Why a Galvanic Cell’s Anode Is Negative (HSC Chemistry)

A zinc strip releases electrons when it reacts in a galvanic cell. So why is the zinc electrode called negative when the zinc ions entering the solution are positive?

Because the anode is the electrode where oxidation supplies electrons to the external circuit. In a galvanic cell, that makes it the negative terminal relative to the cathode, where electrons are accepted. The sign describes the electrical potential of one electrode compared with the other. It does not describe the charge of the ions leaving the metal.

Here’s a useful prediction. If you connect a voltmeter’s red lead to zinc and its black lead to copper in a working zinc-copper cell, will it show a positive reading because zinc is producing electrons?

No. It will show a negative reading because zinc is at the lower electrical potential. Understanding that result explains both why the galvanic anode is negative and why the same rule cannot be applied blindly to every electrochemical cell.

01Testing the sign with a zinc-copper cell

Imagine trying to determine which electrode is negative without being told which metal is the anode.

Set up two beakers.

The left beaker contains a zinc strip immersed in zinc sulfate solution, \(\ce{ZnSO4(aq)}\). The right beaker contains a copper strip immersed in copper sulfate solution, \(\ce{CuSO4(aq)}\).

Connect the solutions using a U-shaped tube containing potassium nitrate solution, \(\ce{KNO3(aq)}\), held in place by porous plugs. This is the salt bridge.

Finally, connect the metal strips to a high-resistance digital voltmeter. Attach the black lead to zinc and the red lead to copper.

What should the voltmeter show?

Under ideal standard conditions, the expected cell potential is approximately \(1.10\ \text{V}\).

A real classroom reading may differ. Consider this set of illustrative results from the same cell, with nothing changed except the voltmeter connections.

These are invented readings to demonstrate the measurement, not results from a performed experiment.

The magnitude remains the same, but the sign changes.

Why?

A voltmeter displays the electrical potential at its red input minus the potential at its black input:

\[
V_{\text{meter}}=V_{\text{red}}-V_{\text{black}}
\]

Here, \(V\) represents electrical potential, measured in volts (\(\text{V}\)).

Copper is at the higher potential. Therefore, connecting the red lead to copper gives a positive reading.

Connecting it to zinc gives a negative reading.

The chemical reaction hasn’t reversed. Only the instrument’s reference direction has changed.

Why the voltmeter test isn’t enough

The meter establishes a potential difference. It does not directly show which atoms are losing electrons.

A high-resistance voltmeter also draws very little current. You cannot use a voltage reading alone as evidence that a substantial redox reaction is continuously occurring.

To identify the electrode reactions, revise the test.

Keep the two half-cells and salt bridge, but replace the voltmeter’s external connection with a resistor and ammeter in series. The resistor allows a controlled current, and the ammeter measures that current.

During sustained operation, the predicted chemical changes are:

• The zinc strip gradually loses mass as zinc atoms enter the solution.
• The copper strip gains mass as copper ions are reduced and deposited onto its surface.
• The ammeter detects current while both the external electrical path and the internal ionic path are complete.

The mass changes provide independent chemical evidence. Unlike the sign on a voltmeter, they do not depend on which way the meter leads were connected.

The earlier measurement tells us which electrode is at higher potential. The revised test helps identify the reactions that maintain the current.

02Why oxidation makes the galvanic anode negative

Start at the surface of the zinc strip.

A zinc atom can become a zinc ion by losing two electrons:

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

The zinc ion enters the surrounding solution. The electrons remain in the conducting metal and can travel through the external circuit.

This process is oxidation, so zinc is the anode.

At the copper electrode, a different process occurs:

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

Copper(II) ions accept electrons and become copper atoms, which join the solid electrode.

This is reduction, so copper is the cathode.

Now follow what happens when the cell supplies current.

At the zinc electrode, oxidation continually makes electrons available. Those electrons leave through the connecting wire.

At the copper electrode, reduction consumes electrons arriving from that wire.

The galvanic reaction therefore maintains an electron flow from zinc to copper.

The zinc electrode is the electron-supplying terminal, at lower electrical potential. The copper electrode is the electron-receiving terminal, at higher potential.

That is the physical meaning of the negative and positive labels.

Does zinc become more and more negatively charged?

It might seem that way. Zinc atoms keep releasing electrons, so perhaps the metal should eventually accumulate an enormous negative charge.

But that isn’t what happens.

When the external circuit is conducting, electrons leave the zinc electrode as oxidation supplies them. At the copper electrode, arriving electrons are consumed by reduction.

A small redistribution of charge at the metal-solution interfaces establishes the electrical potential difference. The electrodes do not need to store a growing pile of excess electrons.

The bulk metals and solutions remain approximately electrically neutral during normal operation.

This is the limit of saying that the anode is negative because it produces electrons. Electron production explains its role in the circuit, but the actual electrode potential depends on the electrochemical conditions at both electrodes.

When the circuit is open, the cell can still have a potential difference even though there is essentially no net electron flow through the external connection.

03How can the solutions remain electrically neutral?

Follow the charge changes inside the two beakers.

At the zinc anode, every zinc atom that oxidises produces one \(\mathrm{Zn}^{2+}\) ion. Positive ions are being added to the solution.

At the copper cathode, \(\mathrm{Cu}^{2+}\) ions leave the solution as they become copper metal. Positive ions are being removed.

If these changes continued without any way to compensate for charge, an electrical imbalance would quickly oppose further reaction.

The salt bridge provides that compensation.

For a potassium nitrate bridge:

The nitrate ions help compensate for the production of positive zinc ions. The potassium ions help compensate for the removal of positive copper ions.

These arrows describe the net charge-compensating migration. Real ions can also diffuse because of concentration differences.

Notice the apparent contradiction.

The negative anode is surrounded by a solution into which positive zinc ions are being produced. Meanwhile, negative nitrate ions move towards that compartment.

There’s no contradiction once you separate three different things:

• Electrode polarity describes electrical potential relative to the other electrode.
• The half-equation describes the electrons lost or gained in a chemical reaction.
• Ion migration maintains charge balance inside the cell.

The negative sign on the anode does not mean that every ion nearby is negative.

For a broader method of identifying electrode roles from reactions, see Anode or Cathode? A Reliable Sign Check for Galvanic Cells.

04How standard reduction potentials explain the sign

We’ve established the physical reason for zinc being negative. Now we can predict the polarity before connecting a voltmeter.

Chemists use standard reduction potentials, written \(E^\circ\), to compare how strongly different half-reactions favour reduction under standard conditions.

A reduction potential is tabulated for the reaction written with electrons on the left.

Consider these values.

All three values are standard reduction potentials, not measured charges on the metals.

Under standard conditions, the more positive reduction potential identifies the half-reaction that is more favourable as a reduction when the two half-cells are paired.

For zinc and copper, copper has the more positive reduction potential.

Therefore, in a spontaneously operating galvanic cell:

• Copper(II) ions are reduced at the cathode.
• Zinc metal is oxidised at the anode.

The standard cell potential is calculated using:

\[
E^\circ_{\text{cell}}
=
E^\circ_{\text{cathode}}
–
E^\circ_{\text{anode}}
\]

Both values on the right are taken from the reduction-potential table. Do not reverse the anode value before subtracting it, or you will change its sign twice.

For zinc-copper:

\[
\begin{aligned}
E^\circ_{\text{cell}}
&=0.34-(-0.76)\\
&=+1.10\ \text{V}
\end{aligned}
\]

The positive result means that the reaction is thermodynamically spontaneous in the direction written under standard conditions.

The overall reaction is:

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

Why is zinc’s standard reduction potential negative?

The value \(-0.76\ \text{V}\) is not the electrical charge on a zinc strip.

Standard electrode potentials are measured relative to a reference electrode. Chemists assign the standard hydrogen electrode a reduction potential of \(0.00\ \text{V}\).

Zinc’s standard reduction potential is lower than that reference value. Copper’s is higher.

The reference is necessary because a voltmeter measures differences between electrodes, not an isolated electrode’s absolute potential.

This convention is explained more fully in Reduction Potentials of Galvanic Half-Cells: HSC Chemistry.

05A copper electrode can be positive in one cell and negative in another

It’s tempting to think of zinc as a naturally negative metal and copper as a naturally positive metal.

That prediction works for zinc-copper cells, but it fails when copper is paired with a different electrode.

Keep a copper electrode in copper sulfate solution.

First, connect it to a zinc half-cell. Copper is the cathode because it has the higher reduction potential.

Now replace the zinc half-cell with a silver half-cell containing a silver electrode and silver nitrate solution.

Silver has a more positive standard reduction potential than copper.

So which metal is now the anode?

Copper.

The copper electrode now undergoes oxidation:

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

Silver ions undergo reduction:

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

Copper has changed from a positive cathode to a negative anode without changing what metal it is.

The partner electrode determines which reaction is spontaneous under the specified conditions.

06Standard conditions are not the same as actual measurements

The familiar zinc-copper value of \(1.10\ \text{V}\) is a standard cell potential.

The standard reduction potentials used here are tabulated at \(25^\circ\text{C}\). For these aqueous half-cells, standard states mean unit activity of the dissolved ions, commonly approximated as \(1.0\ \text{mol L}^{-1}\) concentrations in HSC calculations.

Pure solids are in their standard states, and a gas, where involved, has a standard pressure of \(1\ \text{bar}\), or \(100\ \text{kPa}\).

Real solutions do not always behave ideally, so concentration and activity are not exactly interchangeable.

Now consider a working zinc-copper cell.

As current flows, \(\mathrm{Zn}^{2+}\) is produced and \(\mathrm{Cu}^{2+}\) is consumed. These changes affect the chemical driving force.

In an idealised cell at constant temperature, increasing the amount of zinc ions relative to copper ions makes the forward reaction less favourable, so the equilibrium cell potential decreases.

The voltage measured across a cell supplying current can also be lower than its near-open-circuit voltage because of internal resistance, electrode polarisation, and limitations on ion transport.

This distinction matters when interpreting results.

A lower measured voltage does not automatically mean the electrodes have swapped roles.

And a negative meter reading does not automatically mean the spontaneous cell reaction is running backwards.

You need the conditions, the connections, and the reaction evidence.

07When the anode really is positive

There’s one final comparison that prevents a common mistake.

A galvanic cell generates electrical energy from a spontaneous redox reaction.

An electrolytic cell uses an external power supply to force a reaction in a direction that would not occur spontaneously under those conditions.

The electrode names still depend on the chemistry:

Imagine taking the zinc-copper cell and connecting a sufficiently powerful DC supply with its positive terminal attached to copper and its negative terminal attached to zinc.

If the supply drives the reverse reaction, copper dissolves:

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

Zinc ions gain electrons and deposit on the zinc electrode:

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

The copper electrode is now the positive anode. Zinc is the negative cathode.

The external source is forcing the reaction against its spontaneous direction.

This is not just a change of labels. The actual electrode reactions have reversed.

The exact operating threshold depends on the concentrations, temperature, resistance, and electrode losses. A power supply must overcome the opposing cell potential and any additional losses to sustain reverse current.

The important rule is that anode means oxidation, not negative.

In a normally operating galvanic cell, the oxidation electrode is negative relative to the reduction electrode. In an electrolytic cell driven by a power supply, the oxidation electrode is positive.

08Questions and solutions

These questions move from ordinary electrode identification to experiments where signs, chemical reactions, and electrical measurements give apparently conflicting clues.

For each question, use the conditions stated. Where voltages are idealised, ignore electrode polarisation and unwanted side reactions unless the question says otherwise.

Question 1

A galvanic cell is constructed from iron and silver half-cells under standard conditions.

The standard reduction potentials are:

\[
\begin{aligned}
E^\circ_{\ce{Fe^2+/Fe}}&=-0.44\ \text{V}\\
E^\circ_{\ce{Ag+/Ag}}&=+0.80\ \text{V}
\end{aligned}
\]

Identify the negative electrode and positive electrode. Write the oxidation and reduction half-equations, calculate the standard cell potential, and state the direction of electron flow when the cell supplies current.

Solution 1

Iron is the negative anode, silver is the positive cathode, and the standard cell potential is \(+1.24\ \text{V}\). When the cell operates, electrons travel from iron to silver.

Silver has the more positive reduction potential, so silver ions undergo reduction:

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

Iron has the lower reduction potential, so iron undergoes oxidation:

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

The silver half-equation must be doubled to balance the electrons:

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

Calculate the standard cell potential:

\[
\begin{aligned}
E^\circ_{\text{cell}}
&=0.80-(-0.44)\\
&=+1.24\ \text{V}
\end{aligned}
\]

The positive result confirms that the written reaction is spontaneous under standard conditions.

A common mistake is to call silver the negative electrode because silver ions gain negatively charged electrons. But gaining electrons is reduction, and reduction occurs at the positive cathode in a galvanic cell.

Question 2

A zinc-copper galvanic cell contains a potassium nitrate salt bridge and operates under non-standard conditions.

The red voltmeter lead is attached to zinc, while the black lead is attached to copper. The voltmeter reads \(-1.06\ \text{V}\) when the cell is effectively open-circuit.

The voltmeter is then replaced by a resistor so that the cell supplies current. After a period of operation, the zinc electrode has lost \(0.0981\ \text{g}\), and the copper electrode has gained \(0.0953\ \text{g}\).

The molar masses are:

\[
\begin{aligned}
M(\ce{Zn})&=65.4\ \text{g mol}^{-1}\\
M(\ce{Cu})&=63.5\ \text{g mol}^{-1}
\end{aligned}
\]

Assume the measured mass changes result only from the intended electrode reactions.

Determine whether the voltage reading and mass changes are consistent with ordinary galvanic operation. Calculate the amount of each metal reacted, explain the electron transfer, and state the direction of migration of the salt-bridge ions.

Solution 2

The observations are consistent with normal zinc-copper galvanic operation. Zinc is being oxidised at the negative anode, copper ions are being reduced at the positive cathode, and the negative voltmeter reading is caused by the lead orientation.

First, calculate the amount of zinc lost:

\[
\begin{aligned}
n(\ce{Zn})
&=\frac{m}{M}\\
&=\frac{0.0981}{65.4}\\
&=1.50\times10^{-3}\ \text{mol}
\end{aligned}
\]

Here, \(n\) is amount in moles, \(m\) is mass in grams, and \(M\) is molar mass in grams per mole.

The zinc half-equation is:

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

Each mole of zinc releases two moles of electrons.

Therefore:

\[
\begin{aligned}
n(e^-)
&=2n(\ce{Zn})\\
&=2(1.50\times10^{-3})\\
&=3.00\times10^{-3}\ \text{mol}
\end{aligned}
\]

Now calculate the amount of copper deposited:

\[
\begin{aligned}
n(\ce{Cu})
&=\frac{0.0953}{63.5}\\
&\approx1.50\times10^{-3}\ \text{mol}
\end{aligned}
\]

Copper deposition follows:

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

The two processes require equal amounts of zinc oxidation and copper deposition because both half-reactions involve two electrons.

The measured mole amounts agree to the precision of the data. The mass measurements therefore support the proposed electron transfer.

The electrons travel through the external circuit from zinc to copper.

Meanwhile, nitrate ions migrate towards the zinc compartment, where positive zinc ions are being produced. Potassium ions migrate towards the copper compartment, where copper ions are being consumed.

Why is the voltmeter reading negative?

Because it measures the potential at zinc relative to copper. The red lead is attached to the lower-potential electrode.

If the leads were reversed without changing the chemistry, the meter would read \(+1.06\ \text{V}\).

The tempting interpretation is that a negative voltage means a non-spontaneous reaction. That confuses an instrument reading with the cell potential defined for the spontaneous reaction.

Question 3

An unknown metal \(\ce{X}\) forms \(\mathrm{X}^{2+}\) ions.

Two separate galvanic cells are assembled under standard conditions. Each contains an identical \(\ce{X^2+/X}\) half-cell and a potassium nitrate salt bridge.

In both experiments, the red voltmeter lead is attached to the \(\ce{X}\) electrode.

Given:

\[
\begin{aligned}
E^\circ_{\ce{Cu^2+/Cu}}&=+0.34\ \text{V}\\
E^\circ_{\ce{Zn^2+/Zn}}&=-0.76\ \text{V}
\end{aligned}
\]

Determine the standard reduction potential of \(\ce{X^2+/X}\).

Then identify the anode and cathode in both experiments, write the electrode reaction involving \(\ce{X}\) in each, and explain how the same metal can give readings of opposite signs without any reversed voltmeter leads.

State which potassium nitrate ion migrates towards the \(\ce{X}\) half-cell in each experiment.

Solution 3

The standard reduction potential of the unknown half-cell is \(+0.16\ \text{V}\). The unknown metal is the negative anode when paired with copper but the positive cathode when paired with zinc.

This happens because electrode polarity is relative to the other half-cell.

Test A: Copper and X

The red lead is on X, and the black lead is on copper.

Therefore:

\[
\begin{aligned}
E^\circ_{\ce{X^2+/X}}-0.34
&=-0.18\\
E^\circ_{\ce{X^2+/X}}
&=+0.16\ \text{V}
\end{aligned}
\]

The potential of X is lower than that of copper.

Copper is therefore the cathode, and X is the negative anode.

The reaction at X is oxidation:

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

Electrons move from X to copper when the cell supplies current.

Because positive \(\mathrm{X}^{2+}\) ions are being produced, nitrate ions migrate towards the X half-cell.

Test B: Zinc and X

Using the potential found above:

\[
\begin{aligned}
E^\circ_{\ce{X^2+/X}}
–
E^\circ_{\ce{Zn^2+/Zn}}
&=0.16-(-0.76)\\
&=+0.92\ \text{V}
\end{aligned}
\]

This agrees with the second reading.

X has the higher reduction potential, so it is the positive cathode. Zinc is the negative anode.

The reaction at X is now reduction:

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

Electrons move from zinc to X.

Because X ions are being consumed, potassium ions migrate towards the X half-cell.

The important constraint is that changing the partner metal changes which reduction is favourable.

X has not changed its chemical identity, and the meter leads have not been reversed. The relative electrochemical potentials have changed because a different half-cell is present.

That explains both the opposite voltage signs and the reversal of X’s electrode reaction.

Question 4

An idealised zinc-copper galvanic cell has an open-circuit electromotive force of \(1.10\ \text{V}\).

For this question, represent the cell using an ideal \(1.10\ \text{V}\) source with a fixed internal resistance of \(2.0\ \Omega\), where \(\Omega\) represents ohms.

A variable external resistor is connected between the electrodes. A voltmeter is connected across the cell, with its red lead on copper and black lead on zinc.

First, the external resistance is \(10.0\ \Omega\).

Then consider the mathematical limit where the external resistance is reduced to \(0\ \Omega\), creating an ideal short circuit.

Assume the cell’s emf and internal resistance remain constant during these initial measurements. Ignore polarisation, heating, and changes in solution composition.

1. Calculate the current and terminal voltage when the external resistance is \(10.0\ \Omega\).
2. Calculate the initial current and terminal voltage in the ideal short-circuit limit.
3. A student argues: “If the voltmeter reads zero in the short circuit, there can be no electrochemical driving force. Zinc must have stopped being the anode.” Evaluate this claim.
4. Explain how the phrase “negative anode” must be interpreted when the two external terminals are held at the same potential.

Solution 4

The cell supplies approximately \(0.0917\ \text{A}\) at \(10.0\ \Omega\), with a terminal voltage of \(+0.917\ \text{V}\). In the ideal short-circuit limit, current increases to \(0.55\ \text{A}\) even though the terminal voltage becomes zero.

Zinc can therefore continue undergoing oxidation while a voltmeter reads zero across the shorted terminals.

Part 1: External resistance of \(10.0\ \Omega\)

The total resistance is:

\[
R_{\text{total}}=10.0+2.0=12.0\ \Omega
\]

Using Ohm’s law, \(I=V/R\), where \(I\) is current in amperes, \(V\) is voltage in volts, and \(R\) is resistance in ohms:

\[
\begin{aligned}
I&=\frac{1.10\ \text{V}}{12.0\ \Omega}\\
&=0.0917\ \text{A}
\end{aligned}
\]

The terminal voltage equals the voltage across the external resistor:

\[
\begin{aligned}
V_{\text{terminal}}
&=IR\\
&=(0.0917)(10.0)\\
&=+0.917\ \text{V}
\end{aligned}
\]

Copper remains at the higher terminal potential, so the voltmeter displays a positive value.

Part 2: Short-circuit limit

When the external resistance becomes zero, only the modelled internal resistance limits the current.

\[
\begin{aligned}
I_{\text{short}}
&=\frac{1.10\ \text{V}}{2.0\ \Omega}\\
&=0.55\ \text{A}
\end{aligned}
\]

The terminal voltage is:

\[
\begin{aligned}
V_{\text{terminal}}
&=IR_{\text{external}}\\
&=(0.55)(0)\\
&=0.00\ \text{V}
\end{aligned}
\]

The entire modelled \(1.10\ \text{V}\) emf is balanced by the voltage drop associated with the internal resistance.

This short circuit is a theoretical limit, not a recommended experiment on a real battery.

Part 3: Does zero terminal voltage mean the reaction has stopped?

No.

A terminal voltage of zero does not imply that the cell’s chemical driving force has disappeared.

Under the question’s assumptions, the cell continues to support a current of \(0.55\ \text{A}\).

Zinc still undergoes oxidation:

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

Copper ions still undergo reduction:

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

The current is actually larger than it was with the \(10.0\ \Omega\) external resistor.

The student’s mistake is confusing the terminal potential difference under load with the internal electromotive force.

Part 4: Is zinc still the negative anode?

Zinc is still the anode because oxidation occurs there.

However, when the terminals are connected by an ideal zero-resistance wire, their measured potential difference is zero. Neither external terminal is lower in potential than the other.

So “negative anode” is a description of normal galvanic terminal polarity, not an unchangeable claim that a negative terminal voltage must exist under every circuit condition.

The chemical definition remains valid even when the usual voltage-sign description loses its usefulness.

This is the key surprise: a cell can be actively transferring electrons while the terminal voltage reads zero.

Question 5

Two separate electrochemical cells are initially assembled under standard conditions at \(25^\circ\text{C}\).

Cell A is a zinc-copper cell:

\[
E^\circ_A=+1.10\ \text{V}
\]

Cell B is a copper-silver cell:

\[
E^\circ_B=+0.46\ \text{V}
\]

Each cell has its own potassium nitrate salt bridge.

When operated independently as a galvanic cell, Cell A has zinc as its negative terminal and copper as its positive terminal. Cell B has copper as its negative terminal and silver as its positive terminal.

Now connect the two cells in opposition:

• Connect Cell A’s positive copper terminal directly to Cell B’s positive silver terminal using a wire.
• Connect Cell A’s negative zinc terminal to Cell B’s negative copper terminal through a \(32.0\ \Omega\) resistor.
• Make no other external connections.

Assume ideal electrode behaviour, negligible internal resistance, negligible polarisation, and no significant change in composition during the initial interval.

The two cells are allowed to operate as one connected system.

Determine:

1. The initial current through the resistor and the direction of electron flow in each external connection.
2. Which electrode is the anode and which is the cathode in each cell while they are connected.
3. The electrode reactions occurring in Cell B, and whether Cell B is operating galvanically or electrolytically.
4. The direction of migration of potassium and nitrate ions in each salt bridge.
5. Whether a positive voltmeter reading with its red lead on Cell B’s silver electrode and black lead on its copper electrode proves that silver is Cell B’s cathode.

A student argues that Cell B’s silver electrode must remain its cathode because silver was its positive electrode before the cells were connected.

Explain exactly what is wrong with that reasoning.

Solution 5

Cell A drives Cell B backwards. Cell A operates galvanically with zinc as its negative anode, while Cell B operates electrolytically with silver as its positive anode and copper as its negative cathode.

The initial current in the ideal model is \(0.020\ \text{A}\), or \(20\ \text{mA}\), where \(\text{mA}\) means milliamperes.

The essential insight is that a stronger cell can force a weaker cell to undergo the reverse of its spontaneous reaction, without requiring its terminal polarity to reverse.

Part 1: Which cell controls the current?

Cell A has the larger standard emf:

\[
E^\circ_A=1.10\ \text{V}
\]

Cell B has a smaller opposing emf:

\[
E^\circ_B=0.46\ \text{V}
\]

Because their positive terminals are connected together and their negative terminals are connected through the resistor, the emfs oppose one another.

The net driving voltage is:

\[
\begin{aligned}
E_{\text{net}}
&=1.10-0.46\\
&=+0.64\ \text{V}
\end{aligned}
\]

The positive value means Cell A can drive current against Cell B’s natural galvanic direction.

With negligible internal resistance:

\[
\begin{aligned}
I&=\frac{E_{\text{net}}}{R}\\
&=\frac{0.64\ \text{V}}{32.0\ \Omega}\\
&=0.020\ \text{A}
\end{aligned}
\]

Now track the electrons.

Through the resistor, electrons move from Cell A’s zinc electrode to Cell B’s copper electrode.

Through the direct wire joining the positive terminals, electrons move from Cell B’s silver electrode to Cell A’s copper electrode.

Both paths are necessary for the connected system to operate.

Part 2: Identify the actual anodes and cathodes

In Cell A, the usual zinc-copper reaction proceeds spontaneously.

At zinc:

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

At copper:

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

So Cell A remains a galvanic cell.

Its zinc electrode is the negative anode, and its copper electrode is the positive cathode.

Cell B is different.

When Cell B operates independently as a galvanic cell, the spontaneous reaction is:

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

But the stronger Cell A forces electrons into Cell B’s copper electrode and removes electrons from Cell B’s silver electrode.

That reverses Cell B’s net reaction.

Part 3: The reactions in Cell B

At Cell B’s copper electrode, copper(II) ions gain electrons:

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

Copper is therefore the cathode.

At Cell B’s silver electrode, silver atoms lose electrons:

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

Silver is therefore the anode.

The overall reaction in Cell B is:

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

This reaction is the reverse of Cell B’s spontaneous galvanic reaction under standard conditions.

Cell B is being driven electrolytically by Cell A.

Consequently, Cell B has a positive anode at silver and a negative cathode at copper.

Part 4: The ion movements

In Cell A, zinc ions are produced at the anode, so nitrate ions migrate towards the zinc half-cell.

Copper ions are consumed at the cathode, so potassium ions migrate towards the copper half-cell.

In Cell B, the reversed reaction changes which solution gains or loses positive ions.

Silver ions are produced at the silver anode, so nitrate ions migrate towards the silver compartment.

Copper ions are consumed at the copper cathode, so potassium ions migrate towards the copper compartment.

Part 5: Why the positive silver reading does not prove that silver is a cathode

Silver can remain at a higher electrical potential than copper while silver is undergoing oxidation.

With the ideal cell model, a voltmeter connected red-to-silver and black-to-copper across Cell B gives a positive voltage, approximately \(+0.46\ \text{V}\).

That reading tells us which terminal is at higher potential. It does not independently identify which electrode is undergoing reduction.

Cell B’s operating reaction is being controlled by the stronger external source.

The student’s reasoning assumes that Cell B must continue behaving as an independent galvanic cell merely because its terminal signs are unchanged.

But the current direction is determined by the connected system, not by Cell B in isolation.

Once Cell A drives the current backwards through Cell B, silver oxidises and copper ions reduce.

The critical constraint is the competition between the two cell emfs:

\[
E_A>E_B
\]

Their opposing connection leaves a positive net driving voltage for Cell A’s forward reaction and Cell B’s reverse reaction.

In a real setup, internal resistance, electrode polarisation, and changing concentrations would affect the current and measured voltages. The exact \(20\ \text{mA}\) result belongs to the stated ideal model.

The conclusion about electrode identity follows from the actual oxidation and reduction reactions, not from the voltmeter sign alone.

That is what makes this situation different from merely swapping voltmeter leads: the chemistry in one cell has genuinely reversed.

09The decision that prevents sign mistakes

When identifying an anode, start with the reaction, not the sign.

Oxidation defines the anode in every electrochemical cell. In a galvanic cell delivering electrical energy, that electrode supplies electrons to the external circuit and is normally the negative terminal relative to the cathode.

The negative sign does not mean positive ions cannot leave the anode, and it does not mean a metal is intrinsically negative.

If an experimental reading appears to contradict the electrode reactions, check the voltmeter leads, whether the cell is under load, and whether an external source is driving the reaction backwards.

The next useful skill is to distinguish a standard cell potential calculated from a reduction-potential table from the actual voltage a connected circuit can deliver. That distinction becomes essential when predicting cell behaviour from measurements rather than from an ideal diagram.