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Chemistry » Chemistry 111: Electrochemical Reactions » Standard Electrode Potentials

# EMF of a Cell

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## EMF of a cell

Using the example of the zinc and copper half-cells, we know that when these two half-cells are combined, zinc will be the oxidation half-reaction and copper will be the reduction half-reaction. A voltmeter connected to this cell will show that the zinc electrode is more negative than the copper electrode.

$$\color{red}{\text{Cu}^{2+}\text{(aq) + 2e}^{-} \rightleftharpoons \text{Cu(s)}}$$ (E° = $$\text{+0.34}$$ $$\text{V}$$)

$$\color{blue}{\text{Zn}^{2+}\text{(aq) + 2e}^{-} \rightleftharpoons \text{Zn(s)}}$$ (E° = $$-\text{0.76}$$ $$\text{V}$$)

The reading on the meter will show the potential difference between the two half-cells. This is known as the EMF of the cell. The higher the EMF, the greater amount of energy released per unit charge.

### Definition: EMF of a cell

The EMF of a cell is defined as the maximum potential difference between two electrodes or half-cells in a galvanic cell.

#### Fact:

The EMF of a cell is a the same as the voltage across a disconnected cell (electric circuit theory). A voltmeter is effectively a high resistance ammeter, so a very small current will flow when a voltmeter reading is taken (although this is too small to be noticeable).

#### Tip:

Remember:

Standard conditions are:

p = $$\text{101.3}$$ $$\text{kPa}$$

C = $$\text{1}$$ $$\text{mol.dm^{-3}}$$

T = $$\text{298}$$ $$\text{K}$$.

$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_$$

In standard cell notation the $$\color{blue}{\textbf{anode half-cell}}$$ is always written on the $$\color{blue}{\text{left}}$$ and the $$\color{red}{\textbf{cathode half-cell}}$$ is always written on the $$\color{red}{\text{right}}$$.

$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_$$

The $$\color{blue}{\text{reducing agent}}$$ is being $$\color{blue}{\text{oxidised}}$$. $$\color{blue}{\text{Oxidation}}$$ is a loss of electrons at the $$\color{blue}{\text{anode}}$$.

The $$\color{red}{\text{oxidising agent}}$$ is being $$\color{red}{\text{reduced}}$$. $$\color{red}{\text{Reduction}}$$ is a gain of electrons at the $$\color{red}{\text{cathode}}$$.

It is important to be able to calculate the EMF of an electrochemical cell. To calculate the EMF of a cell:

• take the E° of the atom that is being $$\color{red}{\text{reduced}}$$

• subtract the E° of the atom that is being $$\color{blue}{\text{oxidised}}$$

The reason for defining a reference electrode becomes obvious now. Potential differences can be calculated from the electrode potentials (determined relative to the hydrogen half-cell) without having to construct the cells themselves each time.

You can use any one of the following equations:

• E°$$_{\text{(cell)}}=$$ E°($$\color{red}{\text{reduction half-reaction}}$$) – E°($$\color{blue}{\text{oxidation half-reaction}}$$)

• E°$$_{\text{(cell)}}=$$ E°($$\color{red}{\text{oxidising agent}}$$) – E°($$\color{blue}{\text{reducing agent}}$$)

• E°$$_{\text{(cell)}}=$$ E°($$\color{red}{\text{cathode}}$$) – E°($$\color{blue}{\text{anode}}$$)

So, for the $$\text{Zn}$$-$$\text{Cu}$$ cell:

E°$$_{\text{(cell)}} =$$ $$\text{0.34}$$ – ($$-\text{0.76}$$) = $$\text{0.34}$$ + $$\text{0.76}$$ = $$\text{1.10}$$ $$\text{V}$$

### Definition: Standard EMF

Standard EMF (E°$$_{\text{cell}}$$) is the EMF of a galvanic cell operating under standard conditions. The symbol ° denotes standard conditions.

## Example: Calculating the EMF of a Cell

### Question

A cell contains a solid lead anode in a gold ion solution.

1. Represent the cell using standard notation.

2. Calculate the cell potential (EMF) of the electrochemical cell.

### Step 1: Find appropriate reactions on the table of standard electrode potentials

The reaction involves lead and gold.

$$\text{Pb}^{2+}(\text{aq}) + 2\text{e}^{-}$$ $$\rightleftharpoons$$ $$\text{Pb}(\text{s})$$ (E° V = $$-\text{0.13}$$ $$\text{V}$$)

$$\text{Au}^{3+}(\text{aq}) + 3\text{e}^{-}$$ $$\rightleftharpoons$$ $$\text{Au}(\text{s})$$ (E° V = $$\text{+1.50}$$ $$\text{V}$$)

### Step 2: Which metal is more likely to be reduced, which is more likely to be oxidised?

The E° of lead is a small, negative value and the E° of gold is a large, positive number. Therefore lead is more easily oxidised than gold, and gold is more easily reduced than lead.

### Step 3: Determine which metal is the cathode and which is the anode

Oxidation is loss at the anode, therefore lead is the anode.

Reduction is gain at the cathode, therefore gold is the cathode.

### Step 4: Represent the cell using standard cell notation

The anode is always written on the left, the cathode on the right.

$$\text{Pb}(\text{s})|\text{Pb}^{2+}(\text{aq})||\text{Au}^{3+}(\text{aq})|\text{Au}(\text{s})$$

### Step 5: Calculate the cell potential

E°$$_{\text{(cell)}}=$$ E°$$_\text{(cathode)}$$ – E°$$_\text{(anode)}$$

E°$$_{\text{(cell)}}=$$ E°$$_\text{(gold)}$$ – E°$$_\text{(lead)}$$

= $$\text{+1.50}$$ – ($$-\text{0.13}$$)

= $$\text{+1.63}$$ $$\text{V}$$

## Example: Calculating the EMF of a Cell

### Question

Calculate the cell potential of the electrochemical cell in which the following reaction takes place, and represent the cell using standard notation.

$$\text{Mg}(\text{s}) + 2\text{H}^{+}(\text{aq})$$ $$\to$$ $$\text{Mg}^{2+}(\text{aq}) + \text{H}_{2}(\text{g})$$

### Step 1: Find appropriate reactions on the table of standard electrode potentials

$$\text{Mg}^{2+}(\text{aq}) + 2\text{e}^{-}$$ $$\rightleftharpoons$$ $$\text{Mg}(\text{s})$$ (E° = $$-\text{2.37}$$ $$\text{V}$$)

$$2\text{H}^{+}(\text{aq}) + 2\text{e}^{-}$$ $$\rightleftharpoons$$ $$\text{H}_{2}(\text{g})$$ (E° = $$\text{0.00}$$ $$\text{V}$$)

### Step 2: Which element is more likely to be reduced, which is more likely to be oxidised?

The E° of magnesium is a more negative value than the E° of hydrogen. Therefore magnesium is more easily oxidised than hydrogen, and hydrogen is more easily reduced than magnesium.

### Step 3: Determine which metal is the cathode and which is the anode

Oxidation is loss at the anode, therefore magnesium is the anode.

Reduction is gain at the cathode, therefore hydrogen is the cathode reaction.

### Step 4: Represent the cell using standard notation

The anode is always written on the left, the cathode on the right. The hydrogen electrode includes an inert platinum plate.

$$\text{Mg}(\text{s})|\text{Mg}^{2+}(\text{aq})||\text{H}^{+}(\text{aq}),\text{H}_{2}(\text{g})|\text{Pt}(\text{s})$$

### Step 5: Calculate the cell potential

E$$_{\text{(cell)}} =$$ E°$$_\text{(cathode)}$$ – E°$$_\text{(anode)}$$

E$$_{\text{(cell)}} =$$ E°$$_\text{(hydrogen)}$$ – E°$$_\text{(magnesium)}$$

= $$\text{0.00}$$ – ($$-\text{2.37}$$)

=$$\text{+2.37}$$ $$\text{V}$$