How to Calculate Cell Potential from Reduction Potentials

A tin electrode and a copper electrode are connected in a galvanic cell. A standard reduction potential table gives \(-0.14\ \mathrm{V}\) for tin and \(+0.34\ \mathrm{V}\) for copper. Would the opposite signs partly cancel, giving a cell potential of \(0.20\ \mathrm{V}\)?

No. When both values are standard reduction potentials, subtract the anode’s value from the cathode’s:

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

For this cell:

\[ \begin{aligned} E^\circ_{\mathrm{cell}} &=(+0.34)-(-0.14)\\ &=+0.48\ \mathrm{V} \end{aligned} \]

The negative sign on tin’s reduction potential makes the difference larger, not smaller. The calculation predicts a standard cell potential of \(+0.48\ \mathrm{V}\). To understand why, we need to look at what the electrodes actually do, how the voltmeter is connected, and what the tabulated numbers mean.

01What is the voltmeter actually comparing?

Imagine building the tin-copper cell from scratch.

You have two beakers:

• The left contains a tin electrode immersed in a solution of tin(II) ions, \(\mathrm{Sn}^{2+}\).
• The right contains a copper electrode immersed in a solution of copper(II) ions, \(\mathrm{Cu}^{2+}\).

Both metal strips can conduct electrons. Both solutions contain mobile ions.

If you connect the strips to a voltmeter, which metal should become the negative electrode?

Predict it using what you know about oxidation and reduction. One metal must supply electrons, while ions in the other half-cell accept them.

Tin is more readily oxidised than copper under standard conditions. Tin atoms lose electrons and enter the solution:

\[ \ce{Sn(s) -> Sn^2+(aq) + 2e-} \]

This occurs at the anode.

At the copper electrode, copper(II) ions accept electrons and become copper metal:

\[ \ce{Cu^2+(aq) + 2e- -> Cu(s)} \]

This occurs at the cathode.

In this galvanic cell, electrons travel through the external circuit from the negative tin anode towards the positive copper cathode when the cell is allowed to supply current.

The overall reaction is:

\[ \ce{Sn(s) + Cu^2+(aq) -> Sn^2+(aq) + Cu(s)} \]

Tin gradually dissolves, while copper is deposited onto the copper electrode.

Why the first experimental design is incomplete

Suppose you put the electrodes in separate beakers and connect them only through an external wire.

Could the electron transfer continue indefinitely?

No. Oxidation produces additional \(\mathrm{Sn}^{2+}\) ions in the tin solution, making it more positively charged. Meanwhile, reduction removes \(\mathrm{Cu}^{2+}\) ions from the copper solution, leaving excess negative charge from its remaining anions.

That charge imbalance opposes further electron transfer.

You might detect a transient electrical response, but you haven’t built a complete ionic circuit capable of sustaining the intended galvanic reaction.

The solution is to connect the beakers with a salt bridge containing a suitable electrolyte, such as potassium nitrate.

The bridge allows ions to move between the half-cells without requiring the solutions to be mixed directly.

The two kinds of charge movement have different jobs.

Electrons do not travel through the salt bridge. Ions do.

A high-resistance voltmeter is used to measure the potential difference while drawing very little current. This avoids substantially changing the cell during measurement.

That distinction matters: the calculated standard cell potential is an ideal, near-zero-current value. The terminal voltage of a cell supplying a significant current can be lower.

02Why do we subtract reduction potentials?

The experimental problem doesn’t end when the salt bridge works.

A voltmeter measures the potential difference between two electrodes. It cannot measure the potential of an isolated half-cell without comparing it to something else.

Chemists solved this comparison problem by adopting a common reference electrode.

The standard hydrogen electrode

The standard hydrogen electrode, or SHE, uses a platinum electrode in contact with hydrogen gas and a solution containing hydrogen ions.

Its reduction half-equation is:

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

Its standard reduction potential is assigned:

\[ E^\circ=0.00\ \mathrm{V} \]

This is a convention, not a measurement showing that the hydrogen electrode has no electrical potential.

Other standard reduction potentials are determined relative to the same reference. Consequently, two tabulated values can be compared directly.

For a fuller explanation of how individual half-cell values are established, see Reduction Potentials of Galvanic Half-Cells: HSC Chemistry.

Reading the data table correctly

Here are selected standard reduction potentials at \(25^\circ\mathrm{C}\), rounded to two decimal places.

These are reference values for the worked scenarios, not measurements from an experiment performed for this article.

Every half-equation is written as a reduction, even when that half-cell will undergo oxidation in the galvanic cell.

A more positive reduction potential means a greater tendency for the reduction reaction to occur relative to another half-cell under the same standard conditions.

For a galvanic cell operating spontaneously under standard conditions:

1. Compare the two reduction potentials.
2. Identify the more positive value. That half-cell undergoes reduction at the cathode.
3. Reverse the less positive half-equation to represent oxidation at the anode.
4. Subtract the anode’s tabulated reduction potential from the cathode’s reduction potential.

The formula is:

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

Here:

• \(E^\circ_{\mathrm{cell}}\) is the standard cell potential, in volts (\(\mathrm{V}\)).
• \(E^\circ_{\mathrm{cathode}}\) is the tabulated standard reduction potential for the cathode half-cell.
• \(E^\circ_{\mathrm{anode}}\) is the tabulated standard reduction potential for the anode half-cell, even though oxidation occurs there.
• The superscript \(\circ\) indicates standard-state conditions.

There’s another valid way to calculate the same result: reverse the anode reaction, change the sign of its potential to obtain an oxidation potential, and add that to the cathode’s reduction potential.

Don’t mix the two methods. If you’ve already changed the sign to an oxidation potential, you must add it, not subtract it again.

Also, never multiply an electrode potential just because you’ve multiplied its half-equation to balance electrons. Voltage is energy per unit charge, not the total energy released by a chosen quantity of reactants.

05Standard cell potential versus actual measured voltage

There is one more limitation to the subtraction method.

The tabulated values are standard reduction potentials. They aren’t guaranteed voltages for a cell assembled using any concentration, temperature, or operating conditions.

For the values used here, standard-state conditions are commonly represented in HSC calculations by:

• A temperature of \(25^\circ\mathrm{C}\), or \(298.15\ \mathrm{K}\).
• Aqueous species at approximately \(1.0\ \mathrm{mol\,L^{-1}}\).
• Gases at a standard pressure of \(100\ \mathrm{kPa}\).
• Pure solids and liquids in their standard states.

Strictly, standard electrode potentials use chemical activities rather than simply the numerical concentrations. Treating \(1.0\ \mathrm{mol\,L^{-1}}\) as the standard aqueous condition is a useful school-level approximation.

Now return to the tin-copper cell.

Suppose the copper(II) ion solution becomes much more dilute while the tin(II) ion concentration stays the same.

Would the actual cell voltage necessarily remain \(+0.48\ \mathrm{V}\)?

No.

Lowering the copper(II) ion concentration changes the conditions for the reduction:

\[ \ce{Cu^2+(aq) + 2e- -> Cu(s)} \]

With fewer copper(II) ions available relative to the standard state, copper reduction becomes less favourable. The actual cell voltage decreases, assuming the other conditions remain unchanged.

The standard reduction potential \(E^\circ(\ce{Cu^2+/Cu})\) is still \(+0.34\ \mathrm{V}\). What has changed is the actual electrode potential under the new conditions.

For an idealised cell at \(25^\circ\mathrm{C}\), decreasing the copper(II) ion concentration from \(1.0\) to \(0.010\ \mathrm{mol\,L^{-1}}\) would give an actual cell potential of approximately \(+0.42\ \mathrm{V}\).

The difference comes from concentration, not from replacing either standard reduction potential in the table.

In a real laboratory, measured voltage can also be affected by electrode surfaces, temperature, liquid-junction effects, and the current drawn from the cell.

A high-resistance voltmeter reduces the effect of current draw. A cell driving a lamp or another substantial electrical load may show a lower terminal voltage because of internal resistance and changes near the electrodes.

How to interpret a disagreement with the calculated voltage

If a calculated standard potential is \(+0.48\ \mathrm{V}\), but a meter shows \(+0.42\ \mathrm{V}\), don’t immediately change a tabulated electrode potential.

First establish what was actually measured.

Was the cell prepared under standard conditions? Was the voltmeter connected with the expected polarity? Was the salt bridge intact? Was the cell delivering significant current?

A measured voltage is evidence about a particular cell under particular conditions.

A standard cell potential is a reference prediction.

For more on using the sign of a standard potential to assess redox reactions, see Calculate Standard Cell Potential and Predict Redox Reactions.

Syllabus context: As of November 2026, NSW students are studying the Chemistry Stage 6 Syllabus (2017). The Chemistry 11-12 Syllabus (2025) begins implementation in 2028. The final concentration-cell problem below is an optional extension beyond routine HSC standard-potential calculations.

06Questions and solutions

Question 1

An iron half-cell is connected to a copper half-cell under standard conditions.

The standard reduction potentials are:

\[ \begin{aligned} E^\circ(\ce{Fe^2+/Fe})&=-0.44\ \mathrm{V}\\ E^\circ(\ce{Cu^2+/Cu})&=+0.34\ \mathrm{V} \end{aligned} \]

Calculate the standard cell potential, identify the anode and cathode, and write the overall reaction.

Solution 1

The standard cell potential is \(+0.78\ \mathrm{V}\). Iron is the negative anode, and copper is the positive cathode.

Copper has the more positive reduction potential, so copper(II) ions gain electrons:

\[ \ce{Cu^2+(aq) + 2e- -> Cu(s)} \]

Iron undergoes oxidation:

\[ \ce{Fe(s) -> Fe^2+(aq) + 2e-} \]

Calculate the potential using both values as reductions:

\[ \begin{aligned} E^\circ_{\mathrm{cell}} &=E^\circ_{\mathrm{cathode}} -E^\circ_{\mathrm{anode}}\\ &=(+0.34)-(-0.44)\\ &=+0.78\ \mathrm{V} \end{aligned} \]

The balanced overall reaction is:

\[ \ce{Fe(s) + Cu^2+(aq) -> Fe^2+(aq) + Cu(s)} \]

The positive potential means this reaction is spontaneous under standard conditions. When the cell supplies current, electrons travel from the iron electrode towards the copper electrode.

The tempting error is to add \(-0.44\) and \(+0.34\) directly. That would give \(-0.10\ \mathrm{V}\), but it treats the iron half-cell as though its tabulated reduction reaction were occurring. Iron is actually being oxidised.

Question 2

A galvanic cell is constructed using two half-cells.

The first contains a tin electrode immersed in \(\mathrm{Sn}^{2+}\) solution.

The second contains a platinum electrode immersed in a solution containing both \(\mathrm{Fe}^{3+}\) and \(\mathrm{Fe}^{2+}\). Platinum provides a conducting surface but does not participate in the overall reaction.

A potassium nitrate salt bridge connects the solutions.

All conditions are initially standard.

The reduction potentials are:

\[ \begin{aligned} E^\circ(\ce{Fe^3+/Fe^2+})&=+0.77\ \mathrm{V}\\ E^\circ(\ce{Sn^2+/Sn})&=-0.14\ \mathrm{V} \end{aligned} \]

The molar mass of tin is \(118.7\ \mathrm{g\,mol^{-1}}\).

Determine:

• The standard cell potential and balanced overall reaction.
• The mass change of the tin electrode when \(0.0200\ \mathrm{mol}\) of \(\mathrm{Fe}^{3+}\) is reduced.
• The direction of potassium and nitrate ion movement through the bridge.
• Whether the platinum electrode should gain mass.

Assume sufficient reactants, no competing reactions, and that all stated electron transfer occurs through the intended cell reaction.

Solution 2

The standard cell potential is \(+0.91\ \mathrm{V}\). Tin loses approximately \(1.19\ \mathrm{g}\), while the platinum electrode gains no mass.

The less obvious point is that reduction does not always produce a metal deposit.

The iron(III)/iron(II) half-cell has the more positive reduction potential, so iron(III) ions accept electrons at the platinum surface:

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

Tin undergoes oxidation:

\[ \ce{Sn(s) -> Sn^2+(aq) + 2e-} \]

Calculate the standard potential:

\[ \begin{aligned} E^\circ_{\mathrm{cell}} &=(+0.77)-(-0.14)\\ &=+0.91\ \mathrm{V} \end{aligned} \]

To balance electrons, multiply the iron reduction half-equation by two:

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

The overall reaction is:

\[ \ce{Sn(s) + 2Fe^3+(aq) -> Sn^2+(aq) + 2Fe^2+(aq)} \]

Notice that the iron reduction potential is still \(+0.77\ \mathrm{V}\). Multiplying the half-equation does not multiply its potential.

Now calculate the change in tin mass.

Each mole of tin supplies two moles of electrons, reducing two moles of \(\mathrm{Fe}^{3+}\).

Therefore:

\[ \begin{aligned} n(\ce{Sn}) &=\frac{0.0200}{2}\\ &=0.0100\ \mathrm{mol} \end{aligned} \]

Using \(m=nM\), where \(m\) is mass, \(n\) is amount in moles, and \(M\) is molar mass:

\[ \begin{aligned} m(\ce{Sn}) &=0.0100\times118.7\\ &=1.187\ \mathrm{g}\\ &\approx1.19\ \mathrm{g} \end{aligned} \]

The tin electrode loses \(1.19\ \mathrm{g}\), because tin atoms become dissolved ions.

At the tin anode, positive ions are produced, so nitrate ions migrate towards that half-cell.

At the platinum cathode, \(\mathrm{Fe}^{3+}\) ions become \(\mathrm{Fe}^{2+}\), reducing the positive ionic charge in that solution. Potassium ions migrate towards the platinum half-cell to help maintain charge balance.

Platinum remains an inert conducting surface. Its mass does not increase because no iron metal is produced.

The tempting misconception is that every cathode must acquire a metal coating. Here, reduction changes the oxidation state of dissolved iron ions instead.

Question 3

Three separate standard half-cells are available:

• Tin: \(\ce{Sn^2+/Sn}\)
• Copper: \(\ce{Cu^2+/Cu}\)
• Silver: \(\ce{Ag+/Ag}\)

Each pair is connected using a salt bridge and a high-resistance voltmeter. The same reference conditions are maintained for the initial measurements.

Two measurements are recorded:

Without looking up another electrode potential, predict the reading when the red lead is connected to silver and the black lead to tin.

Then the silver solution is diluted while the other half-cell conditions are unchanged. A new silver-tin measurement gives \(+0.88\ \mathrm{V}\), with red on silver.

Explain whether this observation invalidates the first two measurements.

Finally, explain how copper can act as the cathode in one comparison and the anode in another.

Solution 3

The predicted standard reading is \(+0.94\ \mathrm{V}\). The later \(+0.88\ \mathrm{V}\) measurement does not invalidate it, because the silver half-cell was no longer under the original standard conditions.

The key is to treat voltmeter readings as signed differences.

Let \(V_{\mathrm{Sn}}\), \(V_{\mathrm{Cu}}\), and \(V_{\mathrm{Ag}}\) represent electrode potentials on the same reference scale.

Measurement A gives:

\[ V_{\mathrm{Cu}}-V_{\mathrm{Sn}} =+0.48\ \mathrm{V} \]

Measurement B gives:

\[ V_{\mathrm{Cu}}-V_{\mathrm{Ag}} =-0.46\ \mathrm{V} \]

Rearranging the second measurement:

\[ V_{\mathrm{Ag}}-V_{\mathrm{Cu}} =+0.46\ \mathrm{V} \]

The silver-tin difference is the sum of the two successive differences:

\[ \begin{aligned} V_{\mathrm{Ag}}-V_{\mathrm{Sn}} &=0.46+0.48\\ &=+0.94\ \mathrm{V} \end{aligned} \]

Therefore, a voltmeter with red on silver and black on tin should read \(+0.94\ \mathrm{V}\) under the original standard conditions.

The spontaneous overall reaction is:

\[ \ce{Sn(s) + 2Ag+(aq) -> Sn^2+(aq) + 2Ag(s)} \]

Tin is oxidised, silver ions are reduced, and the calculated cell potential is positive.

Now consider the later measurement of \(+0.88\ \mathrm{V}\).

The silver solution was diluted, changing the conditions for:

\[ \ce{Ag+(aq) + e- -> Ag(s)} \]

The actual silver electrode potential may therefore differ from its standard value.

A lower reading is consistent with a decrease in silver-ion concentration. However, the two original measurements alone cannot quantitatively predict the new value, and a single changed reading cannot establish that dilution was the only experimental influence.

Because the meter still reads positively with red on silver, silver remains the positive electrode relative to tin in the new measurement.

Finally, copper’s electrode role depends on what it is paired with.

Compared with tin, copper has the greater reduction tendency, so copper acts as the cathode.

Compared with silver, copper has the lower reduction tendency, so copper acts as the anode.

Anode and cathode are roles in a particular electrochemical cell. They are not permanent identities assigned to metals.

The deeper constraint here is that voltage differences can be combined only when the compared electrode states remain consistent. Once the silver solution changes, the earlier measurements no longer describe the same complete set of half-cell conditions.

Question 4

Two tin-copper cells are prepared at \(25^\circ\mathrm{C}\).

Both use the same metal electrodes, potassium nitrate salt bridges, and high-resistance voltmeters.

The red lead is connected to copper in both cases.

These are illustrative experimental results.

A student makes three claims:

1. The standard reduction potential of copper in Cell B must have decreased.
2. The different copper-ion concentrations could explain the difference in measured voltage.
3. Because the voltmeter still shows a positive reading in Cell B, copper remains the positive electrode.

Evaluate all three claims.

Then predict what should happen qualitatively if Cell B’s copper-ion concentration is restored to \(1.0\ \mathrm{mol\,L^{-1}}\), assuming the other conditions remain unchanged.

Solution 4

Claim 1 is incorrect, while Claims 2 and 3 are justified. Restoring Cell B’s copper-ion concentration towards its standard value should increase the measured voltage towards \(+0.48\ \mathrm{V}\), under the stated assumptions.

First, calculate the standard prediction.

The tabulated potentials are:

\[ \begin{aligned} E^\circ(\ce{Cu^2+/Cu})&=+0.34\ \mathrm{V}\\ E^\circ(\ce{Sn^2+/Sn})&=-0.14\ \mathrm{V} \end{aligned} \]

Therefore:

\[ \begin{aligned} E^\circ_{\mathrm{cell}} &=(+0.34)-(-0.14)\\ &=+0.48\ \mathrm{V} \end{aligned} \]

This standard value doesn’t change simply because Cell B has a different copper-ion concentration.

Claim 1 confuses a standard reduction potential with the actual potential under nonstandard conditions.

Claim 2 is reasonable because copper(II) ions are the reactants in the cathode reaction:

\[ \ce{Cu^2+(aq) + 2e- -> Cu(s)} \]

Diluting the copper solution makes this reduction less favourable relative to the standard state, reducing the cell’s actual potential.

The observed difference is:

\[ \begin{aligned} \Delta E &=0.48-0.42\\ &=0.06\ \mathrm{V} \end{aligned} \]

That change is consistent with the effect of the stated dilution at \(25^\circ\mathrm{C}\). It is not evidence that the tabulated value \(E^\circ(\ce{Cu^2+/Cu})\) has changed.

Claim 3 follows from the measured polarity. The red lead is attached to copper and the reading is positive, so copper is at the higher electrical potential. In the intended spontaneously operating galvanic cell, copper is the cathode and tin is the anode.

The corresponding bridge-ion movements remain nitrate towards tin and potassium towards copper.

If Cell B is returned to standard copper-ion concentration while the other conditions remain fixed, its voltage should rise towards the standard prediction of \(+0.48\ \mathrm{V}\).

An experimental qualification is important: if a real measured voltage differs from the standard prediction, concentration is not automatically the only possible cause. Electrode condition, junction effects, and measurement conditions must also be considered.

The useful distinction is between a reference potential, which belongs to defined standard conditions, and an actual reading, which belongs to the particular cell being measured.

Question 5

Extension challenge – beyond routine HSC standard-potential calculations.

Two copper electrodes are placed in separate solutions containing copper(II) ions.

The first beaker contains:

• \(10.0\ \mathrm{mL}\) of \(1.00\ \mathrm{mol\,L^{-1}}\) \(\mathrm{Cu}^{2+}\).

The second beaker contains:

• \(2.00\ \mathrm{L}\) of \(0.0100\ \mathrm{mol\,L^{-1}}\) \(\mathrm{Cu}^{2+}\).

The half-cells are connected by a potassium nitrate salt bridge and an external conducting circuit.

The standard reduction potential for both electrodes is:

\[ E^\circ(\ce{Cu^2+/Cu})=+0.34\ \mathrm{V} \]

A student argues:

“Both electrodes are copper, and both standard reduction potentials are identical. Therefore, the cell voltage must be zero and no reaction can occur.”

Another student disagrees:

“The large, dilute beaker contains more copper(II) ions in total, so it must become the cathode.”

Investigate both claims.

Determine:

• Which electrode initially acts as the anode and which acts as the cathode, and whether the initial cell voltage can be nonzero.
• The direction of electron flow and the changes occurring at both copper electrodes.
• The final common copper(II) ion concentration when the cell reaches equilibrium.
• The mass gained or lost by each electrode.
• Whether the cell can spontaneously continue until its electrode polarities reverse.

Use \(M(\ce{Cu})=63.55\ \mathrm{g\,mol^{-1}}\).

Assume \(25^\circ\mathrm{C}\), ideal solution behaviour, constant solution volumes, well-mixed solutions within each beaker, negligible liquid-junction effects, and no competing reactions. Potassium and nitrate ions may travel through the salt bridge, but copper(II) ions do not cross it. Both copper electrodes have sufficient metal for the process to reach equilibrium without an external power source.

Solution 5

The concentrated copper half-cell is initially the cathode, even though the larger, dilute beaker contains more copper(II) ions in total. The cell can produce a nonzero voltage despite both standard reduction potentials being \(+0.34\ \mathrm{V}\).

Under the stated ideal assumptions, the final copper(II) ion concentration is approximately \(0.0149\ \mathrm{mol\,L^{-1}}\). The concentrated-side copper electrode gains approximately \(0.626\ \mathrm{g}\), while the dilute-side electrode loses the same mass.

This is a concentration cell. Its behaviour is determined by the difference in actual electrode conditions, not by a difference in their standard reduction potentials.

Step 1: Explain why equal standard potentials do not guarantee zero actual voltage.

If both copper half-cells were under identical conditions, their potentials would be equal.

The standard cell potential calculated from the table would be:

\[ \begin{aligned} E^\circ_{\mathrm{cell}} &=(+0.34)-(+0.34)\\ &=0.00\ \mathrm{V} \end{aligned} \]

But these two half-cells have very different copper(II) ion concentrations.

The reduction reaction is:

\[ \ce{Cu^2+(aq) + 2e- -> Cu(s)} \]

At higher copper(II) ion concentration, reduction is more favourable than at lower concentration, assuming the same other conditions.

Therefore, the concentrated half-cell has the higher actual reduction potential.

It acts as the cathode.

The dilute half-cell acts as the anode.

The standard potentials are identical, but the actual electrode potentials are not.

Step 2: Identify the electron flow and electrode reactions.

At the dilute anode, copper metal dissolves:

\[ \ce{Cu(s) -> Cu^2+(aq) + 2e-} \]

At the concentrated cathode, copper ions are deposited:

\[ \ce{Cu^2+(aq) + 2e- -> Cu(s)} \]

Electrons flow through the external circuit from the dilute-side electrode towards the concentrated-side electrode.

Nitrate ions from the bridge migrate towards the dilute anode, while potassium ions migrate towards the concentrated cathode.

The unusual feature is that the same metal is oxidised in one compartment and deposited in the other.

The process transfers copper from the electrode in the dilute compartment to the electrode in the concentrated compartment. At the same time, it transfers copper(II) ions in the opposite sense: they are produced in the dilute solution and consumed in the concentrated solution.

The concentration difference gradually decreases.

Step 3: Test the claim about the number of copper(II) ions.

Calculate the initial amount of dissolved copper(II) in each beaker.

Use:

\[ n=cV \]

where \(n\) is amount in moles, \(c\) is concentration in \(\mathrm{mol\,L^{-1}}\), and \(V\) is solution volume in litres.

For the concentrated half-cell:

\[ \begin{aligned} n_{\mathrm{conc}} &=1.00\times0.0100\\ &=0.0100\ \mathrm{mol} \end{aligned} \]

For the dilute half-cell:

\[ \begin{aligned} n_{\mathrm{dilute}} &=0.0100\times2.00\\ &=0.0200\ \mathrm{mol} \end{aligned} \]

The dilute beaker actually contains twice as many moles of copper(II) ions.

Yet it is still the anode.

Why?

Because electrode potential depends on the relevant chemical activities, approximated here by concentrations, not simply on the total number of dissolved ions in an entire container.

A large volume of dilute solution can contain more ions overall than a small volume of concentrated solution while still having a lower reduction potential.

The second student’s argument therefore fails.

Step 4: Calculate the final common concentration.

As the galvanic cell operates, the concentration in the dilute beaker rises, while the concentration in the concentrated beaker falls.

With identical copper electrodes, ideal solutions, and negligible junction effects, the driving voltage reaches zero when the two copper(II) ion concentrations become equal.

The overall number of dissolved copper(II) ions is conserved across the two beakers.

For every mole of copper(II) ions consumed at the concentrated cathode, one mole is produced at the dilute anode.

Initially:

\[ \begin{aligned} n_{\mathrm{total}} &=0.0100+0.0200\\ &=0.0300\ \mathrm{mol} \end{aligned} \]

The combined solution volume is:

\[ \begin{aligned} V_{\mathrm{total}} &=0.0100+2.00\\ &=2.010\ \mathrm{L} \end{aligned} \]

At equilibrium, both solutions have the same concentration, so:

\[ \begin{aligned} c_{\mathrm{final}} &=\frac{n_{\mathrm{total}}}{V_{\mathrm{total}}}\\ &=\frac{0.0300}{2.010}\\ &=0.0149\ \mathrm{mol\,L^{-1}} \end{aligned} \]

The final common concentration is approximately \(0.0149\ \mathrm{mol\,L^{-1}}\).

A tempting mistake would be to average \(1.00\) and \(0.0100\) directly. That ignores the enormous difference in solution volumes.

The correct calculation conserves the total amount of copper(II) ions.

Step 5: Calculate the electrode mass changes.

The small, originally concentrated beaker initially contained \(0.0100\ \mathrm{mol}\) of copper(II) ions.

At equilibrium, its amount is:

\[ \begin{aligned} n_{\mathrm{final}} &=0.0149254\times0.0100\\ &=0.0001493\ \mathrm{mol} \end{aligned} \]

The amount consumed at its cathode is:

\[ \begin{aligned} n_{\mathrm{consumed}} &=0.0100-0.0001493\\ &=0.0098507\ \mathrm{mol} \end{aligned} \]

Each mole of copper(II) ions reduced deposits one mole of copper metal.

Therefore:

\[ \begin{aligned} m_{\mathrm{gained}} &=nM\\ &=0.0098507\times63.55\\ &=0.626\ \mathrm{g} \end{aligned} \]

The concentrated-side copper electrode gains approximately \(0.626\ \mathrm{g}\).

Because the same amount of copper dissolves from the other electrode, the dilute-side copper electrode loses \(0.626\ \mathrm{g}\).

This agrees with conservation of copper across the complete system.

Step 6: Decide whether the cell can reverse its own polarity.

It cannot spontaneously pass through equilibrium and continue running in the opposite direction under the stated assumptions.

As the two solution concentrations approach one another, their reduction potentials approach equality.

The driving voltage decreases towards zero.

Once the concentrations are equal, the two identical half-cells have no potential difference available to drive net electron transfer.

A spontaneous continuation beyond this point would increase the concentration difference again, rather than reduce it. That would require an external energy input.

The important constraint is equilibrium. The cell doesn’t automatically reverse just because one electrode has been dissolving and the other has been gaining mass.

The first student’s error was treating standard potentials as though they completely described nonstandard conditions. The second student’s error was confusing total ion amount with ion concentration.

Both assumptions sound reasonable until you examine exactly what the electrode potential represents.

This is the boundary of the two-reduction-potential method. When a problem supplies standard conditions, subtract the tabulated values. When concentrations change, the actual voltage also changes, and the Nernst equation provides the quantitative connection between reaction conditions and cell potential. That is the next useful concept when a calculated standard voltage doesn’t match a real measurement.