Deduce cell polarity, electron flow and feasibility
Use E°cell to identify which electrode is positive, the direction of electron flow and whether a cell reaction is feasible.
Section 24 · Electrochemistry
A Level; Cambridge International Chemistry 9701, 2028–2030 syllabus, Topic 24. Treat all tabulated electrode potentials as reduction potentials; state conditions, phases, signs and units. Written practice is self-checked
Before you begin
- Use AS redox half-equations, oxidation states, mole ratios, current and charge.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
In a spontaneous galvanic cell, which way do electrons flow externally?
What you will learn
- Identify cathode and anode polarity in a spontaneous cell
- Deduce electron flow through the external circuit
- Use the sign of E°cell to predict standard feasibility
Electrons flow from oxidation to reduction
In a spontaneous galvanic cell, oxidation at the anode releases electrons and reduction at the cathode consumes them. Electrons therefore flow through the external circuit from the negative anode to the positive cathode.
The salt bridge carries ions to balance charge; it does not carry electrons. Conventional current direction is opposite to electron flow in the external wire.
The sign of E°cell determines the standard direction
For the reaction as written, E°cell > 0 indicates feasibility under standard conditions; E°cell < 0 indicates the reverse direction is favoured. At equilibrium Ecell is zero for the reaction as it proceeds.
This is a thermodynamic prediction, not a statement about reaction speed. Use standard cell potential for standard conditions and distinguish it from a non-standard cell voltage.
Pause and explain
What does E°cell > 0 indicate for the reaction as written under standard conditions?
Worked example
For a Zn/Cu galvanic cell with E°cell = +1.10 V, state electrode polarity and electron-flow direction.
Show the worked solution
- Zn is oxidised at the anode.
- Cu²⁺ is reduced at the cathode.
- Electrons move externally from Zn to Cu.
Answer The Zn anode is negative and the Cu cathode is positive. Electrons flow through the external circuit from Zn to Cu; E°cell > 0 means the reaction as written is feasible under standard conditions.
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
A cell has Fe²⁺/Fe and Ag⁺/Ag half-cells and E°cell is positive for Fe + 2Ag⁺ → Fe²⁺ + 2Ag. State the anode, cathode and electron direction.
Give me a hint
Oxidation is on the left; reduction is on the right in the written reaction.
Compare my reasoning
- Fe forms Fe²⁺ and releases electrons.
- Ag⁺ gains electrons at the other electrode.
- Electrons travel through the wire from Fe to Ag.
Fe is the negative anode, Ag is the positive cathode, and electrons flow from Fe to Ag through the external circuit.
Look for these in your work
- Oxidation side identified
- Reduction side identified
- Polarity and electron flow correct
A proposed cell reaction has E°cell = −0.45 V as written. Which direction is feasible under standard conditions?
Give me a hint
A negative value indicates the reverse reaction has positive E°cell.
Compare my reasoning
- The written direction is not feasible under standard conditions.
- Reverse the reaction and change the sign of E°cell.
- The reverse direction has +0.45 V.
The reverse reaction is feasible under standard conditions and has E°cell = +0.45 V.
Look for these in your work
- Sign interpreted
- Direction reversed
- Potential sign reversed
A student says the positive electrode potential proves a reaction will finish quickly. Give the correction.
Give me a hint
Separate thermodynamic feasibility from kinetic rate.
Compare my reasoning
- Positive E°cell predicts a feasible direction under standard conditions.
- It does not give the activation energy or rate constant.
- The reaction may still be slow.
E°cell > 0 predicts the forward direction is thermodynamically feasible under standard conditions, but it does not determine how quickly the reaction occurs.
Look for these in your work
- Feasibility stated
- Kinetic quantities distinguished
- Rate claim qualified
Common mistakes
- Sending electrons through the salt bridge
- Calling the anode positive in a spontaneous galvanic cell
- Using conventional current direction for electron flow
- Treating positive E° as proof of a fast reaction
What can you explain now?
In a spontaneous galvanic cell, which way do electrons flow externally?
Compare with the explanation
From anode to cathode
Oxidation at the anode releases electrons; reduction at the cathode uses them.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
What do anode/cathode, polarity, electron flow and the sign of E°cell each tell you about a galvanic cell?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.