Chemistry » Electrochemical Reactions » Processes In Electrochemical Cells

# Half-Cells and Half-Reactions

## Half-cells and half-reactions

Galvanic cells are actually made up of two half-cells. One half-cell contains the anode and an electrolyte containing the same metal cations. The other half-cell contains the cathode and an electrolyte containing the same metal cations. These half-cells are connected by a salt-bridge and the electrodes are connected through an external circuit.

Earlier in this tutorial we discussed a zinc-copper cell. This was made up of a zinc half-cell, containing a zinc electrode and a zinc(II) sulphate ($$\text{ZnSO}_{4}$$) electrolyte solution, and a copper half-cell, containing a copper electrode and a copper(II) sulphate solution.

### Definition: Half-cell

A half-cell is a structure that consists of a conductive electrode surrounded by a conductive electrolyte.

In each half-cell a half-reaction takes place:

• Copper plate

At the copper plate, there was an increase in mass. This means that $$\text{Cu}^{2+}$$ ions from the copper(II) sulfate solution were deposited onto the plate as atoms of copper metal. The half-reaction that takes place at the copper plate is:

$$\text{Cu}^{2+}(\text{aq}) + 2\text{e}^{-}$$ $$\to$$ $$\text{Cu}(\text{s})$$

As electrons are gained by the copper ions this is the $$\color{red}{\textbf{reduction half-reaction}}$$.

• Zinc plate

At the zinc plate, there was a decrease in mass. This means that some of the solid zinc goes into solution as $$\text{Zn}^{2+}$$ ions. The electrons remain on the zinc plate, giving it a negative charge. The half-reaction that takes place at the zinc plate is:

$$\text{Zn}(\text{s})$$ $$\to$$ $$\text{Zn}^{2+}(\text{aq}) + 2\text{e}^{-}$$

As electrons are lost by the zinc atoms this is the $$\color{blue}{\textbf{oxidation half-reaction}}$$.

• The overall reaction

You can then combine the two half-reactions from these two half-cells to get the overall reaction:

$$\color{blue}{\text{Zn(s)}} + \color{red}{\text{Cu}^{2+}\text{(aq)}} + \color{red}{\text{2e}^{-}} \to \color{blue}{\text{Zn}^{2+}\text{(aq)}} + \color{red}{\text{Cu(s)}} + \color{blue}{\text{2e}^{-}}$$ or, if we cancel the electrons:

$$\color{blue}{\text{Zn(s)}} + \color{red}{\text{Cu}^{2+}\text{(aq)}} \to \color{blue}{\text{Zn}^{2+}\text{(aq)}} + \color{red}{\text{Cu(s)}}$$

It is possible to look at the half-reaction taking place in a half-cell and determine which electrode is the anode and which is the cathode.

• $$\color{blue}{\textbf{Oxidation}}$$ is loss at the $$\color{blue}{\textbf{anode}}$$, therefore the oxidation half-reaction occurs in the half-cell containing the anode.

• $$\color{red}{\textbf{Reduction}}$$ is gain at the $$\color{red}{\textbf{cathode}}$$ so the reduction half-reaction occurs in the half-cell containing the cathode.

Remember that for an electrochemical cell the standard cell notation is:

$$\color{blue}{\textbf{Zn(s)}}$$$$|$$$$\color{blue}{\textbf{Zn}^{2+}\textbf{(aq)}}$$$$||$$$$\color{red}{\textbf{Cu}^{2+}\textbf{(aq)}}$$$$|$$$$\color{red}{\textbf{Cu(s)}}$$

$$|$$ = a phase boundary (solid/aqueous)

$$||$$ = the salt bridge

## Example: Understanding Galvanic Cells

### Question

For the following cell:

$$\text{Zn}(\text{s})|\text{Zn}^{2+}(\text{aq})||\text{Ag}^{+}(\text{aq})|\text{Ag}(\text{s})$$

1. Give the anode and cathode half-reactions.

2. Write the overall equation for the chemical reaction.

3. Give the direction of the current in the external circuit.

### Step 1: Identify the oxidation and reduction reactions

By convention in standard cell notation, the anode is written on the left and the cathode is written on the right. So, in this cell:

Zinc is the anode (solid zinc is oxidised).

Silver is the cathode (silver ions are reduced).

### Step 2: Write the two half-reactions

Oxidation is loss of electrons at the anode: $$\text{Zn}(\text{s})$$ $$\to$$ $$\text{Zn}^{2+}(\text{aq}) + 2\text{e}^{-}$$

Reduction is gain of electrons at the cathode: $$\text{Ag}^{+}(\text{aq}) + \text{e}^{-}$$ $$\to$$ $$\text{Ag}(\text{s})$$

### Step 3: Combine the half-reactions to get the overall equation

Balance the charge by multiplying the reduction half-reaction by $$\text{2}$$.

$$2\text{Ag}^{+}(\text{aq}) + 2\text{e}^{-}$$ $$\to$$ $$2\text{Ag}(\text{s})$$

$$\text{Zn}(\text{s}) + 2\text{Ag}^{+}(\text{aq})$$ $$\to$$ $$\text{Zn}^{2+}(\text{aq}) + \text{Ag}(\text{s})$$

### Step 4: Determine the direction of current flow

The solid zinc is oxidised to form zinc ions. These electrons are left on the zinc electrode (anode) making it negative.

The silver ions take electrons and are reduced to form solid silver. This makes the silver electrode (cathode) positive.

Electron flow is from negative to positive, so from the anode to the cathode. Conventional current is in the opposite direction to electron flow. Therefore current will flow from the cathode (silver) to the anode (zinc).

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