Cells, potentials and spontaneity
Drop a strip of zinc into blue copper(II) sulfate solution in the school lab and the zinc soon becomes coated with a reddish-brown deposit of copper, the blue fades and the beaker warms up. All that energy is released as heat. Could we make those electrons travel along a wire instead, so that they power something, as they do in the battery of an e-scooter? That is exactly what a galvanic cell does.
The galvanic cell: anode, cathode and cell potential
In the zinc/copper(II) sulfate reaction the electrons pass directly from zinc atoms to copper ions. To make them do useful work, we keep the two half-reactions in separate beakers (half-cells) and join them with a wire and a salt bridge.
In the zinc–copper (Daniell) cell, zinc is oxidised at the anode: $\mathrm{Zn(s) \rightarrow Zn^{2+}(aq) + 2e^-}$. Copper(II) ions are reduced at the cathode: $\mathrm{Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)}$. The zinc electrode is the negative terminal (it is where electrons are released) and the copper electrode is the positive terminal.
In the lab you can investigate this yourself: place small pieces of different metals (for example Mg, Zn, Fe, Cu) in solutions of the salts of the others, note which combinations show a deposit or a colour change, and use your results to rank the metals by their tendency to be oxidised. That ranking is the electrochemical series.
In each half-cell the electrolyte is a solution containing ions of the electrode metal. It lets ions move so charge can be balanced, while the electrons travel only through the wire.
The salt bridge (for example a strip of filter paper soaked in potassium nitrate solution) lets ions move between the two half-cells without the solutions mixing. Anions move towards the anode and cations towards the cathode, so each half-cell stays electrically neutral and the circuit is complete. Without it the current stops almost at once, as you will see in the section on the salt bridge.
The same chemistry shows up in displacement reactions, where a more reactive metal displaces a less reactive one from solution. Zinc in copper(II) sulfate solution is coated with copper: $\mathrm{Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)}$. Copper in silver nitrate solution grows silver crystals: $\mathrm{Cu(s) + 2Ag^+(aq) \rightarrow Cu^{2+}(aq) + 2Ag(s)}$. But silver in copper(II) sulfate solution does nothing. The electrochemical series is a guide to the relative tendency of metals to be oxidised: a metal with a lower $E^\circ$ displaces the ions of a metal with a higher $E^\circ$.
The electrochemical series ranks couples by their standard reduction potential, $E^\circ$, measured against the hydrogen half-cell ($E^\circ = 0.00\ \text{V}$). The couple with the higher $E^\circ$ has the greater tendency to gain electrons.
Practice exercise (it uses $E^\circ$ values; the method for calculating $E^\circ_{\text{cell}}$ is explained in the extension box above, and the cell diagram and half-equations are core ideas):
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 $E^\circ(\mathrm{Cu}^{2+}/\mathrm{Cu}) = +0.34\text{ V}$.
- a) Write the conventional cell notation or diagram, indicating the anode and the cathode.
- b) State the sign or polarity assigned to each electrode and briefly justify which half-reaction takes place at each.
Calculating the cell potential
Take the cell based on $\mathrm{Cu(s) + 2Ag^+(aq) \rightarrow Cu^{2+}(aq) + 2Ag(s)}$.
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 $E^\circ(\mathrm{Cu}^{2+}/\mathrm{Cu}) = +0.34\text{ V}$.
- a) Calculate the standard electromotive force of the cell, $E^\circ_{\text{pila}}$.
- 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.
Gibbs energy and spontaneity
The electrical work a cell can do is linked to the Gibbs energy change of the reaction.
Extension exercises (Gibbs energy is beyond the core specification):
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 $F = 96485\text{ C}\times\text{mol}^{-1}$.
- a) Determine the value of the standard Gibbs energy change, $\Delta G^\circ$, in $\text{kJ}\times\text{mol}^{-1}$.
- b) Deduce the criterion for spontaneity from the sign of $\Delta G^\circ$ and $E^\circ_{\text{pila}}$.
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 analysed,…
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 analysed, with reactants under standard conditions. Data: $E^\circ(\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}) = +0.77\text{ V}$; $E^\circ(\mathrm{I_2}/\mathrm{I^-}) = +0.54\text{ V}$; $F = 96485\text{ C}\times\text{mol}^{-1}$.
- a) Write the balanced equation of the spontaneous redox reaction under standard conditions, identify the oxidising and reducing species, and calculate $E^\circ_{\text{pila}}$ and $\Delta G^\circ$.
- 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, analysing the difference between $\Delta G^\circ$ and $\Delta G$.
The salt bridge and the direction of electron flow
Connect a zinc half-cell and a copper half-cell with a wire and a voltmeter. The needle flickers and drops to $0\ \text{V}$. Add a salt bridge between the two solutions and the reading jumps to a steady value (about $1.10\ \text{V}$ under standard conditions, 1.0 mol/L and 25 °C). Why does the current stop without the bridge? The circuit is open, so no charge can flow.
As zinc is oxidised, the zinc half-cell gains positive $\mathrm{Zn^{2+}}$ ions. In the copper half-cell, $\mathrm{Cu^{2+}}$ ions are removed, leaving an excess of negative sulfate ions. Charge builds up and stops the electrons moving.
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 set-up is shown below:
Data: $E^\circ(\mathrm{Zn}^{2+}/\mathrm{Zn}) = -0.76\text{ V}$; $E^\circ(\mathrm{Cu}^{2+}/\mathrm{Cu}) = +0.34\text{ V}$.
- a) Determine the standard electromotive force of the cell and state the correct direction of electron flow through the external metallic conductor.
- b) Explain which species carries the charge through the salt bridge and towards which beaker the $\mathrm{K^+}$ and $\mathrm{NO_3^-}$ ions migrate respectively, stating why electrons do not flow through the solution.
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: $E^\circ(\mathrm{Cr}^{3+}/\mathrm{Cr}) = -0.74\text{ V}$, $E^\circ(\mathrm{Pb}^{2+}/\mathrm{Pb}) = -0.13\text{ V}$, $F = 96485\text{ C}\times\text{mol}^{-1}$.
- a) Write the anode and cathode half-equations and the balanced overall ionic equation, and calculate the standard potential of the cell.
- 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.
Predicting reactions with the electrochemical series
Zinc turnings in $1\ \text{mol/L}$ hydrochloric acid fizz furiously as hydrogen is produced. A copper coin in the same acid does nothing. What decides which metals can give their electrons to $\mathrm{H^+}$ ions?
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 oxidising agent in its standard state that is able to spontaneously oxidise the iodide ion to molecular iodine ($\mathrm{I_2}$), and the bromide ion to liquid bromine ($\mathrm{Br_2}$), but that does not oxidise the chloride ion to chlorine gas ($\mathrm{Cl_2}$).
Standard reduction potential data at $25\text{ }^\circ\mathrm{C}$:
$E^\circ(\mathrm{I_2}/\mathrm{I^-}) = +0.54\text{ V}$; $E^\circ(\mathrm{Br_2}/\mathrm{Br^-}) = +1.07\text{ V}$; $E^\circ(\mathrm{Cl_2}/\mathrm{Cl^-}) = +1.36\text{ V}$.
Oxidising 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 $E^\circ = +1.51\text{ V}$.
- a) Select the most suitable oxidising 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 oxidising agent with each halide, using the lowest whole-number coefficients.
- b) Write the balanced overall reaction that takes place between the selected oxidising agent and the bromide ion in acidic medium. Data: $F = 96485\text{ C}\times\text{mol}^{-1}$.
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 $F = 96485\text{ C}\times\text{mol}^{-1}$.
- 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.
- 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.
Back to the start: by separating the two half-reactions of zinc and copper(II) sulfate with an external wire and a salt bridge, the electrons are forced through the circuit. The standard cell potential is $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.34 - (-0.76) = +1.10\ \text{V}$.