Cells, potentials and spontaneity · Chemistry 6th Year HL

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 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.

Calculating the cell potential

Extension practice: the first exercise below uses the cell notation and polarity.

Practice★☆☆

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.

Solve

Take the cell based on $\mathrm{Cu(s) + 2Ag^+(aq) \rightarrow Cu^{2+}(aq) + 2Ag(s)}$.

Practice★☆☆

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.

Solve

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):

Practice★☆☆

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 $E^\circ_{\text{pila}}$.

Solve

Practice★★☆

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 $\Delta G$.

Solve

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.

Practice★★☆

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.

Solve

Practice★★☆

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.

Solve

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?

Using the electrochemical series

The oxidised form of a couple with a higher $E^\circ$ will spontaneously oxidise (under standard conditions) the reduced form of a couple with a lower $E^\circ$. In this table, the series is listed with the highest $E^\circ$ at the top (the opposite of the usual reactivity order, where the most reactive metal is at the top). Under standard conditions, a metal with a positive $E^\circ$ does not displace hydrogen from a dilute non-oxidising acid (such as hydrochloric acid); a metal with a negative $E^\circ$ does. Likewise, a metal displaces from solution the ions of any metal that has a higher $E^\circ$ than its own; that is, a metal with a lower $E^\circ$ displaces the ions of a metal with a higher $E^\circ$. The bigger the difference in $E^\circ$, the more favourable the reaction.

Why? The couple with the high $E^\circ$ holds electrons more strongly, so it takes them from any reduced species lower in the series.

Choosing an oxidising agent that oxidises I⁻ but not Br⁻

$E^\circ(\mathrm{I_2/I^-}) = +0.54\ \text{V}$, $E^\circ(\mathrm{Br_2/Br^-}) = +1.07\ \text{V}$, $E^\circ(\mathrm{Cl_2/Cl^-}) = +1.36\ \text{V}$. To oxidise iodide, the oxidising agent needs $E^\circ$ above $0.54\ \text{V}$. To leave bromide alone, it needs $E^\circ$ below $1.07\ \text{V}$. To oxidise bromide as well but not chloride, it needs $E^\circ$ between $1.07\ \text{V}$ and $1.36\ \text{V}$.

Among candidates, $\mathrm{MnO_4^-/Mn^{2+}}$ ($+1.51\ \text{V}$) would also oxidise chloride, so it is too strong. $\mathrm{Fe^{3+}/Fe^{2+}}$ ($+0.77\ \text{V}$) oxidises iodide but is too weak to oxidise bromide. A couple with $E^\circ = +1.20\ \text{V}$, such as $\mathrm{IO_3^-/I_2}$, lies between $1.07\ \text{V}$ and $1.36\ \text{V}$, so it oxidises both iodide and bromide but not chloride.

A species that can act as oxidising or reducing agent

Hydrogen peroxide appears in two couples: $E^\circ(\mathrm{H_2O_2/H_2O}) = +1.78\ \text{V}$ (here $\mathrm{H_2O_2}$ is the oxidised form, so it acts as an oxidising agent) and $E^\circ(\mathrm{O_2/H_2O_2}) = +0.68\ \text{V}$ (here $\mathrm{H_2O_2}$ is the reduced form, so it acts as a reducing agent). Take the couple $\mathrm{Fe^{3+}/Fe^{2+}}$ ($+0.77\ \text{V}$) as a test.

Can $\mathrm{H_2O_2}$ oxidise $\mathrm{Fe^{2+}}$? The oxidising couple has $E^\circ = +1.78\ \text{V}$, higher than $+0.77\ \text{V}$, so yes: $E^\circ_{\text{cell}} = 1.78 - 0.77 = +1.01\ \text{V} > 0$.

Can $\mathrm{H_2O_2}$ reduce $\mathrm{Fe^{3+}}$? The reducing couple has $E^\circ = +0.68\ \text{V}$, lower than $+0.77\ \text{V}$, so yes: $E^\circ_{\text{cell}} = 0.77 - 0.68 = +0.09\ \text{V} > 0$.

Because the couple in which it is the oxidised form has a higher $E^\circ$ than the couple in which it is the reduced form ($+1.78 > +0.68$), $\mathrm{H_2O_2}$ can react with itself: one molecule oxidises another (disproportionation), $\mathrm{2H_2O_2 \rightarrow 2H_2O + O_2}$, with $E^\circ_{\text{cell}} = 1.78 - 0.68 = +1.10\ \text{V}$. So it is unstable. To decide what a reagent does with a given species, compare the $E^\circ$ of the relevant couples in the same way.

Practice★★★

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: $F = 96485\text{ C}\times\text{mol}^{-1}$.

Solve

Practice★★★

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.

Solve

Practice★★★

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: $F = 96485\text{ C}\times\text{mol}^{-1}$.

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 $25\text{ }^\circ\mathrm{C}$.
b) Justify whether the $\mathrm{Sn}^{2+}$ ion can be oxidised 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 oxidise $25.0\text{ mL}$ 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

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 $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.34 - (-0.76) = +1.10\ \text{V}$.

Summary

Let's practice! 3 easy exercises in this section

★☆☆ You have a Daniell galvanic cell made up of a zinc electrode immersed in a $1.0\text{ M}$…★☆☆ Consider a galvanic cell under standard conditions ($25\text{ }^\circ\mathrm{C}$,…