Electrochemical Cells and Applications

Lessons

Senior 6 · Chemistry

Electrochemical Cells and Applications

5 lessons

Lesson 2 of 5

The Standard Hydrogen Electrode and Electrode Potentials

25 min

Measuring from sea level

The height of Mount Karisimbi is quoted as a number of metres above sea level, because a height only has meaning relative to some agreed zero. Electrode potentials have the same problem. A voltmeter always needs two terminals, so it can only ever measure the difference between two half-cells; the potential of a single half-cell cannot be measured on its own.

Chemists therefore chose one half-cell as the zero, the standard hydrogen electrode, and measure every other half-cell against it. The resulting list of standard electrode potentials lets you predict the emf of any cell, and which redox reactions can happen, without doing the experiment.

Definition

Standard conditions (electrochemistry): temperature 298 K (25 °C), every dissolved species at 1.00 mol dm⁻³, and every gas at 100 kPa (1 bar).

Standard hydrogen electrode (SHE): hydrogen gas at 100 kPa bubbling over a platinum electrode coated with finely divided platinum (platinum black), in a solution of $\mathrm{H^{+}(aq)}$ at 1.00 mol dm⁻³, at 298 K. Its potential is defined as exactly 0.00 V: $\mathrm{2H^{+}(aq) + 2e^{-} \rightleftharpoons H_2(g)}$ , $E^{\circ} = 0.00\,\mathrm{V}$ .

Standard electrode potential, $E^{\circ}$ : the emf of a cell in which the left-hand electrode is the SHE and the right-hand electrode is the half-cell in question, all under standard conditions. It is quoted for the reduction half-equation. Example: $\mathrm{Zn^{2+}(aq) + 2e^{-} \rightleftharpoons Zn(s)}$ , $E^{\circ} = -0.76\,\mathrm{V}$ .

Electrochemical series: the list of half-equations in order of their standard electrode potentials.

How $E^{\circ}$ is measured

Figure 2.1 shows a zinc half-cell connected to the SHE. The voltmeter reads 0.76 V, and the zinc is the negative terminal: zinc pushes electrons into the external circuit more strongly than hydrogen does. So $E^{\circ}(\mathrm{Zn^{2+}/Zn}) = -0.76\,\mathrm{V}$ . With a copper half-cell instead, the reading is 0.34 V with copper positive, so $E^{\circ}(\mathrm{Cu^{2+}/Cu}) = +0.34\,\mathrm{V}$ .

Figure 2.1: The standard hydrogen electrode (left) connected to a standard zinc half-cell (right).

The SHE is awkward to use: it needs a supply of pure hydrogen, and the platinum is easily "poisoned". In practice a secondary reference such as the silver–silver chloride electrode, whose potential against the SHE is accurately known, is used instead. The SHE also gives the potentials of half-cells with no metal at all, such as $\mathrm{Fe^{3+}/Fe^{2+}}$ or $\mathrm{Cl_2/Cl^{-}}$ , which use a platinum electrode dipping into a solution containing both forms.

The electrochemical series

Half-equation (reduction)	$E^{\circ}$ / V
$\mathrm{Li^{+} + e^{-} \rightleftharpoons Li}$	−3.04
$\mathrm{K^{+} + e^{-} \rightleftharpoons K}$	−2.93
$\mathrm{Na^{+} + e^{-} \rightleftharpoons Na}$	−2.71
$\mathrm{Mg^{2+} + 2e^{-} \rightleftharpoons Mg}$	−2.37
$\mathrm{Al^{3+} + 3e^{-} \rightleftharpoons Al}$	−1.66
$\mathrm{Zn^{2+} + 2e^{-} \rightleftharpoons Zn}$	−0.76
$\mathrm{Fe^{2+} + 2e^{-} \rightleftharpoons Fe}$	−0.44
$\mathrm{Ni^{2+} + 2e^{-} \rightleftharpoons Ni}$	−0.26
$\mathrm{Sn^{2+} + 2e^{-} \rightleftharpoons Sn}$	−0.13
$\mathrm{Pb^{2+} + 2e^{-} \rightleftharpoons Pb}$	−0.13
$\mathrm{2H^{+} + 2e^{-} \rightleftharpoons H_2}$	0.00
$\mathrm{Cu^{2+} + 2e^{-} \rightleftharpoons Cu}$	+0.34
$\mathrm{O_2 + 2H_2O + 4e^{-} \rightleftharpoons 4OH^{-}}$	+0.40
$\mathrm{I_2 + 2e^{-} \rightleftharpoons 2I^{-}}$	+0.54
$\mathrm{Fe^{3+} + e^{-} \rightleftharpoons Fe^{2+}}$	+0.77
$\mathrm{Ag^{+} + e^{-} \rightleftharpoons Ag}$	+0.80
$\mathrm{Br_2(l) + 2e^{-} \rightleftharpoons 2Br^{-}}$	+1.07
$\mathrm{O_2 + 4H^{+} + 4e^{-} \rightleftharpoons 2H_2O}$	+1.23
$\mathrm{Cr_2O_7^{2-} + 14H^{+} + 6e^{-} \rightleftharpoons 2Cr^{3+} + 7H_2O}$	+1.33
$\mathrm{Cl_2 + 2e^{-} \rightleftharpoons 2Cl^{-}}$	+1.36
$\mathrm{MnO_4^{-} + 8H^{+} + 5e^{-} \rightleftharpoons Mn^{2+} + 4H_2O}$	+1.51

The species on the left are oxidising agents, strongest at the bottom ( $\mathrm{MnO_4^{-}}$ , $\mathrm{Cl_2}$ ). The species on the right are reducing agents, strongest at the top (Li, K, Mg). A metal with a very negative $E^{\circ}$ gives up electrons easily, which is why the order of the metals matches the reactivity series you met in earlier years.

Calculating the emf of a cell

The half-cell with the more positive $E^{\circ}$ is the positive electrode (cathode, where reduction happens); the other is the negative electrode (anode).

$E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} = E^{\circ}_{\text{right}} - E^{\circ}_{\text{left}}$

For a cell that works as written, $E^{\circ}_{\text{cell}}$ is positive.

Worked example

Problem. Calculate the standard emf of the Daniell cell, $\mathrm{Zn(s)\,|\,Zn^{2+}(aq)\,\|\,Cu^{2+}(aq)\,|\,Cu(s)}$ , using $E^{\circ}(\mathrm{Zn^{2+}/Zn}) = -0.762\,\mathrm{V}$ and $E^{\circ}(\mathrm{Cu^{2+}/Cu}) = +0.337\,\mathrm{V}$ .

Solution. Copper has the more positive potential, so it is the cathode (right):

$E^{\circ}_{\text{cell}} = (+0.337) - (-0.762) = +1.099\,\mathrm{V} \approx +1.10\,\mathrm{V}$

Answer: +1.10 V, with copper as the positive terminal.

Worked example

Problem. A zinc half-cell is connected to a silver half-cell under standard conditions. (a) Which electrode is positive? (b) Write the overall equation. (c) Calculate $E^{\circ}_{\text{cell}}$ .

Solution. $E^{\circ}(\mathrm{Ag^{+}/Ag}) = +0.80\,\mathrm{V}$ is more positive than $E^{\circ}(\mathrm{Zn^{2+}/Zn}) = -0.76\,\mathrm{V}$ .

(a) Silver is the positive electrode (cathode); $\mathrm{Ag^{+}}$ is reduced.

(b) $\mathrm{Zn(s) \rightarrow Zn^{2+}(aq) + 2e^{-}}$ and $\mathrm{Ag^{+}(aq) + e^{-} \rightarrow Ag(s)}$ , doubled: $\mathrm{Zn(s) + 2Ag^{+}(aq) \rightarrow Zn^{2+}(aq) + 2Ag(s)}$ .

Answer: silver is positive; $E^{\circ}_{\text{cell}} = +1.56\,\mathrm{V}$ . The silver half-equation was doubled to balance the electrons, but its $E^{\circ}$ was not doubled.

Warning

Common mistakes

Multiplying $E^{\circ}$ by the number used to balance electrons. $E^{\circ}$ is a property of the half-cell (an intensive quantity), not of the amount of substance; it never changes when a half-equation is doubled.
Adding the two values instead of subtracting: $E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}}$ .
Reversing the sign of $E^{\circ}$ for the anode and then subtracting it, which counts the change of sign twice.
Forgetting that tabulated values apply only under standard conditions.

Tip

Picture the series as a ladder with the most negative $E^{\circ}$ at the top. In any cell, electrons "fall" from the half-cell higher up the ladder (more negative, the anode) to the one lower down (more positive, the cathode), and $E^{\circ}_{\text{cell}}$ is the distance between the two rungs.

Mr Murray-Green Chemistry Tutorials: Standard hydrogen electrode summary. Watch for the three standard conditions and how the SHE is connected to another half-cell to measure its standard electrode potential.

✦

Practice

Describe the standard hydrogen electrode, giving the three standard conditions.
Explain why the potential of a single half-cell cannot be measured directly.
A copper half-cell connected to the SHE gives 0.34 V, with copper as the positive terminal. What is $E^{\circ}(\mathrm{Cu^{2+}/Cu})$ ? Which electrode would be the anode?
Calculate $E^{\circ}_{\text{cell}}$ for a cell made from (a) magnesium and copper half-cells; (b) iron ( $\mathrm{Fe^{2+}/Fe}$ ) and copper half-cells; (c) nickel and silver half-cells. In each case name the positive electrode.
Write the cell diagram and overall equation for cell (c) in question 4.
From the table, choose (a) the strongest oxidising agent; (b) the strongest reducing agent; (c) a metal that can reduce $\mathrm{Pb^{2+}}$ ions but not $\mathrm{Zn^{2+}}$ ions.
A student calculates the emf of the zinc–silver cell as $2 \times 0.80 - (-0.76) = 2.36\,\mathrm{V}$ , "because two silver ions react". Evaluate this reasoning.

Show answersHide answers

Platinum coated with platinum black, dipping into 1.00 mol dm⁻³ $\mathrm{H^{+}}$ , with hydrogen gas at 100 kPa bubbling over it, at 298 K. Its potential is defined as 0.00 V.
A voltmeter needs two terminals, so it always measures a potential difference between two half-cells. One half-cell (the SHE) has to be chosen as the zero of the scale.
+0.34 V. The SHE would be the anode (negative), where $\mathrm{H_2}$ is oxidised to $\mathrm{H^{+}}$ .
(a) $0.34 - (-2.37) = 2.71\,\mathrm{V}$ , copper positive. (b) $0.34 - (-0.44) = 0.78\,\mathrm{V}$ , copper positive. (c) $0.80 - (-0.26) = 1.06\,\mathrm{V}$ , silver positive.
$\mathrm{Ni(s)\,|\,Ni^{2+}(aq)\,\|\,Ag^{+}(aq)\,|\,Ag(s)}$ ; $\mathrm{Ni(s) + 2Ag^{+}(aq) \rightarrow Ni^{2+}(aq) + 2Ag(s)}$ .
(a) $\mathrm{MnO_4^{-}}$ in acid (+1.51 V). (b) Li (−3.04 V). (c) Iron or nickel: each has an $E^{\circ}$ more negative than lead's (−0.13 V), so it can reduce $\mathrm{Pb^{2+}}$ , but more positive than zinc's (−0.76 V), so it cannot reduce $\mathrm{Zn^{2+}}$ .
The reasoning is wrong. Doubling the silver half-equation balances the electrons, but $E^{\circ}$ does not depend on how many ions react, so it is not doubled. The correct emf is $0.80 - (-0.76) = 1.56\,\mathrm{V}$ .

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: Standard Hydrogen Electrode Summary for A-level Chemistry, Mr Murray-Green Chemistry Tutorials, embedded from YouTube.
The text, SHE diagram, table layout, 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.

1Measuring from sea level

2How E∘ E^{\circ} E∘ is measured

3The electrochemical series

4Calculating the emf of a cell

Measuring from sea level

How $E^{\circ}$ is measured

The electrochemical series

Calculating the emf of a cell