How to Choose the Cathode Using a Reduction-Potential Table

Suppose a standard reduction-potential table gives you two half-reactions:

- \(\ce{Ag+ + e- -> Ag}\): \(E^\circ = +0.80\text{ V}\)
- \(\ce{Cu^{2+} + 2e- -> Cu}\): \(E^\circ = +0.34\text{ V}\)

Which electrode is the cathode?

The answer is the silver electrode. It has the more positive standard reduction potential, so it is the half-cell in which reduction occurs.

This gives you the most useful rule for HSC Chemistry:

**The cathode is the electrode associated with the more positive standard reduction potential. The anode is associated with the less positive standard reduction potential.**

You do not need to reverse reactions first, count electrons, or guess from the metals involved. Start by comparing the reduction potentials as they are printed in the table.

But why does this rule work, and what happens when a question gives you actual electrode measurements rather than standard values? To answer that properly, it helps to understand how the potential table was established.

## Why the more positive reduction potential identifies the cathode

Imagine connecting two half-cells with a wire and a salt bridge. Each half-cell contains an electrode in a solution containing its ions.

One electrode supplies electrons to the external circuit. The other receives them. Electrons therefore travel through the wire from the anode to the cathode.

At the cathode, particles gain electrons. This is reduction.

At the anode, particles lose electrons. This is oxidation.

[IMAGE: An accurately labelled galvanic cell containing a silver electrode in Ag+ solution and a copper electrode in Cu2+ solution, connected by an external wire and voltmeter, with a salt bridge between the solutions. Show electrons travelling through the wire from copper to silver. Show Ag+ ions being reduced at the silver electrode and copper atoms forming Cu2+ ions at the copper electrode. Label anode, cathode, oxidation, reduction, and electron direction.]

The reduction-potential table tells you how strongly each half-cell favours reduction under standard conditions.

A more positive reduction potential means reduction is more favourable relative to the standard hydrogen electrode. When two half-cells are connected in a spontaneous galvanic cell, the half-cell with the more positive reduction potential becomes the cathode.

Think of the table as giving both half-cells the same opportunity: each is written as a reduction reaction, and you compare their values on that common basis.

The half-cell with the higher value undergoes reduction. The other undergoes oxidation, so its listed reaction must be reversed.

### The decision rule

| Comparison | Electrode | Reaction |
|---|---|---|
| More positive \(E^\circ\) | Cathode | Reduction |
| Less positive \(E^\circ\) | Anode | Oxidation |

The word *less* matters. A potential of \(-0.20\text{ V}\) is more positive than \(-0.80\text{ V}\), even though both numbers are negative.

For example, if the table gives:

- \(E^\circ_A = -0.20\text{ V}\)
- \(E^\circ_B = -0.80\text{ V}\)

then electrode A is the cathode. Do not choose B just because its value has the larger negative magnitude.

01Start with the two half-cells, not the whole table

A standard reduction-potential table might contain twenty or more half-reactions. You should not automatically choose the row with the highest value.

First, identify the two half-cells actually connected in the question. Then find their corresponding reduction potentials.

Consider these illustrative values:

The values are rounded illustrative data. Always use the values supplied in your question or examination data sheet.

Suppose the cell contains a copper electrode in a copper(II) solution and a platinum electrode in a solution containing iron(III) and iron(II) ions.

Which values should you compare?

Only \(+0.34\text{ V}\) and \(+0.77\text{ V}\). The silver entry is irrelevant because neither silver ions nor silver metal is involved in the cell.

Therefore, the platinum half-cell is the cathode, where iron(III) ions are reduced to iron(II) ions.

There is another detail worth noticing. The iron half-cell needs a platinum electrode because both iron species are dissolved ions. There is no solid iron electrode in this arrangement.

Platinum acts as an inert electrode. It provides a conducting surface where electrons can enter or leave the solution, but it is not consumed in the reaction.

02Where do the reduction potentials come from?

There is an experimental problem behind the numbers in the table.

You cannot connect a voltmeter to just one half-cell and measure its reduction potential. A voltmeter needs two electrical connections, so it measures a potential difference between two electrodes.

Chemists solve this by comparing an unknown half-cell with a reference electrode called the standard hydrogen electrode (SHE).

The standard hydrogen electrode

The standard hydrogen electrode consists of:

• A platinum electrode immersed in an acidic solution.
• Hydrogen gas passed over the platinum surface.
• Hydrogen ions at standard activity, represented at HSC level by a concentration of \(1.0\text{ mol L}^{-1}\).
• Hydrogen gas at a standard pressure of \(100\text{ kPa}\).
• A temperature of \(298\text{ K}\) when measuring the commonly tabulated values.

The platinum provides a conducting surface for the reaction. Hydrogen gas and hydrogen ions are the reacting species.

The reduction half-equation is:

\[ \ce{2H+(aq) + 2e- -> H2(g)} \]

Chemists assign this electrode a standard reduction potential of exactly:

\[ E^\circ_{\mathrm{red}}=0.00\text{ V} \]

This is a reference convention. It does not mean the hydrogen electrode has no electrical potential.

How the experiment establishes an unknown reduction potential

Imagine connecting an unknown \(\mathrm{M}^{2+}/\mathrm{M}\) half-cell to the standard hydrogen electrode.

The two half-cells are connected by a salt bridge, while their electrodes are connected to a high-resistance voltmeter.

Make a prediction before looking at the reading.

If the unknown half-cell has a more positive reduction potential than the hydrogen electrode, which side should become the cathode?

The unknown half-cell becomes the cathode. Reduction occurs there, and electrons travel from the hydrogen electrode towards the unknown electrode.

If the red voltmeter lead is attached to the unknown electrode and the black lead to the hydrogen electrode, the reading is positive.

For example, suppose the voltmeter reads \(+0.77\text{ V}\).

The hydrogen electrode has a potential of \(0.00\text{ V}\), so:

\[ \begin{aligned} E^\circ_{\mathrm{unknown}} &= E^\circ_{\mathrm{unknown}}-0.00\\\ &=+0.77\text{ V} \end{aligned} \]

The unknown standard reduction potential is therefore \(+0.77\text{ V}\).

What if the unknown electrode is less positive than hydrogen? The hydrogen electrode becomes the cathode, and the unknown electrode becomes the anode. With the same voltmeter lead arrangement, the reading is negative.

The important experimental limitation is that the voltmeter measures a difference, not an isolated electrode potential. The standard hydrogen electrode provides the agreed reference needed to assign the values in the table.

This is also why swapping the voltmeter leads reverses the sign of a reading without changing the underlying chemical reaction.

For a more detailed treatment of this measurement process, see Reduction Potentials of Galvanic Half-Cells: HSC Chemistry.

Standard conditions versus actual measurements

A standard reduction potential, written \(E^\circ_{\mathrm{red}}\), refers to standard-state conditions.

For typical HSC calculations, these are represented by:

Strictly, thermodynamic standard states are defined using unit activity. The concentrations above are the usual school-level representation for aqueous solutions.

An actual voltmeter reading does not have to equal the standard cell potential.

For example, a zinc-silver cell has a standard cell potential of \(+1.56\text{ V}\). If the solutions have different concentrations from their standard values, the actual reading may be \(+1.49\text{ V}\).

That does not mean the standard reduction potentials have changed. It means the actual electrode potentials differ from their standard-state values.

There are two different questions you might be asked:

For a galvanic cell, the electrode connected to the positive voltmeter terminal is the cathode when the meter reads positive with that electrode on its red lead. A negative reading means the electrode on the red lead is at the lower potential.

Remember that the cathode is positive and the anode negative in a galvanic cell. The cathode is always the site of reduction, including in electrolytic cells, although the electrode signs differ in electrolysis.

03A practical method for choosing the cathode

Use this sequence whenever a question gives you a standard reduction-potential table.

Step 1: Identify the two actual half-cells.

Write down the species present in each half-cell. Pay attention to oxidation states. For example, \(\ce{Fe^3+/Fe^2+}\) and \(\ce{Fe^2+/Fe}\) are different redox couples.

Step 2: Find their reduction potentials.

Find the matching half-equations in the table. Keep both reactions written as reductions.

Step 3: Compare the values.

The more positive standard reduction potential identifies the cathode under standard conditions.

Step 4: Identify the anode reaction.

The other half-equation must be reversed because oxidation occurs at the anode.

Step 5: Calculate the standard cell potential if required.

Use:

\[ E^\circ_{\mathrm{cell}} = E^\circ_{\mathrm{red,cathode}} - E^\circ_{\mathrm{red,anode}} \]

Both values on the right are the reduction potentials as printed in the table.

A positive standard cell potential means the reaction is thermodynamically spontaneous in the direction you have written.

Step 6: Determine electron and ion movement.

Electrons travel through the external wire from anode to cathode.

In a typical galvanic cell with a potassium nitrate salt bridge:

• \(\mathrm{K}^{+}\) ions migrate towards the cathode compartment.
• \(\ce{NO3-}\) ions migrate towards the anode compartment.

The salt bridge maintains electrical neutrality as the reactions change the ionic composition of each solution. Its ions do not travel through the external wire.

04Questions and solutions

Question 1

Two half-cells are connected under standard conditions:

\[ \ce{Zn^2+ + 2e- -> Zn} \qquad E^\circ=-0.76\text{ V} \]

\[ \ce{Fe^2+ + 2e- -> Fe} \qquad E^\circ=-0.44\text{ V} \]

Identify the cathode and anode. Calculate the standard cell potential and state the direction of electron flow.

Solution 1

The iron half-cell is the cathode, and the zinc half-cell is the anode. The standard cell potential is \(+0.32\text{ V}\).

The key comparison is:

\[ -0.44\text{ V}>-0.76\text{ V} \]

Although both potentials are negative, the iron reduction potential is more positive.

Therefore, iron(II) ions are reduced:

\[ \ce{Fe^2+ + 2e- -> Fe} \]

Zinc is oxidised:

\[ \ce{Zn -> Zn^2+ + 2e-} \]

Calculate the standard cell potential:

\[ \begin{aligned} E^\circ_{\mathrm{cell}} &=E^\circ_{\mathrm{cathode}} -E^\circ_{\mathrm{anode}}\\\ &=-0.44-(-0.76)\\\ &=\boxed{+0.32\text{ V}} \end{aligned} \]

The balanced overall reaction is:

\[ \ce{Zn + Fe^2+ -> Zn^2+ + Fe} \]

Electrons travel from the zinc anode to the iron cathode through the external wire.

The common mistake is to assume that two negative reduction potentials cannot produce a spontaneous galvanic cell. What matters is their difference, not whether either individual value is positive.

Question 2

A galvanic cell consists of:

• A magnesium electrode immersed in a \(1.0\text{ mol L}^{-1}\) \(\mathrm{Mg}^{2+}\) solution.
• A copper electrode immersed in a \(1.0\text{ mol L}^{-1}\) \(\mathrm{Cu}^{2+}\) solution.
• A potassium nitrate salt bridge connecting the solutions.

The standard reduction potentials are:

\[ \ce{Mg^2+ + 2e- -> Mg} \qquad E^\circ=-2.36\text{ V} \]

\[ \ce{Cu^2+ + 2e- -> Cu} \qquad E^\circ=+0.34\text{ V} \]

Determine:

(a) Which electrode is the cathode and which is the anode?

(b) The standard cell potential.

(c) The balanced overall reaction.

(d) The direction of electron movement and the direction of migration of each salt bridge ion.

Solution 2

Copper is the cathode, magnesium is the anode, and the standard cell potential is \(+2.70\text{ V}\).

(a) Identify the electrodes

Compare the reduction potentials:

\[ +0.34\text{ V}>-2.36\text{ V} \]

Copper has the more positive reduction potential, so reduction occurs at the copper electrode.

Therefore:

• Cathode: copper.
• Anode: magnesium.

(b) Calculate the standard cell potential

\[ \begin{aligned} E^\circ_{\mathrm{cell}} &=E^\circ_{\mathrm{cathode}} -E^\circ_{\mathrm{anode}}\\\ &=0.34-(-2.36)\\\ &=\boxed{+2.70\text{ V}} \end{aligned} \]

(c) Write the overall reaction

At the cathode:

\[ \ce{Cu^2+ + 2e- -> Cu} \]

At the anode:

\[ \ce{Mg -> Mg^2+ + 2e-} \]

Adding the equations gives:

\[ \ce{Mg + Cu^2+ -> Mg^2+ + Cu} \]

Magnesium dissolves into the solution, while copper metal is deposited on the copper electrode.

(d) Determine electron and ion movement

Electrons travel through the external wire from magnesium to copper.

The salt bridge ions move in opposite directions:

• \(\mathrm{K}^{+}\) migrates towards the copper cathode compartment.
• \(\ce{NO3-}\) migrates towards the magnesium anode compartment.

Why?

At the copper cathode, \(\mathrm{Cu}^{2+}\) ions are removed from solution as copper metal forms. Positive charge is consumed, so \(\mathrm{K}^{+}\) ions enter to help maintain electrical neutrality.

At the magnesium anode, \(\mathrm{Mg}^{2+}\) ions enter the solution. Positive charge increases, so \(\ce{NO3-}\) ions migrate into that compartment.

Question 3

An unknown metal half-cell, \(\mathrm{M}^{2+}/\mathrm{M}\), is connected to a standard hydrogen electrode under standard conditions.

The voltmeter has:

• Its red lead connected to the unknown metal electrode.
• Its black lead connected to the platinum electrode of the standard hydrogen electrode.

The voltmeter displays \(-0.28\text{ V}\).

Determine:

(a) The standard reduction potential of the unknown metal half-cell.

(b) Which electrode is the cathode and which is the anode.

(c) The direction of electron flow.

(d) The voltmeter reading if the two leads are swapped, assuming nothing else changes.

Solution 3

The unknown metal has a standard reduction potential of \(-0.28\text{ V}\). The standard hydrogen electrode is the cathode, and the unknown metal electrode is the anode.

(a) Determine the unknown reduction potential

The standard hydrogen electrode has a potential of \(0.00\text{ V}\).

The voltmeter reading is negative because the red lead is connected to the lower-potential electrode.

Therefore:

\[ \begin{aligned} E^\circ_{\mathrm{unknown}} &=E^\circ_{\mathrm{SHE}}-0.28\\\ &=0.00-0.28\\\ &=\boxed{-0.28\text{ V}} \end{aligned} \]

(b) Identify the cathode and anode

Compare the reduction potentials:

\[ 0.00\text{ V}>-0.28\text{ V} \]

The standard hydrogen electrode has the more positive reduction potential.

Therefore:

• Cathode: standard hydrogen electrode.
• Anode: unknown metal electrode.

(c) Determine electron movement

At the unknown metal electrode, oxidation occurs:

\[ \ce{M -> M^2+ + 2e-} \]

At the hydrogen electrode, reduction occurs:

\[ \ce{2H+ + 2e- -> H2} \]

Electrons travel from the unknown metal electrode towards the platinum electrode through the external wire.

The overall reaction is:

\[ \ce{M + 2H+ -> M^2+ + H2} \]

(d) Swap the voltmeter leads

The reading becomes:

\[ \boxed{+0.28\text{ V}} \]

The chemical reaction does not change. Only the voltmeter’s reference polarity changes.

This is an important experimental distinction: a negative voltmeter reading does not mean that a galvanic cell cannot operate. It means the electrode connected to the red lead is at a lower potential than the electrode connected to the black lead.

Question 4

A galvanic cell contains the following half-cells:

• A zinc electrode immersed in a \(\mathrm{Zn}^{2+}\) solution.
• A graphite electrode immersed in a chloride solution, with chlorine gas bubbled over its surface.

The relevant standard reduction potentials are:

\[ \ce{Cl2(g) + 2e- -> 2Cl-(aq)} \qquad E^\circ=+1.36\text{ V} \]

\[ \ce{Zn^2+ + 2e- -> Zn} \qquad E^\circ=-0.76\text{ V} \]

The supplied table also contains:

\[ \ce{Ag+ + e- -> Ag} \qquad E^\circ=+0.80\text{ V} \]

Assume standard conditions and a potassium nitrate salt bridge.

(a) Identify the cathode and anode.

(b) Explain why the silver reduction potential is irrelevant when choosing the cathode.

(c) Calculate the standard cell potential and write the balanced overall reaction.

(d) Explain why graphite is suitable as an electrode even though it does not appear in the half-equation.

(e) State the direction of electron movement and the direction of salt bridge ion migration.

Solution 4

The graphite electrode is the cathode, and the zinc electrode is the anode. The standard cell potential is \(+2.12\text{ V}\).

(a) Identify the electrodes

The two relevant reduction potentials are:

• Chlorine/chloride: \(+1.36\text{ V}\).
• Zinc ion/zinc: \(-0.76\text{ V}\).

Chlorine has the more positive reduction potential.

Therefore, chlorine is reduced at the graphite electrode, making it the cathode.

Zinc is oxidised at the anode.

(b) Explain why silver is irrelevant

The silver half-equation is not part of either half-cell.

Although its reduction potential is positive, there are no silver ions or silver metal involved in this cell.

You must compare the half-reactions that are physically available in the cell, not simply choose a large number from somewhere in the table.

(c) Calculate the standard cell potential

\[ \begin{aligned} E^\circ_{\mathrm{cell}} &=E^\circ_{\mathrm{cathode}} -E^\circ_{\mathrm{anode}}\\\ &=1.36-(-0.76)\\\ &=\boxed{+2.12\text{ V}} \end{aligned} \]

The half-equations are:

Anode:

\[ \ce{Zn -> Zn^2+ + 2e-} \]

Cathode:

\[ \ce{Cl2 + 2e- -> 2Cl-} \]

The overall reaction is:

\[ \ce{Zn + Cl2 -> Zn^2+ + 2Cl-} \]

The positive standard cell potential indicates a thermodynamically spontaneous reaction under standard conditions.

(d) Explain the role of graphite

Graphite is an inert conducting electrode.

Chlorine gas and chloride ions cannot provide a solid metal surface for electrical connection. Graphite provides a surface where electron transfer can occur.

It conducts electrons into the half-cell but is not consumed in the net ionic reaction.

(e) Explain electron and ion movement

Electrons travel from the zinc anode towards the graphite cathode.

The salt bridge ions move as follows:

• \(\mathrm{K}^{+}\) migrates towards the graphite cathode compartment.
• \(\ce{NO3-}\) migrates towards the zinc anode compartment.

The cathode reaction produces chloride ions, increasing negative charge in the cathode solution. Potassium ions help balance this charge.

The anode reaction produces zinc ions, increasing positive charge in the anode solution. Nitrate ions help balance that charge.

Question 5

This final problem tests the difference between a standard reduction-potential prediction and the behaviour of an actual galvanic cell.

A cell is constructed using:

• A silver electrode immersed in a solution containing \(\mathrm{Ag}^{+}\).
• A platinum electrode immersed in a solution containing both \(\mathrm{Fe}^{3+}\) and \(\mathrm{Fe}^{2+}\).
• A potassium nitrate salt bridge.
• A high-resistance voltmeter.

The standard reduction potentials are:

\[ \ce{Ag+ + e- -> Ag} \qquad E^\circ=+0.80\text{ V} \]

\[ \ce{Fe^3+ + e- -> Fe^2+} \qquad E^\circ=+0.77\text{ V} \]

The cell is initially not under standard conditions.

The voltmeter’s red lead is connected to the silver electrode, and its black lead is connected to the platinum electrode.

Initially, the meter reads:

\[ -0.02\text{ V} \]

Next, the following changes are made:

• Silver nitrate is added to the silver half-cell, increasing the \(\mathrm{Ag}^{+}\) concentration.
• An iron(II) solution is added to the platinum half-cell, increasing the \(\mathrm{Fe}^{2+}\) concentration relative to \(\mathrm{Fe}^{3+}\).

The new stable voltmeter reading is:

\[ +0.04\text{ V} \]

All readings are illustrative supplied measurements. Assume the cell operates as a galvanic cell, the salt bridge is inert, and no competing reactions occur.

Answer the following:

(a) Under standard conditions, identify the predicted cathode and calculate the standard cell potential.

(b) Before the concentration changes, identify the actual cathode and anode using the voltmeter reading. Write the balanced overall reaction and determine the electron direction.

(c) After the concentration changes, identify the new cathode and anode. Write the new overall reaction and determine the electron direction.

(d) Explain the direction of \(\mathrm{K}^{+}\) and \(\ce{NO3-}\) migration before and after the concentration changes.

(e) Explain why the actual electrode roles can reverse even though the standard reduction-potential values have not changed.

Solution 5

Initially, platinum is the cathode and silver is the anode. After the concentration changes, silver becomes the cathode and platinum becomes the anode.

The key is that the standard table predicts behaviour under standard conditions, while the voltmeter tells you the actual potential difference in the experiment.

(a) Determine the standard-state prediction

Compare the two standard reduction potentials:

\[ +0.80\text{ V}>+0.77\text{ V} \]

Silver has the more positive standard reduction potential.

Therefore, under standard conditions:

• Cathode: silver.
• Anode: platinum.

Calculate the standard cell potential:

\[ \begin{aligned} E^\circ_{\mathrm{cell}} &=E^\circ_{\mathrm{Ag^+/Ag}} -E^\circ_{\mathrm{Fe^{3+}/Fe^{2+}}}\\\ &=0.80-0.77\\\ &=\boxed{+0.03\text{ V}} \end{aligned} \]

The standard-state reaction is spontaneous in the direction where silver ions are reduced and iron(II) ions are oxidised.

(b) Determine the actual electrode roles before the changes

The voltmeter reads \(-0.02\text{ V}\), with its red lead connected to silver and its black lead connected to platinum.

This means the silver electrode is at a lower potential than the platinum electrode.

Therefore:

• Silver is the actual anode.
• Platinum is the actual cathode.

The reaction at the silver anode is oxidation:

\[ \ce{Ag -> Ag+ + e-} \]

The reaction at the platinum cathode is reduction:

\[ \ce{Fe^3+ + e- -> Fe^2+} \]

Combining these gives:

\[ \boxed{\ce{Ag + Fe^3+ -> Ag+ + Fe^2+}} \]

Electrons travel from the silver electrode to the platinum electrode.

Notice that this is the reverse of the standard-state prediction.

(c) Determine the actual electrode roles after the changes

The voltmeter now reads \(+0.04\text{ V}\), with the red lead still connected to silver.

Silver is now at a higher potential than platinum.

Therefore:

• Silver is the cathode.
• Platinum is the anode.

At the silver cathode, reduction occurs:

\[ \ce{Ag+ + e- -> Ag} \]

At the platinum anode, iron(II) ions are oxidised:

\[ \ce{Fe^2+ -> Fe^3+ + e-} \]

The overall reaction is:

\[ \boxed{\ce{Ag+ + Fe^2+ -> Ag + Fe^3+}} \]

Electrons now travel from the platinum electrode to the silver electrode.

The direction of electron flow has reversed.

(d) Explain the salt bridge ion movement

The salt bridge contains \(\mathrm{K}^{+}\) and \(\ce{NO3-}\) ions. Their migration also reverses when the electrode reactions reverse.

Initially:

• Platinum is the cathode, so \(\mathrm{K}^{+}\) migrates towards the platinum compartment.
• Silver is the anode, so \(\ce{NO3-}\) migrates towards the silver compartment.

After the concentration changes:

• Silver is the cathode, so \(\mathrm{K}^{+}\) migrates towards the silver compartment.
• Platinum is the anode, so \(\ce{NO3-}\) migrates towards the platinum compartment.

The salt bridge maintains electrical neutrality in both compartments throughout the reaction.

(e) Explain why the standard prediction and actual measurements differ

This is the most important part of the question.

The standard reduction potentials are fixed reference values for their specified standard states. They have not changed.

However, actual electrode potentials depend on the conditions in each half-cell, including the concentrations of the reacting species.

Initially, the actual conditions favour the iron(III)/iron(II) reduction relative to silver reduction. The voltmeter confirms this because the platinum electrode is at the higher potential.

Adding silver nitrate increases the concentration of \(\mathrm{Ag}^{+}\), making silver-ion reduction more favourable.

Adding iron(II) ions increases the relative availability of the reduced iron species, favouring its oxidation to iron(III).

Together, these changes favour the reverse reaction, which is confirmed by the positive voltmeter reading.

The standard potential difference of \(+0.03\text{ V}\) still describes the standard-state comparison. It does not force the silver electrode to remain the cathode under every possible set of concentrations.

This is the trap: the more positive standard reduction potential predicts the cathode under standard conditions, not necessarily the cathode in every real experiment.

If the voltmeter leads were swapped, both readings would change sign, but neither chemical reaction would change direction.

05What to learn next

Once you can identify the cathode correctly, the next step is using both half-cell potentials to calculate the cell voltage and predict whether a redox reaction is spontaneous.

For more practice connecting electrode selection with metal displacement reactions, continue with Standard Reduction Potentials for Metal Reactions in HSC Chemistry.

The essential habit to carry into your HSC exam is simple: identify the actual half-cells first, compare their reduction potentials second, and only then assign the cathode, anode, and electron direction.