ca (s)$

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

$Cu (s) │ {Cu}^{2+} (a q) ║ {Au}^{3+} (a q) │ Au (s)$

$3 Cu (s) + {2Au}^{3+} (a q) ⟶ {3Cu}^{2+} (a q) + 2Au (s);$ +1.16 V; spontaneous

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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?

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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?

$3 Cd (s) + {2Al}^{3+} (a q) ⟶ {3Cd}^{2+} (a q) + 2Al (s);$ −1.259 V; nonspontaneous

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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 Br ₂ ( l ) is the same as for Br ₂ ( aq ).
$Pt (s) │ H_{2} (g) │ H^{+} (a q) ║ {Br}_{2} (a q) │ {Br}^{−} (a q) │ Pt (s)$

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Source: OpenStax, Ut austin - principles of chemistry. OpenStax CNX. Mar 31, 2016 Download for free at http://legacy.cnx.org/content/col11830/1.13

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