Concentration cells and emf from solubility

A cell with nothing to react

Put a spoonful of salt in one glass of water and a pinch of salt in another. Join the two glasses with a wet thread. Salt slowly moves from the strong glass to the weak one, until both have the same strength. Nature pushes the two strengths to match.

A cell can use the same push. Take two nickel rods. Dip one in a dilute solution of nickel sulphate and the other in a concentrated solution. Join the solutions with a salt bridge and the rods with a wire. A voltmeter shows a small voltage, although both electrodes are the same metal in the same ion. This is a concentration cell.

Which rod is the anode? Think about what each side wants.

• The dilute side has few $\text{Ni}^{2+}$ ions. Nickel metal dissolves there to make more: $\text{Ni} \rightarrow \text{Ni}^{2+} + 2e^-$. That is oxidation, so the dilute side is the anode.
• The concentrated side has many ions. Nickel ions take electrons and leave the solution as metal: $\text{Ni}^{2+} + 2e^- \rightarrow \text{Ni}$. That is reduction, so the concentrated side is the cathode.

Electrons go through the wire from the dilute side to the concentrated side. Ions move through the salt bridge. The net result is that nickel ions are carried from the strong solution to the weak one, and the cell does electrical work on the way.

The emf of a concentration cell

Both electrodes have the same $E^\circ$, so $E^\circ_{cell} = 0$. All the emf comes from the Nernst term. Write the cell with the dilute solution, $c_1$, on the left:

The cell reaction is $\text{Ni}^{2+}(c_2) \rightarrow \text{Ni}^{2+}(c_1)$, so $Q = \dfrac{c_1}{c_2}$. For an ion with $n$ positive charges:

Here $c_2$ is the larger concentration, so the emf is positive, as it must be for a cell that runs. Three facts follow.

• Only the ratio of the two concentrations matters. A cell with $0.001\ \text{mol/L}$ and $0.1\ \text{mol/L}$ has the same emf as one with $0.01\ \text{mol/L}$ and $1\ \text{mol/L}$.
• The metal matters only through $n$. Every metal ion with $n = 2$ gives the same emf for the same ratio.
• The emf is small. A hundredfold ratio gives $0.118\ \text{V}$ for $n = 1$ and $0.0591\ \text{V}$ for $n = 2$.

A worked example

The cell in the picture is $\text{Ni} \mid \text{Ni}^{2+}(0.0040\ \text{mol/L}) \parallel \text{Ni}^{2+}(0.80\ \text{mol/L}) \mid \text{Ni}$ at $298\ \text{K}$.

1. The ratio is $\dfrac{0.80\ \text{mol/L}}{0.0040\ \text{mol/L}} = 200$, and $\log 200 = 2.301$.
2. Nickel ions have $n = 2$.
3. $E = \dfrac{0.0591\ \text{V}}{2} \times 2.301 = 0.0680\ \text{V}$.

The dilute side is the anode. The reading is $0.068\ \text{V}$, which a good voltmeter shows easily.

Try this. Keep the ratio the same but move both concentrations up. Does the emf change? Now change $n$ from 2 to 1 and read the new emf. Then make the ratio 1, with both sides equal. What does the cell give?

A cell that goes flat

When the cell delivers current, the dilute side gains ions and the concentrated side loses them. The ratio falls towards 1, and the emf falls towards zero. When the two sides are equal, the emf is exactly zero and the cell is flat.

This is equilibrium for the change $\text{Ni}^{2+}(c_2) \rightarrow \text{Ni}^{2+}(c_1)$. It stops when the two concentrations are equal, so $K = 1$ and $E^\circ = 0$. The Gibbs energy released is $\Delta G = -nFE$, and it is the work that the cell can do while it levels the two solutions.

Gas concentration cells

The same idea works for a gas. Take two hydrogen electrodes in the same acid. One has hydrogen at $4.0\ \text{bar}$ and the other at $0.25\ \text{bar}$. The cell reaction is $\text{H}_2(4.0\ \text{bar}) \rightarrow \text{H}_2(0.25\ \text{bar})$, with $n = 2$:

Which electrode is the anode here? The one with the higher pressure. Write each electrode as a reduction and look at where the gas stands.

• For hydrogen, $2\text{H}^+ + 2e^- \rightarrow \text{H}_2$. The gas is a product. More gas lowers the potential, so the high-pressure electrode is the anode.
• For chlorine, $\text{Cl}_2 + 2e^- \rightarrow 2\text{Cl}^-$. The gas is a reactant. More gas raises the potential, so the high-pressure electrode is the cathode.

So the rule is not "high pressure is always the anode". It depends on which side of the half-reaction the gas is on.

A concentration cell made by nature

A piece of steel standing in sea water has no salt bridge and no wire, but it can still behave as a concentration cell, with oxygen as the thing that differs. Near the surface, oxygen from the air dissolves easily, so that part of the steel is a cathode where oxygen is reduced. A little below the surface there is less oxygen, so that part is the anode, where iron dissolves. This is why steel piles rust fastest just below the waterline.

Hunting a very small solubility

Lead iodide is a pale yellow solid that dissolves so little that it is hard to weigh what dissolves. A concentration cell can find the few ions in a saturated solution.

Build the cell $\text{Pb} \mid \text{Pb}^{2+}(\text{saturated PbI}_2) \parallel \text{Pb}^{2+}(0.100\ \text{mol/L}) \mid \text{Pb}$. The saturated solution has the lower concentration of lead ions, so it is the anode side. At $298\ \text{K}$ the emf is $0.0579\ \text{V}$.

1. Lead gives $n = 2$, and $E = \dfrac{0.0591\ \text{V}}{2}\log\dfrac{0.100\ \text{mol/L}}{s}$, where $s$ is $[\text{Pb}^{2+}]$ in the saturated solution.
2. $\log\dfrac{0.100}{s} = \dfrac{0.0579}{0.02955} = 1.959$, so $\dfrac{0.100}{s} = 91.0$ and $s = 1.10 \times 10^{-3}\ \text{mol/L}$.
3. One unit of $\text{PbI}_2$ gives one $\text{Pb}^{2+}$ and two $\text{I}^-$. So $[\text{Pb}^{2+}] = s$ and $[\text{I}^-] = 2s$.
4. $K_{sp} = [\text{Pb}^{2+}][\text{I}^-]^2 = s\,(2s)^2 = 4s^3 = 4 \times (1.10 \times 10^{-3})^3 = 5.3 \times 10^{-9}\ \text{mol}^3/\text{L}^3$.

The number is tiny, as it should be for a salt that hardly dissolves. If you know $K_{sp}$, you can go the other way: find $s$, then the emf. For a salt of the type $\text{MX}$ the product is $K_{sp} = s^2$, and for the type $\text{MX}_2$ it is $4s^3$.

Practise with new numbers

Try it in Python

The program finds the emf of concentration cells for three ratios. Then it turns the emf of the lead iodide cell into a solubility and a solubility product.

Common mistakes

• Putting the anode on the concentrated side. For a metal and its ion, the dilute side is the anode.
• Taking the log of the difference. The emf depends on the log of the ratio of the concentrations.
• Looking for $E^\circ$. A concentration cell has $E^\circ = 0$, and the emf is all Nernst term.
• Forgetting $n$. An ion with two charges gives half the emf of an ion with one charge, for the same ratio.
• Using the same pressure rule for every gas. Check which side of the half-reaction the gas is on.
• Writing $K_{sp} = s^2$ for $\text{MX}_2$. The ion $\text{X}^-$ is $2s$, and it is squared, so the product is $4s^3$.

Remember these

• A concentration cell has two identical electrodes in solutions of different strength. $E^\circ = 0$.
• $E = \dfrac{0.0591\ \text{V}}{n}\log\dfrac{c_{conc}}{c_{dil}}$ at $298\ \text{K}$. The dilute side is the anode.
• The cell runs until the two strengths are equal, and then $E = 0$.
• For a gas, write the electrode as a reduction. A gas on the left raises the potential when its pressure rises, and a gas on the right lowers it.
• A saturated solution of a sparingly soluble salt can be one half of a concentration cell. From the emf we get $s$, and then $K_{sp}$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A cell has two cobalt electrodes in cobalt sulphate solutions of $0.0050\ \text{mol/L}$ and $0.50\ \text{mol/L}$. Which side is the anode, and what is the emf at $298\ \text{K}$?

2. The ratio of concentrations in a cell with $n = 2$ is made ten times larger. By how much does the emf rise?

3. Why does a concentration cell have no standard emf, and why does it stop working?

4. Two hydrogen electrodes dip into the same acid, one at $2.0\ \text{bar}$ and one at $0.50\ \text{bar}$. Which is the anode, and what is the emf?

5. For which type of salt is $K_{sp} = 4s^3$? Give one example with the right formula.