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Wize University Chemistry Textbook > Electrochemistry

Reduction Potentials

Balancing Redox ReactionsPrevious Section
Galvanic Cells Next 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)

Watch Out!
DO NOT assume you are given REDUCTION potentials in the table!
Here we know they are reduction potentials because we see electrons being added on the left sides of the equations.
Your prof may try to trick you and show you oxidation potentials instead, where electrons would be seen on the right sides of the equations.

Another thing we have to know is that we may need to multiply half reactions by a coefficient (later on in problems). When we do this NEVER multiply the potential by a coefficient.
For example in the table we have:
Li+(aq) + e- --> Li(s) Eo=-3.05V

If I needed to multiply this half reaction by 2, I would get:
2Li+(aq) + 2e- --> 2Li(s) Eo=-3.05V

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How Potentials Relate to Spontaneity


  • The overall cell potential will determine the spontaneity of redox reactions:
Ξ”Go=βˆ’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: Ξ”Go=βˆ’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+β†’Ca(s)+Cu(aq)2+Cu_{(s)}+Ca^{2+}_{(aq)} \to Ca_{(s)} +Cu^{2+} _{(aq)}
Standard reduction potentials
Ca(aq)2++2eβˆ’β†’Ca(s)Eo=βˆ’2.868VCa^{2+}_{(aq)} +2e^- \to Ca_{(s)} \hspace{20pt} E^o=-2.868V
Cu(aq)2++2eβˆ’β†’Cu(s)Eo=βˆ’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?

ReactionEΒ°(V)2HCβ„“O(aq)+2H+(aq)+2eβˆ’β†’Cβ„“2(g)+2H2O(β„“)+1.63Cβ„“2(g)+2eβˆ’β†’2Cβ„“βˆ’(aq)+1.36Br2(β„“)+2eβˆ’β†’2Brβˆ’(aq)+1.08I2(s)+2eβˆ’β†’2Iβˆ’(aq)+0.535Ag+(aq)+eβˆ’β†’Ag(s)+0.799Sn4+(aq)+2eβˆ’β†’Sn2+(aq)+0.15\def\arraystretch{2} \begin{array}{c|c} \hline \rm Reaction & EΒ° (V) \\ \hline 2 HCβ„“O (aq) + 2 H^+ (aq) + 2 e^βˆ’ β†’ Cβ„“_2 (g) + 2 H_2O (β„“) & + 1.63 \\ \hline Cβ„“_2(g) + 2 e^βˆ’ β†’ 2 Cβ„“^βˆ’ (aq) &+ 1.36\\ \hline Br_2(β„“) + 2 e^βˆ’ β†’ 2 Br^βˆ’ (aq) & + 1.08\\ \hline I_2 (s) + 2 e^βˆ’ β†’ 2 I^βˆ’ (aq) & + 0.535\\ \hline Ag+ (aq) + e^βˆ’ β†’ Ag (s) &+ 0.799\\ \hline Sn^{4+} (aq) + 2 e^βˆ’ β†’ Sn^{2+} (aq) &+ 0.15\\ \hline \end{array}

Extra Practice
Electrochemistry: Galvanic Cells
The cell pictured below is prepared under standard conditions. In this cell, what will be the anode?
𝐢𝑒2+(π‘Žπ‘ž)+𝐻2(𝑔)β‡ŒπΆπ‘’(𝑠)+2𝐻+(π‘Žπ‘ž)𝐸°=0.339𝑉𝐢𝑒^{2+} (π‘Žπ‘ž)+𝐻_2 (𝑔)\rightleftharpoons 𝐢𝑒 (𝑠)+2 𝐻^+ (π‘Žπ‘ž) \qquad𝐸°=0.339 𝑉
Which one of the species above is the best oxidizing agent?
Zn2+(aq)+2eβˆ’β‡ŒZn(s)Eo=βˆ’0.74VCu2+(aq)+2eβˆ’β‡ŒCu(s)Eo=+0.34VCo2+(aq)+2eβˆ’β‡ŒCo(s)Eo=βˆ’0.28VAg+(aq)+eβˆ’β‡ŒAg(s)Eo=+0.80V\def\arraystretch{1.5}\begin{array}{cc}Zn^{2+}(aq)+2e^-β‡ŒZn(s)&&E^o=-0.74V\\ Cu^{2+}(aq)+2e^-β‡ŒCu(s)&&E^o=+0.34V\\ Co^{2+}(aq)+2e^-β‡ŒCo(s)&&E^o=-0.28V\\ Ag^+(aq)+e^-β‡ŒAg(s)&&E^o=+0.80V\end{array}
A concentration cell consists of two H2/H+ half-cells. Half-cell A has H2 at 0.90 atm bubbling into 0.10 M HCl. Half-cell B has H2 atm bubbling into 2.0 M HCl. Which half-cell houses the anode? What is the voltage of this cell?
Which one of the species below is the best reducing agent?
Use the data below to answer the question:
Zn2+(aq)+2eβˆ’β‡ŒZn(s)                         Eo=βˆ’0.74VZn^{2+}(aq)+2e^-β‡ŒZn(s)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ E^o=-0.74V
Cu2+(aq)+2eβˆ’β‡ŒCu(s)                         Eo=+0.34VCu^{2+}(aq)+2e^-β‡ŒCu(s)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ E^o=+0.34V