Calculating Voltage Revision Notes for HSC SSCE Chemistry

Calculating Voltage

Introduction to voltage calculations

When we work with galvanic cells and redox reactions, we need to be able to calculate the voltage (also called electromotive force or EMF) that these reactions produce. A table of standard electrode potentials gives us the information we need to perform these calculations and determine which substances are stronger oxidising or reducing agents.

The beauty of electrode potentials is that we can combine them mathematically, just as we combine half reactions to create complete redox equations. This makes calculating cell voltages straightforward once you understand the key principles.

The fundamental equation

To calculate the voltage of any redox reaction, we use a simple but powerful equation. The standard voltage of a complete reaction equals the sum of the voltages from its reduction and oxidation half reactions:

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Fundamental Voltage Equation

$E_{\text{total}} = E_{\text{red}} + E_{\text{oxid}}$

This is equation 12.6, and it's the foundation of all voltage calculations.

Key principle: The voltage of a reduction half reaction is simply its standard electrode potential (represented by the symbol $\varepsilon$ ). However, for an oxidation half reaction, we must reverse the sign of the corresponding reduction potential. This is because oxidation is the reverse of reduction.

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Worked Example: Calculating Cell Voltage for Copper and Silver

Consider the reaction:

$\text{Cu(s)} + 2\text{Ag}^+(\text{aq}) \rightarrow \text{Cu}^{2+}(\text{aq}) + 2\text{Ag(s)}$

Step 1: Break this into half reactions

Reduction: $2\text{Ag}^+(\text{aq}) + 2e^- \rightarrow 2\text{Ag(s)}$
Oxidation: $\text{Cu(s)} \rightarrow \text{Cu}^{2+}(\text{aq}) + 2e^-$

Step 2: Find the electrode potentials

For the reduction half reaction: $E_{\text{red}} = \varepsilon_{\text{Ag}} = +0.80\text{ V}$

For the oxidation half reaction: $E_{\text{oxid}} = -\varepsilon_{\text{Zn}} = -(-0.76)\text{ V} = +0.76\text{ V}$

Step 3: Calculate the total voltage

$E_{\text{total}} = 0.80 + 0.76 = 1.56\text{ V}$

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Critical Concept: Electrode Potentials and Stoichiometry

Even if we need to double a half equation to balance electrons (as we did with the silver reaction), we don't change the electrode potential value. This is because potentials are measured per electron, so doubling the equation doesn't affect the voltage.

Spontaneity and voltage sign

The sign of a calculated voltage tells us something crucial about the reaction:

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Voltage Sign and Reaction Direction

If the voltage is positive: The reaction proceeds as written (it is spontaneous)
If the voltage is negative: The reaction proceeds in the reverse direction (the reverse reaction is spontaneous)

A spontaneous reaction is one that occurs naturally without external energy input. For example, when you place copper metal in silver nitrate solution, the reaction happens spontaneously because its voltage is positive (+1.56 V). However, the reverse reaction (silver in copper sulfate) never occurs naturally because it would have a negative voltage.

This principle is fundamental to predicting which redox reactions will occur:

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Spontaneity Rule

A redox reaction is spontaneous if its standard cell voltage is positive.

Case 1: Knowing which electrode is positive

When you know which electrode is positive in a galvanic cell (or when you can work it out from electrode potential data), follow this method:

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Method for Known Positive Electrode

Write the reduction half reaction for the positive electrode
Write the oxidation half reaction for the other electrode
Find the standard electrode potentials from data tables
Remember that the oxidation voltage has the opposite sign to its reduction potential
Apply equation 12.6 to calculate the total voltage

Alternatively: Write reduction at the electrode with the higher electrode potential. The electrode with the higher value will be positive because it has a greater tendency to gain electrons.

Case 2: Not knowing which electrode is positive

Sometimes we don't know which electrode is positive. This happens when we're measuring the electrode potential of a new, untabulated electrode. In this situation, we use a different form of equation 12.6:

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Alternative Form for Unknown Electrodes

$E_{\text{cell}} = \varepsilon_A - \varepsilon_B$

Where:

$E_{\text{cell}}$ is the voltage of electrode A relative to electrode B
$\varepsilon_A$ is the standard electrode potential of electrode A
$\varepsilon_B$ is the standard electrode potential of electrode B

We label the two electrodes A and B, then assume reduction occurs at electrode A (and therefore oxidation at B). The equation then allows us to calculate unknown electrode potentials.

In words: The standard voltage of the cell equals the standard electrode potential of electrode A minus the standard electrode potential of electrode B, where the voltage measured is that of A relative to B.

Understanding what voltage means

When chemists talk about "the voltage of a reaction," they really mean the voltage of the galvanic cell in which that reaction occurs. More specifically, it's the voltage of the piece of metal where reduction is happening, measured relative to the other electrode.

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For instance, if we calculate a voltage of 1.62 V for a nickel-chlorine cell, this means the chlorine electrode (typically a platinum wire in chlorine gas) is 1.62 V positive compared to the nickel electrode.

Investigation 12.2: Measuring electrode potentials

This practical investigation allows you to measure and compare the reduction potentials of several electrodes. You'll use a copper electrode as a reference point to determine the relative reducing strengths of different substances.

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Investigation Aim

To measure and compare reduction potentials using the Cu, Cu²⁺ electrode as a reference, and to determine the relative reducing strengths of the reductants involved.

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Method Overview

Set up galvanic cells using different metal electrodes paired with a copper reference electrode
Use filter paper soaked in potassium nitrate solution as a salt bridge
Measure the voltage of each cell using a voltmeter
Calculate electrode potentials using the equation: $E = \varepsilon_{\text{test}} - \varepsilon_{\text{Cu}} = \varepsilon_{\text{test}} - 0.34$

Analysis: After collecting your voltage measurements, you'll calculate the electrode potential for each metal tested, write the corresponding reduction half reactions, and rank substances by their reducing strength.

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What You'll Discover

This investigation demonstrates that electrode potentials can be measured experimentally and shows how these values relate to the relative strengths of different reducing agents. Any discrepancies between your measured values and tabulated data can arise from non-standard conditions (such as solution concentrations that aren't exactly 1.00 mol L⁻¹ or temperatures that differ from 25°C).

Exam tips

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Key Exam Strategies

Always check the sign of your answer - it tells you about spontaneity
Remember that oxidation voltages have the opposite sign to the reduction potentials
You don't need to multiply electrode potentials when you balance electrons in half equations
When comparing electrode potentials, the substance with the higher value will be reduced (it's the stronger oxidizing agent)
Practice identifying which electrode is positive before you start calculations

Remember!

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Key Points to Remember

Fundamental equation: $E_{\text{total}} = E_{\text{red}} + E_{\text{oxid}}$ - The voltage of a complete reaction equals the sum of its reduction and oxidation half reaction voltages
Sign matters: The voltage of an oxidation half reaction has the opposite sign to its corresponding reduction potential
Spontaneity rule: A positive voltage means the reaction is spontaneous as written; a negative voltage means the reverse reaction is spontaneous
Unknown electrodes: When you don't know which electrode is positive, use $E_{\text{cell}} = \varepsilon_A - \varepsilon_B$ to calculate voltage
Practical application: Electrode potentials can be measured experimentally by comparing an unknown electrode to a reference electrode like Cu/Cu²⁺

Calculating Voltage (HSC SSCE Chemistry): Revision Notes