R3.2.13 Standard cell potential (Higher Level Only)

Calculation of Standard Cell Potential

What is Standard Cell Potential?

Definition

Standard cell potential

The standard cell potential, denoted as $E_{cell}^{\circ}$ , is the voltage produced by an electrochemical cell under standard conditions.

These conditions include:

A temperature of 298 K (25°C),
A pressure of 100 kPa for gases,
A 1.0 M concentration for all aqueous solutions.

Note

But why do we use these specific conditions?

Standard conditions provide a consistent reference point for comparing different electrochemical cells, ensuring that the measured potentials are reliable and reproducible.

The cell potential depends on the difference in the ability of two half-cells to gain or lose electrons, as measured by their standard electrode potentials ( $E^{\circ}$ ).

Analogy

Think of standard conditions as setting the "rules of the game" for electrochemical cells, ensuring a level playing field for comparison, much like standardizing the dimensions of a sports field.

The Formula for Standard Cell Potential

The standard cell potential is calculated using the formula:

$E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ}$

Here’s what each term means:

$E_{cathode}^{\circ}$ : The standard electrode potential of the reduction half-equation occurring at the cathode.
$E_{anode}^{\circ}$ : The standard electrode potential of the oxidation half-equation occurring at the anode.

Tip

Always subtract the (E^\circ) value of the anode (where oxidation occurs) from that of the cathode (where reduction occurs). Remember: Reduction happens at the cathode, and oxidation happens at the anode.

Predicting Spontaneity with $E_{cell}^{\circ}$

The sign of $E_{cell}^{\circ}$ tells us whether the reaction in the electrochemical cell is spontaneous:

If $E_{cell}^{\circ} > 0$ : The reaction is spontaneous, meaning it can generate electrical energy.
If $E_{cell}^{\circ} < 0$ : The reaction is non-spontaneous, meaning it requires an external energy source to proceed.

Analogy

Think of $E_{cell}^{\circ}$ as a hill: a positive value means the reaction "rolls downhill" spontaneously, while a negative value means you have to push it "uphill" with extra energy.

Common Mistake

Do not confuse the signs of $E_{cell}^{\circ}$ . A positive value indicates a spontaneous reaction, not a negative one.

Step-by-Step: Calculating $E_{cell}^{\circ}$

Let’s break down the process of calculating the standard cell potential with an example.

Example

Zinc-Copper Electrochemical Cell

You’re tasked with calculating $E_{cell}^{\circ}$ for a cell consisting of a zinc electrode and a copper electrode. The half-reactions and their standard electrode potentials are:

${Zn}^{2 +} (aq) + 2 e^{-} \to Zn (s) E^{\circ} = - 0.76 V$

${Cu}^{2 +} (aq) + 2 e^{-} \to Cu (s) E^{\circ} = + 0.34 V$

Step 1: Identify the Cathode and Anode

The cathode is where reduction occurs. Since copper has a more positive $E^{\circ}$ , it will be reduced: ${Cu}^{2 +} + 2 e^{-} \to Cu$ .
The anode is where oxidation occurs. Zinc has a more negative $E^{\circ}$ , so it will be oxidized: $Zn \to {Zn}^{2 +} + 2 e^{-}$ .

Step 2: Apply the Formula

$E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ}$

Substitute the values:
$E_{cell}^{\circ} = (+ 0.34 V) - (- 0.76 V)$

Simplify:
$E_{cell}^{\circ} = + 1.10 V$

Step 3: Interpret the Result

Since $E_{cell}^{\circ} = + 1.10 V$ , the reaction is spontaneous. This means the zinc-copper cell can generate electricity.

Note

In the zinc-copper cell, zinc is oxidized to ${Zn}^{2 +}$ , releasing electrons that flow through an external circuit to reduce ${Cu}^{2 +}$ to copper metal.
This flow of electrons is what powers devices.

Common Mistakes and How to Avoid Them

Reversing the Cathode and Anode: Always assign the cathode to the half-reaction with the more positive $E^{\circ}$ .
Ignoring Units: Ensure that all $E^{\circ}$ values are in volts (V) and are taken under standard conditions.
Forgetting to Balance Electrons: While balancing the overall redox equation, ensure the number of electrons lost in oxidation equals those gained in reduction.

Common Mistake

Do not multiply $E^{\circ}$ values by coefficients when balancing half-equations. The potential is an intensive property and does not depend on the amount of substance.

Example question

Calculate $E_{cell}^{\circ}$ for a cell comprising a magnesium electrode and a silver electrode. Use the following data:

${Mg}^{2 +} (aq) + 2 e^{-} \to Mg (s) E^{\circ} = - 2.37 V$
${Ag}^{+} (aq) + e^{-} \to Ag (s) E^{\circ} = + 0.80 V$

Solution

Identify the cathode and anode:
Silver ( $E^{\circ} = + 0.80 V$ ) is the cathode (reduction).
Magnesium ( $E^{\circ} = - 2.37 V$ ) is the anode (oxidation).
Apply the formula:
$E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ}$
$E_{cell}^{\circ} = (+ 0.80, V) - (- 2.37 V) = + 3.17 V$
Interpret the result: $E_{cell}^{\circ} = + 3.17 V$ indicates a highly spontaneous reaction.

Reflection

Self review

What is the significance of a positive $E_{cell}^{\circ}$ ?

Theory of Knowledge

How does the concept of spontaneity in electrochemical cells connect to the idea of entropy in thermodynamics? Can we always predict the behavior of a system based solely on its energy changes?

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