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Standard reduction potentials measured against hydrogen, the cell potential as cathode minus anode, and what its sign predicts.
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By the end of this lesson you will be able to identify the cathode and anode of a cell from their standard reduction potentials, calculate a standard cell potential as cathode minus anode, find an unknown electrode potential from a measured cell, rank oxidising agents, and use the sign of a cell potential to predict whether a reaction is feasible — without multiplying a potential by a coefficient.
Last lesson you built a cell and decided which electrode was the anode by which metal was more reactive. You know a cell pushes electrons from the anode through the wire to the cathode. This lesson measures how hard it pushes, and replaces "more reactive" with a number you can look up.
The standard reduction potential, $E^\circ$, of a half-cell is the voltage it gives when connected to a standard hydrogen electrode, which is defined as $0.00$ V, with every solution at $1$ mol/L, every gas at $100$ kPa and the temperature $25$ °C. It is always quoted for the reduction, written with the electrons on the left. The standard cell potential, $E^\circ_{cell}$, is the voltage of a cell made from two half-cells under standard conditions. A volt is one joule of energy for each coulomb of charge.
Every half-cell has a tendency to take electrons in — to be reduced. A table of standard reduction potentials puts a number on it:
| Half-equation | $E^\circ$ / V |
|---|---|
| $\mathrm{Ag^+ + e^- \rightleftharpoons Ag}$ | $+0.80$ |
| $\mathrm{Cu^{2+} + 2e^- \rightleftharpoons Cu}$ | $+0.34$ |
| $\mathrm{2H^+ + 2e^- \rightleftharpoons H_2}$ | $0.00$ |
| $\mathrm{Fe^{2+} + 2e^- \rightleftharpoons Fe}$ | $-0.44$ |
| $\mathrm{Zn^{2+} + 2e^- \rightleftharpoons Zn}$ | $-0.76$ |
| $\mathrm{Mg^{2+} + 2e^- \rightleftharpoons Mg}$ | $-2.37$ |
The more positive $E^\circ$, the more strongly that half-cell pulls electrons in: silver ions are eager to be reduced. The more negative, the more readily the reverse happens: magnesium gives up electrons easily, which is what "reactive metal" meant all along.
Join two half-cells and they compete. The one with the more positive $E^\circ$ wins and is reduced — it is the cathode. The other is forced to run backwards, as an oxidation — it is the anode. The voltage is the difference in their pulls:
$$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$$
For zinc and copper: $+0.34 - (-0.76) = +1.10$ V, which is exactly what a voltmeter across a zinc–copper cell reads. Subtracting a negative potential adds its size, which is why pairing a very negative metal with a positive one gives a big voltage.
A positive $E^\circ_{cell}$ means the cell works as written: the reaction is feasible under standard conditions. A negative one means the reaction runs the other way.
Another way: picture
Picture a tug of war over electrons, with each half-cell's $E^\circ$ as its strength. The team with the higher number pulls the rope towards itself and is reduced. The voltage is the difference between the two teams' strengths, not their total: two strong teams give a small voltage, a strong team against a weak one a large one.
Another way: steps
To use the table:
A single half-cell's potential cannot be measured on its own: a voltmeter needs two connections, and the second one makes a second half-cell. So chemists chose one half-cell as the reference and called it zero. The standard hydrogen electrode is a platinum plate coated in fine platinum black, dipped in $1$ mol/L hydrogen ions, with hydrogen gas at $100$ kPa bubbling over it at $25$ °C. Every other half-cell is connected to it in turn, and the voltmeter reading, with its sign, is that half-cell's $E^\circ$.
A copper half-cell against hydrogen reads $0.34$ V with copper positive, so copper's $E^\circ$ is $+0.34$ V. A zinc half-cell reads $0.76$ V with zinc negative, so zinc's is $-0.76$ V. Once every half-cell has been measured against the same reference, any two can be compared by subtraction — the way heights above sea level let you find the drop between two mountains without measuring it directly.
The table predicts reactions nobody needs to try. Will chlorine oxidise bromide ions? Chlorine would be reduced ($+1.36$ V) and bromide oxidised (the bromine couple, $+1.07$ V): $1.36 - 1.07 = +0.29$ V, positive, so yes — and chlorine water does turn potassium bromide solution orange. Will iodine oxidise bromide? $0.54 - 1.07$ is negative, so no.
Will copper dissolve in hydrochloric acid, giving hydrogen? Hydrogen ions would be reduced ($0.00$ V) and copper oxidised ($+0.34$ V): $0.00 - 0.34 = -0.34$ V. No. Zinc, at $-0.76$ V, gives $+0.76$ V and fizzes. The reactivity series from earlier chemistry is this table, read without the numbers.
Two cautions. A positive $E^\circ_{cell}$ says a reaction can happen, not that it will happen quickly — that is kinetics again, and some feasible reactions are very slow. And the values are for standard conditions; changing concentrations shifts potentials a little, as Le Chatelier would predict.
In $\mathrm{Cu + 2Ag^+ \rightarrow Cu^{2+} + 2Ag}$ the silver half-equation is doubled to balance the electrons. Its potential is not doubled. A potential is energy per unit of charge — joules per coulomb. Twice as many silver ions reduced carry twice the charge and release twice the energy, and the ratio is unchanged. A car battery twice the size has the same voltage; it just lasts longer. So $E^\circ_{cell} = 0.80 - 0.34 = +0.46$ V, whatever coefficients the balanced equation needs.
Add the two reduction potentials. One half-cell runs as an oxidation, backwards. Cathode minus anode takes care of that without changing any sign in the table.
Subtract the other way round. Anode minus cathode gives the negative of the right answer, and a feasible cell looks impossible.
Multiply by the coefficients. Potentials are per unit of charge and are never multiplied, however the equation is balanced.
The biggest number is the cathode, ignoring its sign. $-2.37$ V is not a strong pull; it is a strong tendency to be oxidised.
A positive $E^\circ_{cell}$ means a fast reaction. It means a feasible one. The rate is a separate question.
Magnesium, $E^\circ = -2.37$ V; silver, $E^\circ = +0.80$ V.
Two reduction potentials from the table.
Silver is more positive, so it is the cathode; magnesium is the anode.
The stronger pull is reduced.
$E^\circ_{cell} = 0.80 - (-2.37) = +3.17$ V.
Cathode minus anode; the minus signs make it a sum of sizes.
A nickel half-cell with a standard copper half-cell gives $0.59$ V, with copper positive.
Copper is the cathode.
$0.59 = 0.34 - E^\circ_{Ni}$.
Cathode minus anode, rearranged.
$E^\circ_{Ni} = 0.34 - 0.59 = -0.25$ V.
Below copper, as an anode's potential must be.
Copper is more positive, so it is the cathode.
The stronger pull.
$E^\circ_{cell} = 0.34 - (-0.44) = \ldots$
…$+0.78$ V, positive, so iron does push electrons to copper.
A cell is made from a half-cell of lead, $E^\circ = -0.13$ V, and a half-cell of copper, $E^\circ = 0.34$ V, both standard reduction potentials. What is the standard cell potential?
Answer: unit: V / mV
A half-cell of zinc ($E^\circ = -0.76$ V) is connected to a half-cell of silver ($E^\circ = 0.80$ V). Which statement is right?
Put these ions in order of strength as oxidising agents, strongest first. Their standard reduction potentials are given.
Number the steps in order (write the number in the box):
Use the standard reduction potentials: $\mathrm{Cl_2/Cl^-}$ $+1.36$, $\mathrm{Br_2/Br^-}$ $+1.07$, $\mathrm{I_2/I^-}$ $+0.54$, $\mathrm{Cu^{2+}/Cu}$ $+0.34$, $\mathrm{H^+/H_2}$ $0.00$, $\mathrm{Zn^{2+}/Zn}$ $-0.76$ V. Will each reaction happen under standard conditions?
| does it react? | |
|---|---|
| chlorine bubbled into potassium bromide solution | |
| iodine added to potassium bromide solution | |
| zinc placed in copper(II) sulfate solution | |
| copper placed in zinc sulfate solution | |
| copper placed in hydrochloric acid | |
| zinc placed in hydrochloric acid |
In the cell reaction $\mathrm{Cu + 2Ag^+ \rightarrow Cu^{2+} + 2Ag}$, the silver half-cell has $E^\circ = +0.80$ V and the copper half-cell $E^\circ = +0.34$ V. What is $E^\circ_{cell}$?
A half-cell of magnesium is connected to a standard copper half-cell, $E^\circ = +0.34$ V. The copper is the positive electrode, and the voltmeter reads $2.71$ V under standard conditions. What is the standard reduction potential of the magnesium half-cell?
Answer: unit: V / mV
A science museum builds a battery from zinc–silver cells, each giving $1.56$ V, to light a lamp that needs at least $4.5$ V. What is the fewest cells it must connect in series?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A cell is made from a chlorine half-cell, $\mathrm{Cl_2 + 2e^- \rightleftharpoons 2Cl^-}$ with $E^\circ = 1.36$ V, and a half-cell of silver with $E^\circ = 0.80$ V. What is the standard cell potential?
Answer: unit: V / mV
You can calculate a cell potential and use its sign. Say out loud why the potential of the silver half-cell is not doubled when the equation has two silver ions. Next: why some changes happen of their own accord at all — entropy.
11. Your turn: what is $E^\circ_{cell}$ for an iron–copper cell, with iron at $-0.44$ V and copper at $+0.34$ V?, step 3