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Chemical transformations · Oxidation and reduction

Galvanic Cells, Standard Potentials and Thermodynamic Favorability

Drop a strip of zinc into a blue copper(II) sulfate solution and a reddish coating of copper forms on the zinc while the beaker warms up. All that energy is released as heat. What if you forced the same electrons to travel through a wire instead? That is exactly what a battery does, and this page shows you how to predict and calculate it.

The galvanic cell: anode, cathode and cell potential

In the beaker, zinc hands electrons directly to Cu²⁺ ions, and nothing useful happens except heating. Separate the two half-reactions into two containers, connect them with a wire, and the electrons must travel through the external circuit. That current can power a device.

Practice★☆☆Not started
You have a Daniell galvanic cell made up of a zinc electrode immersed in a $1.0\text{ M}$ solution of $\mathrm{ZnSO_4}$ and a copper electrode immersed in a…

You have a Daniell galvanic cell made up of a zinc electrode immersed in a $1.0\text{ M}$ solution of $\mathrm{ZnSO_4}$ and a copper electrode immersed in a $1.0\text{ M}$ solution of $\mathrm{CuSO_4}$, at $25\text{ }^\circ\mathrm{C}$. The standard reduction potentials are $E^\circ(\mathrm{Zn}^{2+}/\mathrm{Zn}) = -0.76\text{ V}$ and $E^\circ(\mathrm{Cu}^{2+}/\mathrm{Cu}) = +0.34\text{ V}$.

  1. a) Write the conventional cell notation or diagram, indicating the anode and the cathode.
  2. b) State the sign or polarity assigned to each electrode and briefly justify which half-reaction takes place at each.
Solve

Cell notation and calculating E°cell

Chemists write a cell as a short line diagram instead of a drawing.

Cell notation for the zinc-copper cell, labeling anode, salt bridge, cathode and the two half-cells. Zn | Zn²⁺ ‖ Cu²⁺ | Cu anode oxidation salt bridge cathode reduction Zn half-cell Cu half-cell
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Consider a galvanic cell under standard conditions ($25\text{ }^\circ\mathrm{C}$, concentrations $1.0\text{ M}$) based on the overall reaction:…

Consider a galvanic cell under standard conditions ($25\text{ }^\circ\mathrm{C}$, concentrations $1.0\text{ M}$) based on the overall reaction: $\mathrm{Cu(s)} + 2\,\mathrm{Ag^+}(\text{aq}) \rightarrow \mathrm{Cu}^{2+}(\text{aq}) + 2\,\mathrm{Ag(s)}$. Data: $E^\circ(\mathrm{Ag}^+/\mathrm{Ag}) = +0.80\text{ V}$ and $E^\circ(\mathrm{Cu}^{2+}/\mathrm{Cu}) = +0.34\text{ V}$.

  1. a) Calculate the standard electromotive force of the cell, $E^\circ_{\text{pila}}$.
  2. b) Justify in detail why the standard reduction potential of the $\mathrm{Ag}^+/\mathrm{Ag}$ couple must not be multiplied by the stoichiometric coefficient 2 when calculating the emf of the cell.
Solve

Free energy, K and thermodynamic favorability

Axis of standard cell potential: negative E° means unfavorable (ΔG° positive), positive E° means favorable (ΔG° negative). unfavorable ΔG° > 0 favorable ΔG° < 0 E° < 0 E° > 0 0

The same free energy also connects to the equilibrium constant. Combining $\Delta G^\circ = -RT\ln K$ with $\Delta G^\circ = -nFE^\circ$ gives

$$E^\circ_{\text{cell}} = \frac{RT}{nF}\ln K$$

so a large positive $E^\circ$ means a large $K$: the reaction goes nearly to completion. A cell with $E^\circ = 0$ has $K = 1$.

Practice★☆☆Not started
For a cell made up of the couples $\mathrm{Ni}^{2+}/\mathrm{Ni}$ and $\mathrm{Cd}^{2+}/\mathrm{Cd}$ at $25\text{ }^\circ\mathrm{C}$, the overall spontaneous reaction…

For a cell made up of the couples $\mathrm{Ni}^{2+}/\mathrm{Ni}$ and $\mathrm{Cd}^{2+}/\mathrm{Cd}$ at $25\text{ }^\circ\mathrm{C}$, the overall spontaneous reaction under standard conditions is $\mathrm{Cd(s)} + \mathrm{Ni}^{2+}(\text{aq}) \rightarrow \mathrm{Cd}^{2+}(\text{aq}) + \mathrm{Ni(s)}$, the standard electromotive force being $E^\circ_{\text{pila}} = +0.15\text{ V}$. Data: Faraday constant $F = 96485\text{ C}\times\text{mol}^{-1}$.

  1. a) Determine the value of the standard Gibbs energy change, $\Delta G^\circ$, in $\text{kJ}\times\text{mol}^{-1}$.
  2. b) Deduce the criterion for spontaneity from the sign of $\Delta G^\circ$ and $E^\circ_{\text{pila}}$.
Solve
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The feasibility of a cell based on the standard couples $\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}$ and $\mathrm{I_2}/\mathrm{I^-}$ at $25\text{ }^\circ\mathrm{C}$ is analyzed,…

The feasibility of a cell based on the standard couples $\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}$ and $\mathrm{I_2}/\mathrm{I^-}$ at $25\text{ }^\circ\mathrm{C}$ is analyzed, with reactants under standard conditions. Data: $E^\circ(\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}) = +0.77\text{ V}$; $E^\circ(\mathrm{I_2}/\mathrm{I^-}) = +0.54\text{ V}$; $F = 96485\text{ C}\times\text{mol}^{-1}$.

  1. a) Write the balanced equation of the spontaneous redox reaction under standard conditions, identify the oxidizing and reducing species, and calculate $E^\circ_{\text{pila}}$ and $\Delta G^\circ$.
  2. b) A student claims that, because $\Delta G^\circ < 0$, the reaction will always be spontaneous at any concentration and in any experimental context. Justify, with reasoning, why this claim is false, analyzing the difference between $\Delta G^\circ$ and $\Delta G$.
Solve

The salt bridge and the direction of current

Connect zinc and copper electrodes in two separate beakers with a wire and a voltmeter. The needle jumps and drops to zero almost at once. Add a tube of salt solution between the beakers and the voltmeter reads about 1.10 V when both solutions are 1.0 M. Why does the current stop without the bridge?

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In the laboratory, a Daniell galvanic cell is set up at $25\text{ }^\circ\mathrm{C}$ with metallic zinc and copper electrodes immersed in $1.0\text{ M}$ aqueous…

In the laboratory, a Daniell galvanic cell is set up at $25\text{ }^\circ\mathrm{C}$ with metallic zinc and copper electrodes immersed in $1.0\text{ M}$ aqueous solutions of their respective cations, connected by a salt bridge of $\mathrm{KNO_3}$. The diagram of the experimental setup is shown below:

V e⁻ ? Zn(s) Cu(s) Zn²⁺ (1.0 M) Cu²⁺ (1.0 M) Salt bridge (KNO₃)

Data: $E^\circ(\mathrm{Zn}^{2+}/\mathrm{Zn}) = -0.76\text{ V}$; $E^\circ(\mathrm{Cu}^{2+}/\mathrm{Cu}) = +0.34\text{ V}$.

  1. a) Determine the standard electromotive force of the cell and state the correct direction of electron flow through the external metallic conductor.
  2. b) Explain which species carries the charge through the salt bridge and toward which beaker the $\mathrm{K^+}$ and $\mathrm{NO_3^-}$ ions migrate respectively, stating why electrons do not flow through the solution.
Solve
Practice★★☆Not started
A galvanic cell is built by connecting a half-cell with a chromium rod immersed in $\mathrm{Cr(NO_3)3}\text{ }1.0\text{ M}$ and another with a lead rod in…

A galvanic cell is built by connecting a half-cell with a chromium rod immersed in $\mathrm{Cr(NO_3)_3}\text{ }1.0\text{ M}$ and another with a lead rod in $\mathrm{Pb(NO_3)_2}\text{ }1.0\text{ M}$, at $25\text{ }^\circ\mathrm{C}$, using a salt bridge saturated with $\mathrm{NH_4NO_3}$. Data: $E^\circ(\mathrm{Cr}^{3+}/\mathrm{Cr}) = -0.74\text{ V}$, $E^\circ(\mathrm{Pb}^{2+}/\mathrm{Pb}) = -0.13\text{ V}$, $F = 96485\text{ C}\times\text{mol}^{-1}$.

  1. a) Write the anode and cathode half-equations and the balanced overall ionic equation, and calculate the standard potential of the cell.
  2. b) Explain how electroneutrality is maintained in each half-cell, specifying the movement of the salt bridge ions, and calculate the value of $\Delta G^\circ$ for the overall reaction.
Solve

Predicting reactions with the reduction-potential table

Zinc turns hydrochloric acid into a fizzing cloud of hydrogen, but a copper penny in the same acid does nothing. What measurable property tells you which metals can give electrons to H⁺ ions?

Vertical scale of standard reduction potentials from highest at the top to lowest at the bottom. E° high MnO₄⁻/Mn²⁺ +1.51 V Cl₂/Cl⁻ +1.36 V Fe³⁺/Fe²⁺ +0.77 V Zn²⁺/Zn −0.76 V low
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You have aqueous solutions containing chloride ($\mathrm{Cl^-}$), bromide ($\mathrm{Br^-}$) and iodide ($\mathrm{I^-}$) ions, all at concentration $1.0\text{ M}$ and at…

You have aqueous solutions containing chloride ($\mathrm{Cl^-}$), bromide ($\mathrm{Br^-}$) and iodide ($\mathrm{I^-}$) ions, all at concentration $1.0\text{ M}$ and at $25\text{ }^\circ\mathrm{C}$. You want to select an oxidizing agent in its standard state that is able to spontaneously oxidize the iodide ion to molecular iodine ($\mathrm{I_2}$), and the bromide ion to liquid bromine ($\mathrm{Br_2}$), but that does not oxidize the chloride ion to chlorine gas ($\mathrm{Cl_2}$).

Standard reduction potential data at $25\text{ }^\circ\mathrm{C}$:

$E^\circ(\mathrm{I_2}/\mathrm{I^-}) = +0.54\text{ V}$; $E^\circ(\mathrm{Br_2}/\mathrm{Br^-}) = +1.07\text{ V}$; $E^\circ(\mathrm{Cl_2}/\mathrm{Cl^-}) = +1.36\text{ V}$.

Oxidizing agent candidates: $\mathrm{Fe}^{3+}(\text{aq})$ with $E^\circ(\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}) = +0.77\text{ V}$; $\mathrm{IO_3^-}(\text{aq})/\mathrm{I_2(s)}$ in acidic medium with $E^\circ = +1.20\text{ V}$; $\mathrm{MnO_4^-}(\text{aq})/\mathrm{Mn}^{2+}(\text{aq})$ in acidic medium with $E^\circ = +1.51\text{ V}$.

  1. a) Select the most suitable oxidizing agent, justifying your answer by comparing the potentials of the candidates and calculating $E^\circ$ and $\Delta G^\circ$ for the reaction of the chosen oxidizing agent with each halide, using the lowest whole-number coefficients.
  2. b) Write the balanced overall reaction that takes place between the selected oxidizing agent and the bromide ion in acidic medium. Data: $F = 96485\text{ C}\times\text{mol}^{-1}$.
Solve
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A metal sheet containing a mixture of cadmium, silver and iron is placed in solutions of acids and salts at $25\text{ }^\circ\mathrm{C}$ to evaluate displacement…

A metal sheet containing a mixture of cadmium, silver and iron is placed in solutions of acids and salts at $25\text{ }^\circ\mathrm{C}$ to evaluate displacement reactions.

Data: $E^\circ(\mathrm{Ag}^+/\mathrm{Ag}) = +0.80\text{ V}$; $E^\circ(\mathrm{H}^+/\mathrm{H_2}) = 0.00\text{ V}$; $E^\circ(\mathrm{Pb}^{2+}/\mathrm{Pb}) = -0.13\text{ V}$; $E^\circ(\mathrm{Cd}^{2+}/\mathrm{Cd}) = -0.40\text{ V}$; $E^\circ(\mathrm{Fe}^{2+}/\mathrm{Fe}) = -0.44\text{ V}$. Constant $F = 96485\text{ C}\times\text{mol}^{-1}$.

  1. a) Determine which metals in the sheet will dissolve spontaneously in an aqueous $1.0\text{ M}$ solution of hydrochloric acid ($\mathrm{HCl}$), releasing hydrogen gas under standard conditions, and write the corresponding reactions.
  2. b) If a piece of pure iron metal is immersed in a solution containing $\mathrm{Pb}^{2+}$ and $\mathrm{Cd}^{2+}$ ions, both at standard concentration $1.0\text{ M}$, reason quantitatively whether the iron will displace one, both or neither of these cations from the solution, calculating the standard emf and $\Delta G^\circ$ of the processes that are spontaneous. Use the thermodynamic model of standard potentials, without dissolved oxygen, complexation or passivation; the prediction does not give information about the actual rate.
Solve
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The stability of the peroxide and the oxidation of the $\mathrm{Sn}^{2+}$ cation in acidic medium are studied at $25\text{ }^\circ\mathrm{C}$. Two reagents are…

The stability of the peroxide and the oxidation of the $\mathrm{Sn}^{2+}$ cation in acidic medium are studied at $25\text{ }^\circ\mathrm{C}$. Two reagents are available: potassium permanganate ($\mathrm{KMnO_4}$) and hydrogen peroxide ($\mathrm{H_2O_2}$). The standard reduction potentials are known:

$E^\circ(\mathrm{Sn}^{4+}/\mathrm{Sn}^{2+}) = +0.15\text{ V}$; $E^\circ(\mathrm{MnO_4^-}/\mathrm{Mn}^{2+}) = +1.51\text{ V}$; $E^\circ(\mathrm{O_2}/\mathrm{H_2O_2}) = +0.68\text{ V}$; $E^\circ(\mathrm{H_2O_2}/\mathrm{H_2O}) = +1.78\text{ V}$. Data: $F = 96485\text{ C}\times\text{mol}^{-1}$.

  1. a) Show, with reasoning, using the values of $E^\circ$ and $\Delta G^\circ$, whether $\mathrm{H_2O_2}$ is thermodynamically stable with respect to disproportionation into water and oxygen under standard conditions at $25\text{ }^\circ\mathrm{C}$.
  2. b) Justify whether the $\mathrm{Sn}^{2+}$ ion can be oxidized to $\mathrm{Sn}^{4+}$ by $\mathrm{MnO_4^-}$ in acidic medium, and determine the minimum volume, in $\text{mL}$, of $\mathrm{KMnO_4}\text{ }0.020\text{ M}$ solution needed to completely oxidize $25.0\text{ mL}$ of $\mathrm{Sn}^{2+}\text{ }0.050\text{ M}$. For the titration in b), assume quantitative reduction of permanganate to manganese(II), excess sulfuric acid and no side reactions.
Solve

Beyond standard conditions: the Nernst equation

Real cells are rarely at 1.0 M. As a battery runs, reactants are used up and products build up, so $Q$ changes and the voltage drops. The Nernst equation captures this:

$$E = E^\circ - \frac{RT}{nF}\ln Q \qquad\text{or at 25 °C:}\qquad E = E^\circ - \frac{0.0257}{n}\ln Q$$

Here $Q$ is the reaction quotient written like $K$, with solids and pure liquids left out. When $Q < 1$ (lots of reactants), $E > E^\circ$. When $Q = K$, $E = 0$: the cell is "dead," at equilibrium.

Back to the beginning: when you separate the zinc and copper half-reactions with an external wire and a salt bridge, the electrons are forced through the circuit, and the cell delivers $E^\circ_{\text{cell}} = 0.34 - (-0.76) = +1.10\ \mathrm{V}$ of useful electrical energy instead of just heat.

Summary

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