Galvanic Cells and Cell Diagrams – Electrochemical Cells and Applications (Senior 6 Chemistry) | Ganzaa

Electrochemical Cells and Applications

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Senior 6 · Chemistry

Electrochemical Cells and Applications

5 lessons

Lesson 1 of 5

Galvanic Cells and Cell Diagrams

25 min

Getting electricity from a displacement reaction

Put a strip of zinc into blue copper(II) sulfate solution and, as you saw in earlier years, the zinc becomes coated with brown copper, the blue colour fades and the solution warms up: $\mathrm{Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)}$ . Electrons pass directly from zinc atoms to copper ions, and the energy released is wasted as heat.

Now keep the two halves of the reaction apart: zinc in one beaker, copper ions in another, joined by a wire and a "salt bridge". The same electrons are forced to travel through the wire, and they can light a lamp on the way. This is a galvanic cell (also called a voltaic cell), the basis of every battery. A lemon with two different metals pushed into it is a simple version.

Definition

Galvanic (voltaic) cell: a device in which a spontaneous redox reaction produces an electric current, because oxidation and reduction take place in separate half-cells. Example: the Daniell cell (zinc and copper).

Half-cell: an electrode in contact with a solution containing the oxidised and reduced forms of one redox couple. Example: a zinc rod in $\mathrm{Zn^{2+}(aq)}$ , the $\mathrm{Zn^{2+}/Zn}$ half-cell.

Anode: the electrode where oxidation happens. Cathode: the electrode where reduction happens. In a galvanic cell the anode is the negative terminal and the cathode the positive one.

Salt bridge: a tube or strip of filter paper soaked in a concentrated solution of an unreactive salt, such as $\mathrm{KNO_3}$ , that joins the two solutions and lets ions move between them, completing the circuit.

Cell emf, $E_{\text{cell}}$ : the potential difference between the two electrodes when no current flows, measured in volts with a high-resistance voltmeter. Example: 1.10 V for the standard Daniell cell.

The Daniell cell

Figure 1.1 shows a zinc half-cell joined to a copper half-cell.

Figure 1.1: In the Daniell cell, electrons flow through the wire from the zinc anode to the copper cathode, and ions move through the salt bridge.

At the zinc anode, zinc atoms are oxidised: $\mathrm{Zn(s) \rightarrow Zn^{2+}(aq) + 2e^{-}}$ . The electrons left behind make the zinc negative, and the rod slowly gets thinner.
At the copper cathode, copper(II) ions take electrons and are reduced: $\mathrm{Cu^{2+}(aq) + 2e^{-} \rightarrow Cu(s)}$ . Copper is deposited and the blue colour fades.
In the wire, electrons flow from zinc to copper. (Conventional current is drawn the other way, from + to −.)
In the salt bridge, ions carry the charge: nitrate ions move towards the zinc side, where positive $\mathrm{Zn^{2+}}$ is being produced, and potassium ions move towards the copper side, where $\mathrm{Cu^{2+}}$ is being removed. Without the salt bridge, each solution would at once build up an excess charge and the current would stop.

Adding the two half-equations gives the overall cell reaction, $\mathrm{Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)}$ : the displacement reaction, now made to do electrical work.

Writing cell diagrams

Chemists write a cell in a shorthand called a cell diagram (or cell notation):

$\mathrm{Zn(s)\,|\,Zn^{2+}(aq)\,\|\,Cu^{2+}(aq)\,|\,Cu(s)} \qquad E_{\text{cell}} = +1.10\,\mathrm{V}$

Symbol or rule	Meaning
left-hand electrode	the anode (oxidation); right-hand electrode: the cathode (reduction)
single vertical line, \|	a boundary between phases, e.g. solid metal and solution
double vertical line, ‖	the salt bridge
comma	two species in the same phase, e.g. $\mathrm{Fe^{3+}(aq), Fe^{2+}(aq)}$
Pt(s) at an end	an inert platinum electrode, used when neither form of the couple is a metal that can carry electrons

For example, a cell in which iron(II) ions are oxidised and silver ions are reduced is written $\mathrm{Pt(s)\,|\,Fe^{2+}(aq), Fe^{3+}(aq)\,\|\,Ag^{+}(aq)\,|\,Ag(s)}$ .

Worked example

Problem. A magnesium half-cell is joined to a silver half-cell. Magnesium is the more reactive metal. (a) Write the half-equations at each electrode and name the anode and cathode. (b) Write the overall equation. (c) Write the cell diagram. (d) State the direction of electron flow in the wire.

Solution.

(a) The more reactive metal is oxidised. Anode (−): $\mathrm{Mg(s) \rightarrow Mg^{2+}(aq) + 2e^{-}}$ . Cathode (+): $\mathrm{Ag^{+}(aq) + e^{-} \rightarrow Ag(s)}$ .

(b) Balance the electrons by doubling the silver half-equation: $\mathrm{Mg(s) + 2Ag^{+}(aq) \rightarrow Mg^{2+}(aq) + 2Ag(s)}$ .

(d) From the magnesium electrode to the silver electrode.

Answer: as above. Note that the salt bridge here must not contain chloride: $\mathrm{Ag^{+}}$ would precipitate as AgCl. Potassium nitrate is used.

Worked example

Problem. For the cell $\mathrm{Pt(s)\,|\,I^{-}(aq), I_2(aq)\,\|\,Fe^{3+}(aq), Fe^{2+}(aq)\,|\,Pt(s)}$ , write the reactions at each electrode and the overall reaction, and explain why both electrodes are platinum.

Solution. Left (anode, oxidation): $\mathrm{2I^{-}(aq) \rightarrow I_2(aq) + 2e^{-}}$ . Right (cathode, reduction): $\mathrm{Fe^{3+}(aq) + e^{-} \rightarrow Fe^{2+}(aq)}$ . Doubling the second and adding:

$\mathrm{2Fe^{3+}(aq) + 2I^{-}(aq) \rightarrow 2Fe^{2+}(aq) + I_2(aq)}$

All four species are dissolved ions or molecules; none is a solid conductor. Platinum is unreactive and conducts, so it can pass electrons to and from the solution without taking part in the reaction.

Answer: as above.

Warning

Common mistakes

Saying electrons flow through the salt bridge. Electrons flow only in the wire; the salt bridge carries ions.
Assuming the anode is always positive. In a galvanic cell it is negative; in an electrolytic cell (Senior 3) it is positive. What never changes: oxidation at the anode, reduction at the cathode.
Replacing the salt bridge with a metal wire. A wire would set up two extra electrode reactions, not complete the ionic circuit.
Putting spectator ions into the cell diagram, or writing the cathode on the left.

Tip

"An Ox, Red Cat": ANode = OXidation, REDuction = CAThode. In cell diagrams, think "oxidation on the left, reduction on the right", in the same order as the electrons travel.

Tyler DeWitt: Galvanic cells (voltaic cells). Watch how the electrons move through the wire and the ions through the salt bridge, and why the zinc electrode gets lighter while the copper electrode gets heavier.

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Practice

State the function of the salt bridge in a galvanic cell, and suggest why potassium nitrate is a better choice than potassium chloride when one half-cell contains silver ions.
In the Daniell cell, which electrode gains mass, and which loses mass? Explain with half-equations.
Write the cell diagram for a cell made from a nickel half-cell (anode) and a copper half-cell (cathode), and the overall equation.
For the cell $\mathrm{Al(s)\,|\,Al^{3+}(aq)\,\|\,Ni^{2+}(aq)\,|\,Ni(s)}$ , write the half-equations and the balanced overall equation.
Write a cell diagram for a cell in which chlorine gas is reduced to chloride ions at a platinum electrode and zinc is oxidised.
A student sets up a zinc–copper cell but finds the voltmeter reads zero. Suggest two possible reasons.
Explain why the energy from the reaction of zinc with copper(II) sulfate appears as heat in a beaker but as electrical energy in a Daniell cell.

Show answersHide answers

It completes the circuit by letting ions move between the two solutions, keeping each solution electrically neutral. $\mathrm{Cl^{-}}$ would react with $\mathrm{Ag^{+}}$ to give a precipitate of AgCl, removing silver ions and blocking the bridge; nitrate does not react.
Copper gains mass: $\mathrm{Cu^{2+} + 2e^{-} \rightarrow Cu}$ deposits copper. Zinc loses mass: $\mathrm{Zn \rightarrow Zn^{2+} + 2e^{-}}$ dissolves it.
$\mathrm{Ni(s)\,|\,Ni^{2+}(aq)\,\|\,Cu^{2+}(aq)\,|\,Cu(s)}$ ; $\mathrm{Ni + Cu^{2+} \rightarrow Ni^{2+} + Cu}$ .
$\mathrm{Al \rightarrow Al^{3+} + 3e^{-}}$ (×2); $\mathrm{Ni^{2+} + 2e^{-} \rightarrow Ni}$ (×3); overall $\mathrm{2Al + 3Ni^{2+} \rightarrow 2Al^{3+} + 3Ni}$ .
$\mathrm{Zn(s)\,|\,Zn^{2+}(aq)\,\|\,Cl_2(g)\,|\,Cl^{-}(aq)\,|\,Pt(s)}$ . (The order $\mathrm{Cl_2(g), Cl^{-}(aq)}$ is also seen; Pt is on the outside.)
Any two: no salt bridge, or it has dried out; a broken or loose connection; both electrodes made of the same metal in the same solution; the voltmeter switched off or on the wrong range.
In the beaker, electrons pass directly from Zn atoms to $\mathrm{Cu^{2+}}$ ions at the metal surface, so the energy is released as heat in the solution. In the cell, the two half-reactions are separated, so the electrons must travel through the external circuit, where their energy can do electrical work.

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.
Image: Lemon battery experiment.JPG by ChristianSW, via Wikimedia Commons, licensed under CC BY 3.0.
Video: Galvanic Cells (Voltaic Cells), Tyler DeWitt, embedded from YouTube.
The text, cell diagram drawing, 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.

1Getting electricity from a displacement reaction

2The Daniell cell

3Writing cell diagrams

Getting electricity from a displacement reaction

The Daniell cell

Writing cell diagrams