Anode or Cathode? A Reliable Sign Check for Galvanic Cells
Learn to label anodes and cathodes correctly by starting with oxidation and reduction, not electrode signs. Use a zinc-copper cell, worked examples, and practice questions to check polarity, electron flow, and salt-bridge ion migration.
You’ve balanced the half-equations for a zinc-copper cell, but your diagram labels zinc as the cathode, copper as the negative electrode, and electrons travelling from copper to zinc. The equations, signs, and arrows are contradicting one another.
The mistake usually starts when you memorise a rule about electrode signs and apply it before checking the chemistry. There’s a more dependable way to label the entire diagram, starting with one simple prediction: which electrode loses metal atoms to the solution?
In a spontaneous zinc-copper cell, it’s zinc. Zinc atoms become zinc ions and release electrons. That observation gives you the starting point for identifying the electrodes, their polarity, and the direction of electron flow.
01The diagnosis: you’re using signs to define the electrodes
Two definitions control the entire problem:
- Anode: the electrode where oxidation occurs.
- Cathode: the electrode where reduction occurs.
Oxidation is the loss of electrons. Reduction is the gain of electrons.
A useful memory aid is An Ox, Red Cat: oxidation at the anode, reduction at the cathode.
Notice what these definitions don’t mention. Neither electrode is defined by its electrical sign, its position in a diagram, or the metal it contains.
The signs depend on the type of electrochemical cell.
| Electrode | Reaction | Galvanic cell | Electrolytic cell |
|---|---|---|---|
| Anode | Oxidation | Negative (−) | Positive (+) |
| Cathode | Reduction | Positive (+) | Negative (−) |
A galvanic cell generates electrical energy from a spontaneous chemical reaction. An electrolytic cell uses an external power supply to drive a reaction that would not proceed spontaneously in that direction.
The electrode names stay the same. The signs switch.
The trap: if you memorise “the anode is negative” as a universal rule, you’ll get electrolytic-cell questions wrong. If you memorise “the cathode is negative”, you’ll get galvanic-cell questions wrong.
Identify the reactions first. Assign the signs second.
02Rebuild the zinc-copper cell from the half-equations
A zinc-copper galvanic cell contains two half-cells:
- A zinc electrode immersed in a solution containing \(\mathrm{Zn}^{2+}\) ions.
- A copper electrode immersed in a solution containing \(\mathrm{Cu}^{2+}\) ions.
The electrodes are connected by an external wire, and the solutions are connected by a salt bridge.
Zinc–copper galvanic cell: two distinct pathways
| Zinc anode (−) | Zn(s) → Zn²⁺(aq) + 2e⁻ Zn strip in ZnSO₄(aq); oxidation releases electrons. |
|---|---|
| External wire and voltmeter | Electron flow through the wire: zinc → copper. |
| Copper cathode (+) | Cu²⁺(aq) + 2e⁻ → Cu(s) Cu strip in CuSO₄(aq); reduction deposits copper. |
| Salt bridge below the wire | KNO₃(aq) K⁺ → copper side; NO₃⁻ → zinc side. Ions only. |
Step 1: Work out what happens to zinc
Zinc atoms lose two electrons and enter the solution as zinc ions.
\[ \ce{Zn(s) -> Zn^2+(aq) + 2e^-} \]
The electrons appear on the right-hand side of the equation because they are produced by the oxidation reaction.
Therefore, zinc is oxidised, so the zinc electrode is the anode.
As zinc atoms leave the metal and enter the solution, the zinc electrode loses mass.
Step 2: Work out what happens to copper
Copper ions in the solution gain two electrons and become solid copper.
\[ \ce{Cu^2+(aq) + 2e^- -> Cu(s)} \]
The electrons appear on the left-hand side because they are consumed by the reduction reaction.
Therefore, copper ions are reduced at the copper electrode, making it the cathode.
As copper metal deposits onto the electrode, the copper electrode gains mass.
Step 3: Trace the electrons
Now look at the two half-equations together.
- Zinc oxidation produces electrons.
- Copper reduction consumes electrons.
In a galvanic cell, those electrons travel through the external wire from the zinc electrode to the copper electrode.
\[ \boxed{\ce{Zn -> Zn^2+ + 2e^-}} \]
\[ \boxed{\ce{Cu^2+ + 2e^- -> Cu}} \]
The overall reaction is
\[ \ce{Zn(s) + Cu^2+(aq) -> Zn^2+(aq) + Cu(s)} \]
The electrons released by zinc are transferred through the circuit and used by copper ions. Electrons do not travel through the salt bridge.
Step 4: Assign the electrode signs
Because this is a spontaneous galvanic cell, the zinc anode is the negative terminal, and the copper cathode is the positive terminal.
Why? The oxidation reaction at zinc supplies electrons to the external circuit, while the reduction reaction at copper consumes electrons arriving through that circuit.
The signs describe the relative electrical potentials of the two terminals. They are not the definitions of anode and cathode.
One final detail: conventional current flows in the opposite direction to electron flow. So, in the external wire, electrons travel from zinc to copper, while conventional current is directed from copper to zinc.
03The salt bridge: a separate route for ions
You’ve now identified the electrodes and electron flow. There’s still a problem to solve.
As the cell operates, zinc ions enter one solution and copper ions disappear from the other. If charge continued to build up in each half-cell, the reaction would soon become difficult to sustain.
That’s the job of the salt bridge.
Imagine it as a side lane that lets ions move between the half-cells to maintain electrical neutrality. The analogy has one important limit: this lane carries ions, not electrons.
Suppose the salt bridge contains potassium nitrate, \(\ce{KNO3}\). It supplies \(\mathrm{K}^{+}\) cations and \(\mathrm{NO_{3}}^{-}\) anions.
Two pathways in a zinc–copper cell
| Zinc anode (−): ZnSO₄ solution | Zn → Zn²⁺ + 2e⁻ Oxidation produces Zn²⁺; NO₃⁻ enters this compartment. |
|---|---|
| Separate pathways | KNO₃ salt bridge Wire: electrons Zn → Cu. Bridge: NO₃⁻ → Zn; K⁺ → Cu. |
| Copper cathode (+): CuSO₄ solution | Cu²⁺ + 2e⁻ → Cu Reduction consumes Cu²⁺; K⁺ enters this compartment. |
Which ions move where?
At the zinc anode, nitrate ions migrate into the zinc half-cell.
Oxidation produces \(\mathrm{Zn}^{2+}\) ions in the solution. Nitrate ions help balance the increasing positive charge.
At the copper cathode, potassium ions migrate into the copper half-cell.
Reduction removes \(\mathrm{Cu}^{2+}\) ions from the solution. This leaves a relative excess of negative ions, such as sulfate ions from the copper sulfate solution. Potassium ions help restore charge balance.
For this cell with a potassium nitrate bridge:
- \(\mathrm{NO_{3}}^{-}\) migrates towards the anode compartment.
- \(\mathrm{K}^{+}\) migrates towards the cathode compartment.
Notice the pattern. Anions migrate towards the anode compartment, while cations migrate towards the cathode compartment in this standard salt-bridge model.
Don’t confuse this with electron flow. Electrons travel through the wire; ions migrate through the solutions and salt bridge. The two processes work together to keep the cell operating.
04A reliable sign-check method for exams
When you’re given a galvanic-cell diagram, follow this order rather than trying to remember every arrow and sign separately.
051
Find the half-equations.
Identify which species loses electrons and which species gains them. If the equations aren’t given, use the information about the species and the reaction to work them out.
062
Name the electrodes from the reactions.
Oxidation means anode. Reduction means cathode. Do this before adding the signs.
073
Draw electron flow through the external wire.
For a galvanic cell, electrons flow from the anode to the cathode. They do not travel through the salt bridge.
084
Add the signs.
For a spontaneous galvanic cell, the anode is negative and the cathode is positive.
095
Check the salt bridge and the physical changes.
Anions migrate towards the anode compartment, and cations towards the cathode compartment. A metal being oxidised should lose mass; a metal being deposited by reduction should gain mass.
Electrode labelling decision flow
- Identify half-equations
- Oxidation → anode
- Reduction → cathode
- Identify cell type: galvanic or electrolytic
- Assign signs: galvanic A−/C+; electrolytic A+/C−
- Verify: electrons via wire; ions via salt bridge
The important habit is to treat the signs and arrows as a consistency check. If the half-equations say zinc is oxidised but your diagram labels zinc as the cathode, return to the reactions instead of trying to make the diagram fit.
10Worked example: fix a contradictory zinc-copper diagram
A student draws a spontaneous zinc-copper cell. They label zinc as the positive cathode and copper as the negative anode, then draw electrons travelling from copper to zinc. Correct the diagram and explain each correction.
Step 1: Write the half-equations.
At zinc:
\[ \ce{Zn(s) -> Zn^2+(aq) + 2e^-} \]
At copper:
\[ \ce{Cu^2+(aq) + 2e^- -> Cu(s)} \]
Step 2: Identify the electrodes from the electron transfer.
Zinc loses electrons, so oxidation occurs at zinc. Zinc is the anode.
Copper ions gain electrons, so reduction occurs at copper. Copper is the cathode.
Step 3: Correct the electron-flow arrow.
The electrons produced at zinc travel through the external wire towards copper. The correct arrow is
\[ \ce{Zn -> Cu} \]
This arrow represents electron flow through the wire, not the movement of ions through the salt bridge.
Step 4: Correct the signs.
Because the cell is galvanic, zinc is negative and copper is positive.
The corrected labels are:
- Zinc: anode, negative terminal, oxidation.
- Copper: cathode, positive terminal, reduction.
- Electron flow: zinc to copper through the external circuit.
Step 5: Identify the precise error in the student’s method.
The student started with electrode signs, used those signs to assign the names, and then drew the electron arrow to match the incorrect labels. The corrected method starts with the half-equations. The reactions determine the electrode names; the type of cell determines the polarity.
11Worked example: apply the method to a different cell
A spontaneous magnesium-silver cell contains a magnesium electrode in a solution of \(\mathrm{Mg}^{2+}\) ions and a silver electrode in a solution of \(\mathrm{Ag}^{+}\) ions. A potassium nitrate salt bridge connects the half-cells.
During operation, \(0.243\ \mathrm{g}\) of magnesium is consumed. Determine the electrode labels, signs, electron-flow direction, salt-bridge ion migration, and mass of silver deposited.
Use \(M(\ce{Mg}) = 24.3\ \mathrm{g\\,mol^{-1}}\) and \(M(\ce{Ag}) = 107.9\ \mathrm{g\\,mol^{-1}}\). Assume the only electrode reactions are magnesium oxidation and silver-ion reduction.
Step 1: Write the half-equations.
Magnesium is oxidised:
\[ \ce{Mg(s) -> Mg^2+(aq) + 2e^-} \]
Silver ions are reduced:
\[ \ce{2Ag+(aq) + 2e^- -> 2Ag(s)} \]
Step 2: Label the electrodes and assign their signs.
Magnesium is the anode because it is oxidised. Silver is the cathode because silver ions are reduced.
Since the cell is galvanic, magnesium is negative and silver is positive.
Step 3: Calculate the amount of magnesium consumed.
Use \(n=m/M\), where \(n\) is amount in moles, \(m\) is mass, and \(M\) is molar mass.
\[ \begin{aligned} n(\ce{Mg}) &= \frac{0.243\ \mathrm{g}} {24.3\ \mathrm{g\\,mol^{-1}}}\\\ &= 0.0100\ \mathrm{mol} \end{aligned} \]
Step 4: Calculate the mass of silver deposited.
One mole of magnesium releases two moles of electrons. Those electrons reduce two moles of silver ions.
Therefore, one mole of magnesium consumed corresponds to two moles of silver deposited.
\[ \begin{aligned} n(\ce{Ag}) &= 2(0.0100)\\\ &= 0.0200\ \mathrm{mol} \end{aligned} \]
Convert the amount of silver to mass:
\[ \begin{aligned} m(\ce{Ag}) &= nM\\\ &= 0.0200\ \mathrm{mol} \times 107.9\ \mathrm{g\\,mol^{-1}}\\\ &= 2.158\ \mathrm{g}\\\ &\approx 2.16\ \mathrm{g} \end{aligned} \]
The mass of silver deposited is \(2.16\ \mathrm{g}\), to three significant figures.
Step 5: Complete the flow and charge-balance labels.
Electrons flow through the wire from magnesium to silver. Nitrate ions migrate towards the magnesium anode compartment, and potassium ions migrate towards the silver cathode compartment.
Magnesium loses mass as it oxidises, while silver gains mass as silver ions are reduced to solid silver.
The same decision method works even when the metals, ions, and numerical details change. You don’t need to memorise a separate set of rules for each cell.
Galvanic cell: Mg and Ag
| Mg anode (−) | Mg → Mg²⁺ + 2e⁻ Mg strip in Mg²⁺ solution. Oxidation; the strip loses mass. |
|---|---|
| External wire | Electron flow: Mg → Ag. |
| KNO₃ salt bridge | K⁺ and NO₃⁻ NO₃⁻ → Mg side; K⁺ → Ag side. |
| Ag cathode (+) | 2Ag⁺ + 2e⁻ → 2Ag Ag⁺ gains electrons; silver deposits. The strip gains mass. |
12Wording changes that break memorised shortcuts
A few small changes to an exam question can expose whether you understand the process or have simply learnt a diagram.
The electrodes are drawn on opposite sides
The anode is not automatically the left-hand electrode. The cathode is not automatically the right-hand electrode.
Their positions on the page are a drawing choice. Identify them using oxidation and reduction.
The question asks for current, not electron flow
These directions are opposite in the external wire.
For the spontaneous zinc-copper cell:
- Electron flow: Zn to Cu.
- Conventional current: Cu to Zn.
Read the wording carefully before drawing an arrow.
The electrode signs are given
Treat the signs as evidence about the cell, not as definitions of the electrodes.
If the question describes a spontaneous galvanic cell, check that the negative electrode is the anode and the positive electrode is the cathode. Then verify that these labels agree with the half-equations.
The electrode mass changes
Mass changes can provide useful evidence, provided the relevant electrode reaction involves the metal itself.
If a zinc electrode loses mass as \(\mathrm{Zn}^{2+}\) enters solution, zinc is being oxidised. If copper ions become solid copper on an electrode, that electrode gains mass as reduction occurs.
However, mass change isn’t a universal test. An inert electrode may participate in electron transfer without gaining or losing metal, so use the half-equations whenever possible.
13Why galvanic signs cannot be transferred blindly to electrolytic cells
The anode is always the site of oxidation, and the cathode is always the site of reduction. Those definitions do not change.
What changes is the way the reaction is driven.
In a galvanic cell, a spontaneous reaction generates electrical energy. In an electrolytic cell, an external power supply forces the electrode reactions to occur in the required direction.
| Feature | Galvanic cell | Electrolytic cell |
|---|---|---|
| Energy transfer | Chemical to electrical | Electrical to chemical |
| Anode | Oxidation, negative | Oxidation, positive |
| Cathode | Reduction, positive | Reduction, negative |
| Electron movement | Anode to cathode through the external circuit | The supply removes electrons from the anode and supplies electrons to the cathode |
Galvanic vs electrolytic cells
| Galvanic cell | Anode (−): oxidation; cathode (+): reduction. Electrons flow through the wire from anode to cathode. Spontaneous reaction generates electrical energy. |
|---|---|
| Electrolytic cell | Supply (+) connects to anode: oxidation. Supply (−) connects to cathode: reduction. External power drives the non-spontaneous reaction. |
Consider what happens if an external power supply forces the zinc-copper reaction to run in reverse.
The copper electrode can now be the anode, where copper metal is oxidised to \(\mathrm{Cu}^{2+}\). The zinc electrode can be the cathode, where \(\mathrm{Zn}^{2+}\) is reduced to zinc metal, assuming the intended reactions dominate.
The electrode names have changed because the reactions have changed. The signs have changed because the external power supply is driving the reaction.
This is why “zinc is always the anode” is not a valid rule. Zinc is the anode in the spontaneous zinc-copper galvanic cell, but it can be the cathode when an external power supply drives the reverse reaction.
14A 10-second diagram check
Before submitting a galvanic-cell answer, run through this list.
- Oxidation occurs at the anode.
- Reduction occurs at the cathode.
- The anode is negative, and the cathode is positive.
- Electrons flow through the external wire from anode to cathode.
- Salt-bridge anions migrate towards the anode compartment, and cations towards the cathode compartment.
- The half-equations, electrode signs, arrows, and observed mass changes agree.
If one label fails this check, don’t redraw the entire diagram from memory. Find the half-equation that establishes the reaction, correct the electrode name, and rebuild the other labels from there.
15Questions and solutions
Question 1
A spontaneous galvanic cell contains zinc metal in a solution of \(\mathrm{Zn}^{2+}\) ions and copper metal in a solution of \(\mathrm{Cu}^{2+}\) ions.
Write the two half-equations and identify the anode, cathode, electrode signs, and electron-flow direction.
Solution 1
Zinc is the negative anode, copper is the positive cathode, and electrons flow from zinc to copper through the external wire.
The half-equations are
\[ \ce{Zn(s) -> Zn^2+(aq) + 2e^-} \]
\[ \ce{Cu^2+(aq) + 2e^- -> Cu(s)} \]
Zinc loses electrons, so oxidation occurs at the zinc electrode. That makes zinc the anode. Copper ions gain electrons, so reduction occurs at the copper electrode, making copper the cathode.
Because this is a galvanic cell, the anode is negative and the cathode is positive. Electrons travel from the electrode producing them to the electrode consuming them: zinc to copper.
Question 2
A zinc-copper galvanic cell uses a \(\ce{KNO3}\) salt bridge.
Explain the direction of migration of each salt-bridge ion. Predict what happens if the salt bridge is removed while the external wire remains connected.
Solution 2
Potassium ions migrate towards the copper half-cell, while nitrate ions migrate towards the zinc half-cell. Without the salt bridge, the cell soon slows and stops because charge imbalance prevents sustained operation.
At the zinc anode, the oxidation reaction produces \(\mathrm{Zn}^{2+}\) ions, increasing the positive charge entering the solution. Nitrate ions migrate into this compartment to help maintain charge balance.
At the copper cathode, reduction consumes \(\mathrm{Cu}^{2+}\) ions. Potassium ions migrate into this compartment to help balance the relative excess of negative ions remaining in the solution.
If the salt bridge is removed, ions can no longer migrate between the half-cells through that pathway. Charge imbalance develops, opposing further electron transfer through the wire. The salt bridge is therefore essential to sustained operation, even though it does not carry electrons.
Question 3
A zinc-copper galvanic cell deposits \(0.635\ \mathrm{g}\) of copper. Assume that \(\mathrm{Cu}^{2+}\) is the only species reduced and zinc is the only species oxidised.
Calculate the mass of zinc consumed. Then identify the polarity of the zinc electrode.
Use \(M(\ce{Cu})=63.55\ \mathrm{g\\,mol^{-1}}\) and \(M(\ce{Zn})=65.38\ \mathrm{g\\,mol^{-1}}\).
Solution 3
The cell consumes approximately \(0.653\ \mathrm{g}\) of zinc, and the zinc electrode is the negative anode.
First, calculate the amount of copper deposited:
\[ \begin{aligned} n(\ce{Cu}) &= \frac{m}{M}\\\ &= \frac{0.635\ \mathrm{g}} {63.55\ \mathrm{g\\,mol^{-1}}}\\\ &= 9.99\times10^{-3}\ \mathrm{mol} \end{aligned} \]
The balanced half-equations show that one mole of zinc oxidised supplies the two moles of electrons required to reduce one mole of copper ions. Therefore, the mole ratio of zinc consumed to copper deposited is \(1:1\).
\[ n(\ce{Zn})=9.99\times10^{-3}\ \mathrm{mol} \]
Now calculate the mass of zinc consumed:
\[ \begin{aligned} m(\ce{Zn}) &= nM\\\ &= (9.99\times10^{-3}\ \mathrm{mol})\\\ &\quad\times(65.38\ \mathrm{g\\,mol^{-1}})\\\ &\approx 0.653\ \mathrm{g} \end{aligned} \]
Zinc is oxidised, so it is the anode. Because the cell is galvanic, that anode is the negative terminal. Notice that the zinc mass lost is slightly greater than the copper mass deposited because zinc has a greater molar mass than copper.
Question 4
A voltmeter is connected to a spontaneous zinc-copper galvanic cell. Its red lead is connected to copper, and its black lead is connected to zinc. The voltmeter reads \(+1.10\ \mathrm{V}\).
A student concludes that copper must be the anode and that electrons flow from copper to zinc.
Explain what the voltmeter reading actually means, identify the electrodes, and state the directions of electron flow and conventional current in the external wire.
Solution 4
The reading means that the copper electrode is \(1.10\ \mathrm{V}\) higher in electrical potential than the zinc electrode. Copper is the positive cathode, zinc is the negative anode, electrons flow from zinc to copper, and conventional current flows from copper to zinc.
A positive voltmeter reading with the red lead on copper and the black lead on zinc indicates
\[ V_{\ce{Cu}}-V_{\ce{Zn}}=+1.10\ \mathrm{V} \]
This establishes the relative terminal polarity. It does not define which electrode is the anode.
The half-equations determine the electrode names: zinc is oxidised, so it is the anode; copper ions are reduced, so copper is the cathode.
Electrons travel through the external wire from zinc to copper. Conventional current is defined in the opposite direction, from copper to zinc. The student’s error is confusing electrical polarity and current direction with the definitions of oxidation and reduction.
Question 5
A zinc-copper cell initially operates as a spontaneous galvanic cell. An external power supply is then connected, with its positive terminal connected to copper and its negative terminal connected to zinc. The supply is strong enough to force the reverse metal-ion reactions. Assume that competing reactions do not occur.
After several minutes, the zinc electrode has gained mass and the copper electrode has lost mass.
(a) Identify the anode, cathode, and electrode signs while the power supply is connected.
(b) Describe electron flow through the external circuit and ion migration through a \(\ce{KNO3}\) salt bridge during this forced reaction.
(c) The power supply is removed, and the two electrodes are connected directly by a wire while the salt bridge remains in place. Predict the direction of the reaction. Explain why the electrode labels do not stay attached permanently to the same metals.
Solution 5
While the power supply drives the reverse reaction, copper is the positive anode and zinc is the negative cathode. When the supply is removed and the electrodes are connected directly by a wire, the spontaneous zinc-to-copper reaction resumes.
(a) Electrode names and signs
The observed mass changes show that zinc metal is being deposited while copper metal is dissolving.
At the copper electrode, oxidation occurs:
\[ \ce{Cu(s) -> Cu^2+(aq) + 2e^-} \]
Copper is therefore the anode. In this electrolytic operation, the anode is positive because it is connected to the positive terminal of the external power supply.
At the zinc electrode, reduction occurs:
\[ \ce{Zn^2+(aq) + 2e^- -> Zn(s)} \]
Zinc is the cathode. It is negative because it is connected to the negative terminal of the supply.
The observed mass changes agree with these assignments: copper loses mass as it enters the solution as \(\mathrm{Cu}^{2+}\), while zinc gains mass as \(\mathrm{Zn}^{2+}\) ions become solid zinc.
(b) Electron flow and salt-bridge migration
The power supply forces the reverse reaction by removing electrons from the copper anode and supplying electrons to the zinc cathode.
In the metal leads, electrons flow from the copper anode into the positive terminal of the power supply. Electrons are supplied from the negative terminal to the zinc cathode. Do not draw a direct electron-flow arrow through the salt bridge.
The salt-bridge ions also change which half-cell they enter:
- Nitrate ions, \(\mathrm{NO_{3}}^{-}\), migrate towards the copper anode compartment, where oxidation produces \(\mathrm{Cu}^{2+}\) ions.
- Potassium ions, \(\mathrm{K}^{+}\), migrate towards the zinc cathode compartment, where reduction consumes \(\mathrm{Zn}^{2+}\) ions.
The bridge still maintains charge balance. Its ion movements follow the electrode reactions, not the identity of the metal.
(c) Reaction after the power supply is removed
The original spontaneous reaction resumes: zinc is oxidised, and copper ions are reduced. Zinc becomes the negative anode again, copper becomes the positive cathode again, and electrons flow from zinc to copper through the wire.
During the forced reverse reaction, zinc ions were consumed to produce zinc metal, while copper metal was oxidised to produce copper ions. The solution therefore contains less \(\mathrm{Zn}^{2+}\) and more \(\mathrm{Cu}^{2+}\) than before the reverse reaction began. The reverse operation has increased the availability of the reactants for the original spontaneous reaction and decreased the relative amounts of its products.
Once the external supply is removed, the original reaction can proceed spontaneously in the forward direction, assuming the cell remains connected and the intended reactions continue to dominate.
The key constraint is that anode and cathode are reaction labels, not permanent names for zinc and copper. Zinc is the anode when zinc is oxidised, but it becomes the cathode when zinc ions are reduced. The same reasoning applies to other metals and electrochemical cells.
Once you can label the reactions consistently, the next useful check is the cell potential. Using standard reduction potentials, calculate
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]
A positive standard cell potential supports the proposed spontaneous galvanic direction under standard conditions. Assign the anode and cathode from the half-equations first, then use the potential calculation to check whether your proposed direction is correct.