Standard Electrode Potentials Revision Notes for Grade 12 NSC Matric Physical Sciences

Standard Electrode Potentials

What are standard electrode potentials?

Standard electrode potentials are measurements that show how easily different substances gain or lose electrons in chemical reactions. These values help us predict which reactions will happen spontaneously and calculate the voltage that electrochemical cells can produce.

The position of equilibrium in electrochemical reactions can change based on conditions like concentration, temperature, and pressure. To make fair comparisons between different substances, we use standard conditions:

Pressure: 101.3 kPa (1 atmosphere)
Temperature: 298 K (25°C)
Concentration: 1 mol.dm⁻³

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Using these standard conditions ensures that electrode potential values can be compared directly and used to calculate potential differences between electrodes without constructing specific cells each time.

The standard hydrogen electrode

To measure electrode potentials, we need a reference point. The standard hydrogen electrode (SHE) serves as this reference and is assigned a potential of exactly 0.00 V.

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Definition: Standard hydrogen electrode

The standard hydrogen electrode is a redox electrode which forms the basis of the scale of oxidation-reduction potentials.

The standard hydrogen electrode consists of a platinum electrode in a solution containing 1 mol.dm⁻³ H⁺ ions at 25°C. Hydrogen gas at 1 atmospheric pressure bubbles over the platinum electrode. The reaction occurring is:

$2H^+(aq) + 2e^- \rightleftharpoons H_2(g)$

When other electrodes are connected to the standard hydrogen electrode, we can measure their relative electrode potentials. For example:

If zinc is connected to the SHE, the reading is -0.76 V (zinc is more negative)
If copper is connected to the SHE, the reading is +0.34 V (copper is more positive)

Understanding the standard electrode potential table

The standard electrode potential table lists reduction half-reactions arranged from most negative to most positive E° values. This arrangement tells us about the relative strengths of reducing and oxidising agents.

Key patterns in the table:

Large negative values (e.g. $Li^+ + e^- \rightleftharpoons Li$ , E° = -3.04 V):

The element or compound ionises easily
It releases electrons easily
The substance is easily oxidised
It acts as a good reducing agent

Large positive values (e.g. $Au^{3+} + 3e^- \rightleftharpoons Au$ , E° = +1.50 V):

The element or compound gains electrons easily
The substance is easily reduced
It acts as a good oxidising agent

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Important trends:

Reducing ability decreases as you move down the table (from negative to positive values)
Oxidising ability increases as you move down the table (from negative to positive values)

Using electrode potentials to predict reactions

The electrode potential values help us predict which reactions will occur. The substance with the higher (more positive) electrode potential will be reduced, while the substance with the lower (more negative) electrode potential will be oxidised.

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Worked Example: Predicting reactions

Question: In a cell containing $Cu^{2+}(aq) + 2e^- \rightleftharpoons Cu(s)$ and $Ag^+(aq) + e^- \rightleftharpoons Ag(s)$ , determine which reaction is oxidation and which is reduction.

Solution: Step 1: Find electrode potentials from the table

$Cu^{2+}(aq) + 2e^- \rightleftharpoons Cu(s)$ : E° = +0.34 V
$Ag^+(aq) + e^- \rightleftharpoons Ag(s)$ : E° = +0.80 V

Step 2: Compare values Silver has a higher electrode potential (+0.80 V > +0.34 V), so silver is more easily reduced than copper.

Step 3: Write the half-reactions

Reduction (gain of electrons): $Ag^+(aq) + e^- \rightarrow Ag(s)$
Oxidation (loss of electrons): $Cu(s) \rightarrow Cu^{2+}(aq) + 2e^-$

Calculating EMF of electrochemical cells

The EMF (electromotive force) tells us the maximum voltage an electrochemical cell can produce.

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Definition: EMF of a cell

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

To calculate EMF, use the formula:

$E°(cell) = E°(cathode) - E°(anode)$

Or alternatively:

$E°(cell) = E°(reduction\ half-reaction) - E°(oxidation\ half-reaction)$

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Worked Example: Calculating EMF

Question: A cell contains a solid lead anode in a gold ion solution. Calculate the cell potential (EMF).

Solution: Step 1: Find appropriate reactions from the electrode potential table

$Pb^{2+}(aq) + 2e^- \rightleftharpoons Pb(s)$ : E° = -0.13 V
$Au^{3+}(aq) + 3e^- \rightleftharpoons Au(s)$ : E° = +1.50 V

Step 2: Determine which is oxidised and which is reduced Gold has a higher electrode potential, so it will be reduced (cathode). Lead has a lower electrode potential, so it will be oxidised (anode).

Step 3: Calculate the EMF $E°(cell) = E°(cathode) - E°(anode)$ $E°(cell) = E°(gold) - E°(lead) = +1.50 - (-0.13) = +1.63 V$

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Worked Example: Another EMF calculation

Question: Calculate the cell potential for the reaction $Mg(s) + 2H^+(aq) \rightarrow Mg^{2+}(aq) + H_2(g)$

Solution: Step 1: Find electrode potentials

$Mg^{2+}(aq) + 2e^- \rightleftharpoons Mg(s)$ : E° = -2.37 V
$2H^+(aq) + 2e^- \rightleftharpoons H_2(g)$ : E° = 0.00 V

Step 2: Identify anode and cathode Magnesium has the more negative potential, so it will be oxidised (anode). Hydrogen will be reduced (cathode).

Step 3: Calculate EMF $E°(cell) = E°(hydrogen) - E°(magnesium) = 0.00 - (-2.37) = +2.37 V$

Predicting reaction spontaneity

The sign of the EMF tells us whether a reaction will occur spontaneously:

Positive EMF: The reaction is spontaneous (will occur naturally)
Negative EMF: The reaction is non-spontaneous (will not occur naturally)

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Worked Example: Determining spontaneity

Question: Will copper react with dilute sulfuric acid (H₂SO₄)?

Solution: Step 1: Write the overall equation $Cu(s) + 2H^+(aq) \rightarrow Cu^{2+}(aq) + H_2(g)$

Step 2: Identify what should be oxidised and reduced

Copper → $Cu^{2+}$ (oxidation)
$H^+$ → $H_2$ (reduction)

Step 3: Calculate EMF

$Cu^{2+}(aq) + 2e^- \rightleftharpoons Cu(s)$ : E° = +0.34 V
$2H^+(aq) + 2e^- \rightleftharpoons H_2(g)$ : E° = 0.00 V

$E°(cell) = E°(hydrogen) - E°(copper) = 0.00 V - (+0.34 V) = -0.34 V$

Step 4: Interpret the result The EMF is negative, so the reaction is non-spontaneous. Copper will not react with dilute sulfuric acid.

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Important exam tips

Always check which reaction has the higher electrode potential - this will be reduced
The substance with the lower electrode potential will be oxidised
In standard cell notation, write the anode first (left side), then the cathode (right side)
Remember that electrode potentials in the table are for reduction reactions - reverse them when writing oxidation half-reactions
Positive EMF = spontaneous, negative EMF = non-spontaneous

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Key Points to Remember:

Standard electrode potentials compare how easily substances gain or lose electrons under standard conditions (101.3 kPa, 298 K, 1 mol.dm⁻³)
The standard hydrogen electrode serves as the reference point with E° = 0.00 V
More negative E° values indicate good reducing agents (easily oxidised), while more positive E° values indicate good oxidising agents (easily reduced)
EMF = E°(cathode) - E°(anode) calculates the maximum voltage a cell can produce
Positive EMF means the reaction is spontaneous, negative EMF means it's non-spontaneous

Standard Electrode Potentials (Grade 12 NSC Matric Physical Sciences): Revision Notes