Electrochemical Cells and Applications

Lessons

Senior 6 · Chemistry

Electrochemical Cells and Applications

5 lessons

Lesson 3 of 5

Predicting Feasibility and the Effect of Concentration

25 min

Why not permanganate in hydrochloric acid?

Analysts who titrate iron(II) with potassium permanganate always acidify with sulfuric acid, never hydrochloric acid. Yet the same iron(II) can be titrated with potassium dichromate in hydrochloric acid without trouble. Both are powerful oxidising agents; what is the difference?

Standard electrode potentials answer this in two lines, as you will see below. This lesson shows how to use $E^{\circ}$ values to predict whether any redox reaction can happen, how that links to the free energy change, and why changing concentrations can tip the balance.

Definition

Feasible (thermodynamically spontaneous) reaction: a reaction that can proceed by itself under the stated conditions. A redox reaction is feasible under standard conditions if $E^{\circ}_{\text{cell}}$ for it, written as a cell, is positive.

Faraday constant, $F$ : the charge on one mole of electrons, $96\,485\,\mathrm{C\,mol^{-1}}$ .

Relation to free energy: $\Delta G^{\circ} = -nFE^{\circ}_{\text{cell}}$ , where $n$ is the number of moles of electrons transferred in the equation as written. Positive $E^{\circ}_{\text{cell}}$ means negative $\Delta G^{\circ}$ .

Nernst equation: $E = E^{\circ} - \frac{RT}{nF}\ln Q$ ; at 298 K, $E = E^{\circ} - \frac{0.0592}{n}\log_{10}Q$ , where $Q$ is the reaction quotient.

Concentration cell: a cell made of two half-cells of the same couple at different concentrations. Example: Cu in 0.010 mol dm⁻³ $\mathrm{Cu^{2+}}$ joined to Cu in 1.0 mol dm⁻³ $\mathrm{Cu^{2+}}$ .

Will it react? The $E^{\circ}_{\text{cell}}$ test

Find the two half-equations in the electrochemical series (Lesson 2).
The species being reduced belongs to the cathode half-cell; the species being oxidised belongs to the anode half-cell.
Calculate $E^{\circ}_{\text{cell}} = E^{\circ}_{\text{reduced}} - E^{\circ}_{\text{oxidised}}$ . If it is positive, the reaction is feasible; if negative, the reverse reaction is feasible instead.

A quick picture: an oxidising agent (left side of the series) reacts with any reducing agent (right side) that lies above it in the table, where the table runs from most negative $E^{\circ}$ at the top to most positive at the bottom.

Back to the permanganate puzzle

Possible reaction	$E^{\circ}_{\text{cell}}$ / V	Feasible?
$\mathrm{MnO_4^{-}}$ oxidising $\mathrm{Cl^{-}}$ to $\mathrm{Cl_2}$	$1.51 - 1.36 = + 0.15$	yes: some permanganate is wasted on the acid, so the titre is too large
$\mathrm{Cr_2O_7^{2-}}$ oxidising $\mathrm{Cl^{-}}$ to $\mathrm{Cl_2}$	$1.33 - 1.36 = - 0.03$	no: dichromate leaves chloride alone
$\mathrm{Cr_2O_7^{2-}}$ oxidising $\mathrm{Fe^{2+}}$ to $\mathrm{Fe^{3+}}$	$1.33 - 0.77 = + 0.56$	yes: the titration reaction

Free energy and the equilibrium constant

Because $\Delta G^{\circ} = -nFE^{\circ}_{\text{cell}}$ and $\Delta G^{\circ} = -RT\ln K$ ,

$\ln K = \frac{nFE^{\circ}_{\text{cell}}}{RT} \qquad\text{or, at 298 K,}\qquad \log_{10}K = \frac{nE^{\circ}_{\text{cell}}}{0.0592}$

Even a modest emf means a huge $K$ : for the Daniell cell, $\log_{10}K = \frac{2 \times 1.10}{0.0592} = 37$ . A reaction with $E^{\circ}_{\text{cell}}$ above about +0.3 V goes essentially to completion.

Two limits of the prediction

Rate. $E^{\circ}$ says whether a reaction can happen, not how fast. Acidified permanganate should oxidise water to oxygen ( $1.51 - 1.23 = +0.28\,\mathrm{V}$ ), yet its solutions keep for a long time because that reaction is very slow.
Conditions. $E^{\circ}$ values apply at 1 mol dm⁻³ and 100 kPa. Real concentrations change the potentials, as the Nernst equation shows.

How concentration changes an electrode potential

For a metal half-cell, $\mathrm{M^{n+}(aq) + ne^{-} \rightleftharpoons M(s)}$ , the Nernst equation becomes

$E = E^{\circ} + \frac{0.0592}{n}\log_{10}[\mathrm{M^{n+}}]$

Lowering $[\mathrm{M^{n+}}]$ tenfold makes $E$ less positive by $\frac{0.0592}{n}$ V (Figure 3.1). This fits Le Chatelier's principle: removing the ion that accepts electrons makes the reduction less favourable. For a whole cell, increasing the concentration of a reactant in the cell reaction raises $E_{\text{cell}}$ ; increasing a product lowers it.

Figure 3.1: Electrode potentials of two metal half-cells at 298 K as the ion concentration changes (calculated by Ganzaa from the Nernst equation).

Worked example

Problem. Use $E^{\circ}$ values to predict whether each reaction is feasible under standard conditions. Write the overall equation for any that are. (a) Chlorine with bromide ions. (b) Iodine with iron(II) ions. (c) Copper with dilute hydrochloric acid.

Solution.

(a) $\mathrm{Cl_2}$ reduced, $\mathrm{Br^{-}}$ oxidised: $E^{\circ}_{\text{cell}} = 1.36 - 1.07 = +0.29\,\mathrm{V}$ . Feasible: $\mathrm{Cl_2 + 2Br^{-} \rightarrow 2Cl^{-} + Br_2}$ .

(b) $\mathrm{I_2}$ reduced, $\mathrm{Fe^{2+}}$ oxidised: $E^{\circ}_{\text{cell}} = 0.54 - 0.77 = -0.23\,\mathrm{V}$ . Not feasible; the reverse, $\mathrm{2Fe^{3+} + 2I^{-} \rightarrow 2Fe^{2+} + I_2}$ , is (+0.23 V).

(c) $\mathrm{H^{+}}$ reduced, Cu oxidised: $E^{\circ}_{\text{cell}} = 0.00 - 0.34 = -0.34\,\mathrm{V}$ . Not feasible: copper does not give hydrogen with dilute acids.

Answer: (a) yes; (b) no; (c) no.

Worked example

Problem. (a) Calculate $\Delta G^{\circ}$ for the Daniell cell reaction, $E^{\circ}_{\text{cell}} = +1.099\,\mathrm{V}$ . (b) Calculate the emf of a Daniell cell in which $[\mathrm{Zn^{2+}}] = 1.0\,\mathrm{mol\,dm^{-3}}$ and $[\mathrm{Cu^{2+}}] = 0.010\,\mathrm{mol\,dm^{-3}}$ , at 298 K.

Solution.

(a) $n = 2$ : $\Delta G^{\circ} = -nFE^{\circ}_{\text{cell}} = -2 \times 96\,485 \times 1.099 = -2.12 \times 10^{5}\,\mathrm{J\,mol^{-1}} = -212\,\mathrm{kJ\,mol^{-1}}$ .

(b) For $\mathrm{Zn + Cu^{2+} \rightarrow Zn^{2+} + Cu}$ , $Q = \frac{[\mathrm{Zn^{2+}}]}{[\mathrm{Cu^{2+}}]} = \frac{1.0}{0.010} = 100$ :

$E_{\text{cell}} = 1.099 - \frac{0.0592}{2}\log_{10}100 = 1.099 - 0.059 = 1.040\,\mathrm{V}$

Answer: (a) −212 kJ mol⁻¹; (b) 1.04 V. Lowering the concentration of the reactant $\mathrm{Cu^{2+}}$ lowers the emf, as Le Chatelier's principle predicts.

Worked example

Problem. In the laboratory, chlorine is made by warming manganese(IV) oxide with concentrated hydrochloric acid, although $E^{\circ}(\mathrm{MnO_2/Mn^{2+}}) = +1.22\,\mathrm{V}$ is below $E^{\circ}(\mathrm{Cl_2/Cl^{-}}) = +1.36\,\mathrm{V}$ . Use the Nernst equation to explain why it works. Take $[\mathrm{H^{+}}] = [\mathrm{Cl^{-}}] = 12\,\mathrm{mol\,dm^{-3}}$ , $[\mathrm{Mn^{2+}}] = 1.0\,\mathrm{mol\,dm^{-3}}$ and $p(\mathrm{Cl_2}) = 100\,\mathrm{kPa}$ (ignore non-ideal behaviour).

Solution. $\mathrm{MnO_2 + 4H^{+} + 2e^{-} \rightleftharpoons Mn^{2+} + 2H_2O}$ :

$E = 1.224 + \frac{0.0592}{2}\log_{10}\frac{[\mathrm{H^{+}}]^4}{[\mathrm{Mn^{2+}}]} = 1.224 + 0.0296 \times \log_{10}(12^4) = 1.224 + 0.128 = 1.35\,\mathrm{V}$

$\mathrm{Cl_2 + 2e^{-} \rightleftharpoons 2Cl^{-}}$ :

$E = 1.36 - \frac{0.0592}{2}\log_{10}[\mathrm{Cl^{-}}]^2 = 1.36 - 0.0296 \times \log_{10}(144) = 1.36 - 0.064 = 1.30\,\mathrm{V}$

Answer: under these conditions $E_{\text{cell}} = 1.35 - 1.30 = +0.05\,\mathrm{V}$ , so the reaction becomes feasible. High $[\mathrm{H^{+}}]$ makes $\mathrm{MnO_2}$ a stronger oxidising agent, and high $[\mathrm{Cl^{-}}]$ makes chloride easier to oxidise. With dilute acid the emf is negative and no chlorine forms.

Warning

Common mistakes

Treating a positive $E^{\circ}_{\text{cell}}$ as a promise that a reaction will be fast. It only says the reaction is energetically possible.
Using the wrong $n$ in $\Delta G^{\circ} = -nFE^{\circ}$ or in the Nernst equation. $n$ is the number of electrons in the balanced overall equation (2 for the Daniell cell; 6 for dichromate with iron(II)).
Leaving $\Delta G^{\circ}$ in J mol⁻¹ when kJ mol⁻¹ is asked for, or the reverse.
Putting solids or water into $Q$ . Only dissolved species (and gas pressures) appear.

Tip

Check your $Q$ with Le Chatelier: if you increase a reactant, $Q$ falls, $\log_{10}Q$ becomes smaller (or negative), and $E_{\text{cell}}$ must go up. If your answer goes the other way, you have written $Q$ upside down.

Chad's Prep: The Nernst equation and non-standard cell potentials. Watch how Q is built from the balanced cell reaction and why changing concentrations moves the cell potential.

✦

Practice

Use $E^{\circ}$ values to decide whether each reaction is feasible: (a) zinc with dilute sulfuric acid; (b) bromine with chloride ions; (c) silver ions with copper metal.
Explain why acidified potassium dichromate can be used to titrate iron(II) in the presence of hydrochloric acid, but acidified potassium permanganate cannot.
Calculate $\Delta G^{\circ}$ for $\mathrm{Zn + 2Ag^{+} \rightarrow Zn^{2+} + 2Ag}$ ( $E^{\circ}_{\text{cell}} = +1.56\,\mathrm{V}$ ).
Calculate the electrode potential of a silver electrode in 0.010 mol dm⁻³ $\mathrm{Ag^{+}}$ at 298 K.
Calculate the emf of the cell $\mathrm{Zn\,|\,Zn^{2+}(0.10\,mol\,dm^{-3})\,\|\,Cu^{2+}(2.0\,mol\,dm^{-3})\,|\,Cu}$ at 298 K.
A concentration cell is made from two silver electrodes, one in 0.10 mol dm⁻³ and one in 0.0010 mol dm⁻³ $\mathrm{Ag^{+}}$ . Calculate its emf and state which electrode is positive.
$E^{\circ}_{\text{cell}}$ for the oxidation of water by acidified permanganate is +0.28 V. Evaluate the claim: "Acidified potassium permanganate solution cannot be stored, because it reacts with the water it is dissolved in."

Show answersHide answers

(a) $0.00 - (-0.76) = +0.76\,\mathrm{V}$ : feasible, $\mathrm{Zn + 2H^{+} \rightarrow Zn^{2+} + H_2}$ . (b) $1.07 - 1.36 = -0.29\,\mathrm{V}$ : not feasible. (c) $0.80 - 0.34 = +0.46\,\mathrm{V}$ : feasible, $\mathrm{Cu + 2Ag^{+} \rightarrow Cu^{2+} + 2Ag}$ .
Dichromate cannot oxidise chloride ( $1.33 - 1.36 = -0.03\,\mathrm{V}$ ), so only iron(II) reacts. Permanganate can ( $1.51 - 1.36 = +0.15\,\mathrm{V}$ ), so some of it is used up oxidising chloride to chlorine and the titre is too large.
$n = 2$ : $\Delta G^{\circ} = -2 \times 96\,485 \times 1.56 = -3.01 \times 10^{5}\,\mathrm{J\,mol^{-1}} = -301\,\mathrm{kJ\,mol^{-1}}$ .
$E = 0.80 + 0.0592 \log_{10}0.010 = 0.80 - 0.118 = +0.68\,\mathrm{V}$ .
$Q = \frac{0.10}{2.0} = 0.050$ ; $E = 1.10 - 0.0296 \times \log_{10}0.050 = 1.10 + 0.039 = 1.14\,\mathrm{V}$ .
$E = 0.0592 \log_{10}\frac{0.10}{0.0010} = 0.0592 \times 2 = 0.118\,\mathrm{V}$ . The electrode in the more concentrated solution (0.10 mol dm⁻³) is positive; there $\mathrm{Ag^{+}}$ is reduced.
The claim is only half right. $E^{\circ}_{\text{cell}}$ is positive, so the reaction is thermodynamically feasible, but it is very slow, so the solution can be stored for a long time. It does decompose slowly, faster in light and when $\mathrm{MnO_2}$ is present, which is why such solutions are kept in dark bottles and standardised before use.

Sources and credits

Standard electrode potentials at 25 °C: Standard electrode potential (data page), Wikipedia contributors, Wikipedia, licensed under CC BY-SA 4.0; values compiled there from P. Vanýsek, "Electrochemical Series", CRC Handbook of Chemistry and Physics (CRC Press), and other cited sources; adapted by Ganzaa.
Video: Nernst Equation, Nonstandard Cell Potential, Chad's Prep, embedded from YouTube.
The text, graph (calculated by Ganzaa), table, worked examples and Practice items were written and drawn by Ganzaa.
This lesson adapts CC BY-SA material and is shared under CC BY-SA 4.0.

1Why not permanganate in hydrochloric acid?

2Will it react? The Ecell∘ E^{\circ}_{\text{cell}} Ecell∘​ test

3Back to the permanganate puzzle

4Free energy and the equilibrium constant

5Two limits of the prediction

6How concentration changes an electrode potential

Why not permanganate in hydrochloric acid?

Will it react? The $E^{\circ}_{\text{cell}}$ test

Back to the permanganate puzzle

Free energy and the equilibrium constant

Two limits of the prediction

How concentration changes an electrode potential