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Wize High School Grade 12 Chemistry Textbook > Electrochemistry

Reduction Potentials

Balancing Redox Reactions: Oxidation State MethodPrevious Section
Galvanic CellsNext Section
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Reduction Potentials

  • Each half reaction has a potential.
  • For example, the first half reaction below has a reduction potential of -3.05V!
  • We got these potentials from measuring the voltage when the other half reaction was the reference electrode (H 0V)
  • In the table below, are the reactions showing reductions or oxidations?
    reductions
  • Therefore, these are
    reduction
    potentials!
  • If we flip a reaction, the Eo potential changes signs
  • let's do this for the first half reaction in the table so it is clear:
Li(s) --> Li+(aq) + e- Eo=3.05
  • Is this reaction now describing reduction or oxidation?
    oxidation
    !
  • So do you think the potential written would now be a reduction or oxidation potential?
    oxidation
    potential
  • In general:
  • If we see a more positive reduction potential:
  • It means something is more likely to be oxidized/reduced:
    reduced
  • and will act as a(n) reducing/oxidizing agent:
    oxidizing
    agent
  • If we see a more negative reduction potential:
  • It means something is more likely to be oxidized/reduced:
    oxidized
  • and will act as a(n) reducing/oxidizing agent:
    reducing
    agent

Based on this table, answer the following questions:

1) What is the strongest oxidizing agent?
F2(g)

2) What is the weakest reducing agent?
F-(aq)

3) What is the strongest reducing agent?
Li(s)


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Reduction Potentials Application

  • Below we see a diagram showing mitochondrial cellular respiration.
  • If we just consider the electron carrier NADH:
  • NADH molecules gives up their electrons to the first complex, then the electron goes to the 2nd complex, then the 3rd, then the 4th and finally oxygen is the final electron acceptor.
  • Order NADH, complex 1, complex 2, complex 3, complex 4, and oxygen in order of increasing reduction potential:
  • First think: does increasing reduction potential means that it wants electrons more or less?
    more
  • Order:
  • Which one is the strongest reducing agent?
    NADH
  • Which one is the strongest oxidizing agent?
    O2
  • Which one would have the highest reduction potential?
    O2
  • Which one would have the highest oxidation potential?
    NADH


  • NADH < complex 1 < complex 2 < complex 3 < complex 4 < O2
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How Potentials Relate to Spontaneity


  • The overall cell potential will determine the spontaneity of redox reactions:
ΔGo=−nFEredoxo\Delta G^o=-nFE_{redox}^o
Eoredox=the potential associated with the redox reaction
n=# of electrons
F=Faraday's constant (96 500C/mol e-)
deltaGo=Gibbs free energy


  • According to this equation, when Eoredox is positive....deltaGo is positive or negative:
    negative
  • This means the reaction would be spontaneous/non-spontaneous:
    spontaneous
  • When this is the case we have a galvanic cell (more on this soon!)
  • K
    >
    1
  • When Eoredox is negative...deltaGo is positive or negative:
    positive
  • This means the reaction would be spontaneous/non-spontaneous:
    non-spontaneous
  • When this is the case we have an electrolytic cell (more on this soon!)
  • K
    <
    1
Apply: If we had the following overall reaction, would it be spontaneous or not?
Zn(s) + Cu2+(aq) --> Zn2+(aq) + Cu(s)

1) Let's write out the oxidation and reduction half reactions:
a) Oxidation half reaction:
Zn(s) --> Zn2+(aq) + 2e-

b) Reduction half reaction:
Cu2+(aq) + 2e- --> Cu(s)

2) Given the above reactions that are occuring, write out an oxidation potential and a reduction potential using the table values below:

From the table:
Zn2+ (aq) + 2e- -> Zn(s) Eo=-0.76V
Cu2+(aq) + 2e- -> Cu(s) Eo=+0.34V

For us,
Zn(s)
is being oxidized, so Eoox=
+0.76
(changes sign)
And
Cu2+(aq)
is being reduced, so Eored=
+0.34V
(stays the same)

3) Solve for Eoredox:

Eredoxo=Eoxo + EredoE_{redox}^o=E_{ox}^o\ +\ E_{red}^o

Eoredox= 0.76 + 0.34
Eoredox=1.10V

4) Figure out if the reaction is spontaneous using this equation: ΔGo=−nFEredoxo\Delta G^o=-nFE_{redox}^o
  • Our Eoredox value is positive/negative:
    positive
  • This means our deltaGo value is positive/negative:
    negative
  • Therefore, the reaction is spontaneous/non-spontaneous:
    spontaneous

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Will the following reaction proceed spontaneously or will current need to be added externally?

Cu(s)+Ca(aq)2+→Ca(s)+Cu(aq)2+Cu_{(s)}+Ca^{2+}_{(aq)} \to Ca_{(s)} +Cu^{2+} _{(aq)}
Standard reduction potentials
Ca(aq)2++2e−→Ca(s)Eo=−2.868VCa^{2+}_{(aq)} +2e^- \to Ca_{(s)} \hspace{20pt} E^o=-2.868V
Cu(aq)2++2e−→Cu(s)Eo=−0.340VCu^{2+}_{(aq)} +2e^- \to Cu_{(s)} \hspace{20pt} E^o=-0.340V

  • For reaction 1: -2.868V stays the same since that equation is not flipped.
  • For reaction 2: we get 0.340V since that equation is flipped according to the overall equation.
  • For the Erxn=-2.868+0.340=-2.528 (doing the full calculation isn't necessary since we just need to figure out the sign of E to tell us about the sign of G)
  • We know E is negative now, so deltaG would be positive.
  • The electromotive force (E) is negative so this reaction would be non-spontaneous. We would have to apply at least 2.528V to force this reaction to proceed. (Current needs to be added externally)

Based on the following reduction potentials, which of these species is the strongest reducing agent?

ReactionE°(V)2HCℓO(aq)+2H+(aq)+2e−→Cℓ2(g)+2H2O(ℓ)+1.63Cℓ2(g)+2e−→2Cℓ−(aq)+1.36Br2(ℓ)+2e−→2Br−(aq)+1.08I2(s)+2e−→2I−(aq)+0.535Ag+(aq)+e−→Ag(s)+0.799Sn4+(aq)+2e−→Sn2+(aq)+0.15\def\arraystretch{2} \begin{array}{c|c} \hline \rm Reaction & E° (V) \\ \hline 2 HCℓO (aq) + 2 H^+ (aq) + 2 e^− → Cℓ_2 (g) + 2 H_2O (ℓ) & + 1.63 \\ \hline Cℓ_2(g) + 2 e^− → 2 Cℓ^− (aq) &+ 1.36\\ \hline Br_2(ℓ) + 2 e^− → 2 Br^− (aq) & + 1.08\\ \hline I_2 (s) + 2 e^− → 2 I^− (aq) & + 0.535\\ \hline Ag+ (aq) + e^− → Ag (s) &+ 0.799\\ \hline Sn^{4+} (aq) + 2 e^− → Sn^{2+} (aq) &+ 0.15\\ \hline \end{array}

Extra Practice
Electrochemistry: Reduction potential application
Reduction Potentials Application
Below we see a diagram showing mitochondrial cellular respiration.
If we just consider the electron carrier NADH, NADH molecules gives up their electrons to the first complex, then the electron goes to the 2nd complex, then the 3rd, then the 4th and finally oxygen (O2) is the final electron acceptor.
Electrochemistry: Reduction potential application
Reduction Potentials Application
Below we see a diagram showing mitochondrial cellular respiration.
If we just consider the electron carrier NADH, NADH molecules gives up their electrons to the first complex, then the electron goes to the 2nd complex, then the 3rd, then the 4th and finally oxygen (O2) is the final electron acceptor.
Electrochemistry: Reduction Potentials
Based on the following reduction potentials, which of these species is the strongest reducing agent?
ReactionE°(V)2HCℓO(aq)+2H+(aq)+2e−→Cℓ2(g)+2H2O(ℓ)+1.63Cℓ2(g)+2e−→2Cℓ−(aq)+1.36Br2(ℓ)+2e−→2Br−(aq)+1.08I2(s)+2e−→2I−(aq)+0.535Ag+(aq)+e−→Ag(s)+0.799Sn4+(aq)+2e−→Sn2+(aq)+0.15\def\arraystretch{2} \begin{array}{c|c} \hline \rm Reaction & E° (V) \\ \hline 2 HCℓO (aq) + 2 H^+ (aq) + 2 e^− → Cℓ_2 (g) + 2 H_2O (ℓ) & + 1.63 \\ \hline Cℓ_2(g) + 2 e^− → 2 Cℓ^− (aq) &+ 1.36\\ \hline Br_2(ℓ) + 2 e^− → 2 Br^− (aq) & + 1.08\\ \hline I_2 (s) + 2 e^− → 2 I^− (aq) & + 0.535\\ \hline Ag+ (aq) + e^− → Ag (s) &+ 0.799\\ \hline Sn^{4+} (aq) + 2 e^− → Sn^{2+} (aq) &+ 0.15\\ \hline \end{array}
Electrochemistry: Reduction potential application
Reduction Potentials Application
Below we see a diagram showing mitochondrial cellular respiration.
If we just consider the electron carrier NADH, NADH molecules gives up their electrons to the first complex, then the electron goes to the 2nd complex, then the 3rd, then the 4th and finally oxygen (O2) is the final electron acceptor.
Electrochemical Equilibrium and Cells
Will the following reaction proceed spontaneously or will current need to be added externally?
Cu(s)+Ca(aq)2+→Ca(s)+Cu(aq)2+Cu_{(s)}+Ca^{2+}_{(aq)} \to Ca_{(s)} +Cu^{2+} _{(aq)}
In basic solution Se2- and SO32- ions react spontaneously:
2Se(aq)2−+2SO3(aq)2−+3H2O(l)→2Se(s)+6OH(aq)−+S2O(aq)2−,Ecello=0.35V2Se_{(aq)}^{2-}+2SO_{3(aq)}^{2-}+3H_2O_{(l)} \to 2Se_{(s)}+6OH_{(aq)}^-+S_2O^{2-}_{(aq)}, E^o_{cell}=0.35V
Write balanced half reactions for this process
(Duplicated)
Use the standard reduction potentials below to answer the following question: which species listed is the strongest oxidizing agent?
Reduction half−reactionE°(V)𝐶𝑎2+(𝑎𝑞)+2𝑒−⇌𝐶𝑎(𝑠)−2.87𝑁𝑎+(𝑎𝑞)+𝑒−⇌𝑁𝑎(𝑠)−2.71𝐴𝑙3+(𝑎𝑞)+3𝑒−⇌𝐴𝑙(𝑠)−1.66𝑀𝑛2+(𝑎𝑞)+2𝑒−⇌𝑀𝑛(𝑠)−1.18𝐶𝑢2+(𝑎𝑞)+2𝑒−⇌𝐶𝑢(𝑠)+0.15𝐴𝑔+(𝑎𝑞)+𝑒−⇌𝐴𝑔(𝑠)+0.80\begin{array}{c:c}\rm Reduction\ half-reaction& E° (V)\\ \hline 𝐶𝑎^{2+} (𝑎𝑞)+2 𝑒^−⇌𝐶𝑎 (𝑠)& -2.87\\ \hline 𝑁𝑎^+ (𝑎𝑞)+𝑒^−⇌𝑁𝑎 (𝑠)&-2.71\\ \hline 𝐴𝑙^{3+} (𝑎𝑞)+3 𝑒^−⇌𝐴𝑙(𝑠) &-1.66\\ \hline 𝑀𝑛^{2+} (𝑎𝑞)+2 𝑒^−⇌𝑀𝑛 (𝑠)&-1.18\\ \hline 𝐶𝑢^{2+} (𝑎𝑞)+2 𝑒^− ⇌𝐶𝑢 (𝑠) &+0.15\\ \hline 𝐴𝑔^+ (𝑎𝑞)+𝑒^− ⇌𝐴𝑔 (𝑠)&+0.80\end{array}
Electrochemistry: Reduction potentials in a Galvanic Cell
Use the standard reduction potentials below to answer the following question: which metal could be used to create a sacrificial anode for an aluminum pipe?
Reduction half−reactionE°(V)𝐶𝑎2+(𝑎𝑞)+2𝑒−⇌𝐶𝑎(𝑠)−2.87𝑁𝑎+(𝑎𝑞)+𝑒−⇌𝑁𝑎(𝑠)−2.71𝐴𝑙3+(𝑎𝑞)+3𝑒−⇌𝐴𝑙(𝑠)−1.66𝑀𝑛2+(𝑎𝑞)+2𝑒−⇌𝑀𝑛(𝑠)−1.18𝐶𝑢2+(𝑎𝑞)+2𝑒−⇌𝐶𝑢(𝑠)+0.15𝐴𝑔+(𝑎𝑞)+𝑒−⇌𝐴𝑔(𝑠)+0.80\begin{array}{c:c}\rm Reduction\ half-reaction& E° (V)\\ \hline 𝐶𝑎^{2+} (𝑎𝑞)+2 𝑒^−⇌𝐶𝑎 (𝑠)& -2.87\\ \hline 𝑁𝑎^+ (𝑎𝑞)+𝑒^−⇌𝑁𝑎 (𝑠)&-2.71\\ \hline 𝐴𝑙^{3+} (𝑎𝑞)+3 𝑒^−⇌𝐴𝑙(𝑠) &-1.66\\ \hline 𝑀𝑛^{2+} (𝑎𝑞)+2 𝑒^−⇌𝑀𝑛 (𝑠)&-1.18\\ \hline 𝐶𝑢^{2+} (𝑎𝑞)+2 𝑒^− ⇌𝐶𝑢 (𝑠) &+0.15\\ \hline 𝐴𝑔^+ (𝑎𝑞)+𝑒^− ⇌𝐴𝑔 (𝑠)&+0.80\end{array}
Electrochemistry: Standard Reduction Potentials
A galvanic cell is constructed from chromium and copper electrodes immersed in aqueous solutions of Cr3+and Cu2+ respectively (half-reactions shown below).
Cr3+(aq)+3e−⇌Cr(s)Cr^{3+}(aq)+3e^-⇌Cr(s) E°red=−0.74VE°_{red}=-0.74V
Cu2+(aq)+2e−⇌Cu(s)Cu^{2+}(aq)+2e^-⇌Cu(s) E°red=+0.34VE°_{red}=+0.34V
Electrochemistry: Reduction potentials
Based on the data table, which of these species is the strongest reducing agent?
Based on the following reduction potentials, which of these species is the strongest reducing agent?
ReactionE°(V)2HCℓO(aq)+2H+(aq)+2e−→Cℓ2(g)+2H2O(ℓ)+1.63Cℓ2(g)+2e−→2Cℓ−(aq)+1.36Br2(ℓ)+2e−→2Br−(aq)+1.08I2(s)+2e−→2I−(aq)+0.535Ag+(aq)+e−→Ag(s)+0.799Sn4+(aq)+2e−→Sn2+(aq)+0.15\def\arraystretch{2} \begin{array}{c|c} \hline \rm Reaction & E° (V) \\ \hline 2 HCℓO (aq) + 2 H^+ (aq) + 2 e^− → Cℓ_2 (g) + 2 H_2O (ℓ) & + 1.63 \\ \hline Cℓ_2(g) + 2 e^− → 2 Cℓ^− (aq) &+ 1.36\\ \hline Br_2(ℓ) + 2 e^− → 2 Br^− (aq) & + 1.08\\ \hline I_2 (s) + 2 e^− → 2 I^− (aq) & + 0.535\\ \hline Ag+ (aq) + e^− → Ag (s) &+ 0.799\\ \hline Sn^{4+} (aq) + 2 e^− → Sn^{2+} (aq) &+ 0.15\\ \hline \end{array}
Electrochemistry: Oxidizing agents
Which one of the species above is the best oxidizing agent?
Zn2+(aq)+2e−⇌Zn(s)Eo=−0.74VCu2+(aq)+2e−⇌Cu(s)Eo=+0.34VCo2+(aq)+2e−⇌Co(s)Eo=−0.28VAg+(aq)+e−⇌Ag(s)Eo=+0.80V\def\arraystretch{1.5}\begin{array}{cc}Zn^{2+}(aq)+2e^-⇌Zn(s)&&E^o=-0.74V\\ Cu^{2+}(aq)+2e^-⇌Cu(s)&&E^o=+0.34V\\ Co^{2+}(aq)+2e^-⇌Co(s)&&E^o=-0.28V\\ Ag^+(aq)+e^-⇌Ag(s)&&E^o=+0.80V\end{array}
Electrochemistry: Galvanic Cells
The cell pictured below is prepared under standard conditions. In this cell, what will be the anode?
𝐶𝑢2+(𝑎𝑞)+𝐻2(𝑔)⇌𝐶𝑢(𝑠)+2𝐻+(𝑎𝑞)𝐸°=0.339𝑉𝐶𝑢^{2+} (𝑎𝑞)+𝐻_2 (𝑔)\rightleftharpoons 𝐶𝑢 (𝑠)+2 𝐻^+ (𝑎𝑞) \qquad𝐸°=0.339 𝑉
Use the standard reduction potentials below to answer the following question: which metal could be used as an anode for an aluminum pipe?
Reduction half−reactionE°(V)𝐶𝑎2+(𝑎𝑞)+2𝑒−⇌𝐶𝑎(𝑠)−2.87𝑁𝑎+(𝑎𝑞)+𝑒−⇌𝑁𝑎(𝑠)−2.71𝐴𝑙3+(𝑎𝑞)+3𝑒−⇌𝐴𝑙(𝑠)−1.66𝑀𝑛2+(𝑎𝑞)+2𝑒−⇌𝑀𝑛(𝑠)−1.18𝐶𝑢2+(𝑎𝑞)+2𝑒−⇌𝐶𝑢(𝑠)+0.15𝐴𝑔+(𝑎𝑞)+𝑒−⇌𝐴𝑔(𝑠)+0.80\begin{array}{c:c}\rm Reduction\ half-reaction& E° (V)\\ \hline 𝐶𝑎^{2+} (𝑎𝑞)+2 𝑒^−⇌𝐶𝑎 (𝑠)& -2.87\\ \hline 𝑁𝑎^+ (𝑎𝑞)+𝑒^−⇌𝑁𝑎 (𝑠)&-2.71\\ \hline 𝐴𝑙^{3+} (𝑎𝑞)+3 𝑒^−⇌𝐴𝑙(𝑠) &-1.66\\ \hline 𝑀𝑛^{2+} (𝑎𝑞)+2 𝑒^−⇌𝑀𝑛 (𝑠)&-1.18\\ \hline 𝐶𝑢^{2+} (𝑎𝑞)+2 𝑒^− ⇌𝐶𝑢 (𝑠) &+0.15\\ \hline 𝐴𝑔^+ (𝑎𝑞)+𝑒^− ⇌𝐴𝑔 (𝑠)&+0.80\end{array}
Which one of the species above is the best oxidizing agent?
Zn2+(aq)+2e−⇌Zn(s)Eo=−0.74VCu2+(aq)+2e−⇌Cu(s)Eo=+0.34VCo2+(aq)+2e−⇌Co(s)Eo=−0.28VAg+(aq)+e−⇌Ag(s)Eo=+0.80V\def\arraystretch{1.5}\begin{array}{cc}Zn^{2+}(aq)+2e^-⇌Zn(s)&&E^o=-0.74V\\ Cu^{2+}(aq)+2e^-⇌Cu(s)&&E^o=+0.34V\\ Co^{2+}(aq)+2e^-⇌Co(s)&&E^o=-0.28V\\ Ag^+(aq)+e^-⇌Ag(s)&&E^o=+0.80V\end{array}
The cell pictured is prepared at standard conditions and the cell voltage measured to be +0.339 V. Which of the following changes would result in an increase to the measured cell voltage?
𝐶𝑢2+(𝑎𝑞)+𝐻2(𝑔)⇌𝐶𝑢(𝑠)+2𝐻+(𝑎𝑞)𝐸°=0.339𝑉𝐶𝑢^{2+} (𝑎𝑞)+𝐻_2 (𝑔)\rightleftharpoons 𝐶𝑢 (𝑠)+2 𝐻^+ (𝑎𝑞) \qquad𝐸°=0.339 𝑉