Standard Electrode Potentials and Reactivity of Metals Revision Notes for HSC SSCE Chemistry

Standard Electrode Potentials and Reactivity of Metals

Understanding metal reactivity through electrode potentials

Earlier, you learned about the activity series of metals by observing their reactions with water, dilute acids, oxygen, and through displacement reactions. Now, you can determine the same information using standard electrode potentials from an electrochemical table. This method provides a more quantitative and precise way to predict metal reactivity and the spontaneity of redox reactions.

Comparing reduction half-reactions

When comparing two reduction half-reactions, there is a fundamental principle to remember:

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Statement 12.11: Comparing Reduction Half-Reactions

The reduction half-reaction with the algebraically larger standard electrode potential ( $\varepsilon°$ ) proceeds as written, and it drives the other reaction in the reverse direction.

This means you need to compare the numerical values of the electrode potentials, taking into account whether they are positive or negative. For example, $+0.34$ V is algebraically larger than $-0.40$ V.

Example: Comparing aluminium and magnesium

Let's compare these two reduction half-reactions:

$\text{Al}^{3+}(aq) + 3e^- \rightarrow \text{Al}(s) \qquad \varepsilon°_{\text{Al}} = -1.66 \text{ V}$

$\text{Mg}^{2+}(aq) + 2e^- \rightarrow \text{Mg}(s) \qquad \varepsilon°_{\text{Mg}} = -2.37 \text{ V}$

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Worked Example: Determining Which Reaction Proceeds Forward

Given: Two reduction half-reactions for aluminium and magnesium with their electrode potentials.

Step 1: Compare the electrode potentials algebraically

Aluminium: $\varepsilon°_{\text{Al}} = -1.66$ V
Magnesium: $\varepsilon°_{\text{Mg}} = -2.37$ V
Since $-1.66 > -2.37$ , aluminium has the larger electrode potential

Step 2: Determine the direction of each reaction

The aluminium half-reaction has the larger $\varepsilon°$ , so it proceeds as written (reduction of Al³⁺ ions)
The magnesium reaction is driven in reverse (oxidation of Mg metal)

Step 3: Write the balanced overall reaction

$2\text{Al}^{3+}(aq) + 3\text{Mg}(s) \rightarrow 2\text{Al}(s) + 3\text{Mg}^{2+}(aq)$

Conclusion: Magnesium metal can displace aluminium from a solution containing aluminium ions.

The connection between electrode potential and metal reactivity

For metals, reactivity refers to how easily a metal can be oxidised to form positive ions. In other words, it measures how readily the oxidation half-reaction occurs for that metal.

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Key Relationship: Electrode Potential and Metal Reactivity

The more negative the standard electrode potential, the more reactive the metal.

This makes sense because:

A more negative $\varepsilon°$ means the reduction reaction is less favourable
This means the reverse reaction (oxidation) is more favourable
More favourable oxidation equals higher reactivity

Examples of metal reactivity

Consider these comparisons:

Magnesium ( $\varepsilon° = -2.37$ V) is more reactive than iron ( $\varepsilon° = -0.45$ V)
Aluminium ( $\varepsilon° = -1.66$ V) is more reactive than tin ( $\varepsilon° = -0.14$ V)

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Notice how the numerical value matters! Even though both aluminium and magnesium have negative electrode potentials, magnesium is more reactive because its value is more negative (further from zero).

The activity series and standard electrode potentials

The activity series you studied previously is actually a list of metals arranged in order of increasing electrode potential. When you look at a table of standard electrode potentials:

The section with negative $\varepsilon°$ values represents the metal activity series
Metals become less reactive as their electrode potential increases (becomes less negative or more positive)

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Understanding the Pattern

Reactivity decreases as you move up the electrode potential table:

Zinc ( $\varepsilon° = -0.76$ V) → High reactivity
Lead ( $\varepsilon° = -0.13$ V) → Moderate reactivity
Copper ( $\varepsilon° = +0.34$ V) → Low reactivity
Silver ( $\varepsilon° = +0.80$ V) → Very low reactivity

This pattern directly corresponds to the activity series—it's the same information presented in a quantitative way!

Predicting spontaneous reactions

A spontaneous reaction is one that occurs naturally as written, without requiring an external energy source. For redox reactions, there's a simple criterion to determine spontaneity:

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Criterion for Spontaneous Redox Reactions

A redox reaction is spontaneous if it has a positive standard cell voltage.

This means:

Calculate the overall cell voltage using: $E°_{\text{total}} = E°_1 + E°_2$
If $E°_{\text{total}} > 0$ , the reaction is spontaneous as written
If $E°_{\text{total}} < 0$ , the reverse reaction is spontaneous

Worked example: Predicting spontaneity of the cadmium-copper reaction

Let's determine whether cadmium metal will react spontaneously with copper sulfate solution. There are two methods to solve this problem.

Method 1: Calculating the total cell voltage

This method involves breaking the overall reaction into half-reactions, assigning voltages, and calculating the total.

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Key Points from Method 1:

The oxidation half-reaction (Cd → Cd²⁺) requires reversing the sign: $E° = -(-0.40) = +0.40$ V
The reduction half-reaction (Cu²⁺ → Cu) uses the value as given: $E° = +0.34$ V
Total cell voltage: $E°_{\text{total}} = +0.40 + 0.34 = +0.74$ V
Since $E°_{\text{total}} = +0.74$ V is positive, the reaction is spontaneous as written
Conclusion: Cadmium metal will reduce copper ions to copper metal

Method 2: Comparing electrode potentials

This method compares the two reduction half-reactions directly.

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Key Points from Method 2:

Compare the two reduction potentials directly:
- Copper: $\varepsilon° = +0.34$ V
- Cadmium: $\varepsilon° = -0.40$ V
Copper has the higher $\varepsilon°$ value
Therefore, the copper half-reaction proceeds forward (Cu²⁺ ions are reduced)
The cadmium half-reaction proceeds in reverse (Cd metal is oxidised)
Conclusion: The reaction is spontaneous as written

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Comparing the Two Methods

Both methods give the same conclusion, but Method 2 is often quicker when you simply need to determine the direction of spontaneity. Method 1 is more useful when you need to calculate the actual cell voltage for quantitative analysis.

Oxidising and reducing agents

Standard electrode potentials also tell us about the strength of oxidising and reducing agents. Remember that in a reduction half-reaction:

$\text{oxidised form} + ne^- \rightarrow \text{reduced form}$

The species on the left is the oxidant (it accepts electrons), while the species on the right is the reductant (it can donate electrons in the reverse reaction).

Oxidising agent strength

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Relationship Between $\varepsilon°$ and Oxidising Power

The greater the value of $\varepsilon°$ , the stronger the oxidising power of the oxidised form (left-hand side) of the half-reaction.

Example: Comparing bromine and iodine as oxidants:

Bromine: $\varepsilon° = +1.09$ V
Iodine: $\varepsilon° = +0.54$ V
Bromine is a stronger oxidising agent than iodine

This makes sense because a higher $\varepsilon°$ means the reduction reaction is more favourable, so the species more readily accepts electrons.

Reducing agent strength

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Relationship Between $\varepsilon°$ and Reducing Power

The algebraically smaller (more negative) the value of $\varepsilon°$ , the stronger the reducing power of the reduced form (right-hand side) of the half-reaction.

Example: Comparing aluminium and nickel as reductants:

Aluminium: $\varepsilon° = -1.66$ V
Nickel: $\varepsilon° = -0.26$ V
Aluminium is a stronger reducing agent than nickel

This makes sense because a more negative $\varepsilon°$ means the reduction is less favourable, so the reverse (oxidation) is more favourable—making it a better electron donor.

Key concepts

When comparing two reduction half-reactions, the one with the larger $\varepsilon°$ proceeds as written and drives the other one in the reverse direction
The reactivity of metals decreases as standard electrode potential increases (becomes less negative)
A redox reaction is spontaneous if it has a positive standard cell voltage ( $E°_{\text{total}} > 0$ )
The greater the value of $\varepsilon°$ , the greater the oxidising strength of the oxidised form of the half-reaction
The algebraically smaller (more negative) the value of $\varepsilon°$ , the greater the reducing strength of the reduced form of the half-reaction

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Key Points to Remember:

More negative electrode potential = more reactive metal. Metals with very negative $\varepsilon°$ values (like magnesium at $-2.37$ V) are highly reactive.
Positive total cell voltage = spontaneous reaction. Always calculate $E°_{\text{total}}$ to predict whether a redox reaction will occur naturally.
Higher $\varepsilon°$ = stronger oxidant. Species at the top of the electrode potential table (with large positive values) are powerful oxidising agents.
Lower $\varepsilon°$ = stronger reductant. Species at the bottom of the electrode potential table (with large negative values) are powerful reducing agents.
Two methods, same answer. You can determine spontaneity by either calculating the total cell voltage or comparing electrode potentials directly—both will give you the correct prediction.

Standard Electrode Potentials and Reactivity of Metals (HSC SSCE Chemistry): Revision Notes