Calculate standard cell potentials
Combine tabulated standard reduction potentials to find the voltage for a complete cell.
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
Which tabulated half-cell is the cathode reduction in a spontaneous cell?
What you will learn
- Identify cathode and anode from reduction potentials
- Calculate E°cell using cathode minus anode
- Keep tabulated half-equations as reductions until assigning the cell direction
Subtract reduction potentials
Standard tables list reductions. The more positive reduction potential is the cathode reduction in a spontaneous galvanic cell. The other half-equation runs in reverse at the anode.
Calculate E°cell = E°red(cathode) − E°red(anode). Reversing a half-equation changes its role in the cell, but the subtraction formula already accounts for that; do not reverse a value and subtract it a second time.
The voltage belongs to the balanced cell reaction
A standard cell potential is measured in volts. Multiplying a half-equation by a coefficient does not multiply its electrode potential; E° is an intensive property.
Balance electron transfer when writing the overall reaction, then check the cell potential using the two original tabulated reduction values.
Pause and explain
What formula uses tabulated reduction potentials?
Worked example
Calculate E°cell for Zn|Zn²⁺ || Cu²⁺|Cu using E°(Cu²⁺/Cu)=+0.34 V and E°(Zn²⁺/Zn)=−0.76 V.
Show the worked solution
- Cu²⁺/Cu has the more positive reduction potential and is the cathode.
- Zn²⁺/Zn is the anode reduction value in the subtraction.
- Use cathode minus anode.
Answer E°cell = +0.34 − (−0.76) = +1.10 V.
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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.
Calculate E°cell for Mg|Mg²⁺ || Ag⁺|Ag using E°(Ag⁺/Ag)=+0.80 V and E°(Mg²⁺/Mg)=−2.37 V.
Give me a hint
The more positive reduction is the cathode; subtract the anode reduction value.
Compare my reasoning
- Ag⁺/Ag is the cathode.
- Mg²⁺/Mg is the anode value.
- E°cell = 0.80 − (−2.37).
E°cell = +3.17 V.
Look for these in your work
- Electrodes identified
- Negative value subtracted correctly
- Voltage stated
Find E°cell for a cell with cathode E°=+0.15 V and anode reduction E°=+0.77 V. Is the reaction as assigned spontaneous under standard conditions?
Give me a hint
Use cathode minus anode, then inspect the sign.
Compare my reasoning
- E°cell = 0.15 − 0.77.
- The result is negative.
- The assigned forward direction is not feasible under standard conditions; reverse it for the positive direction.
E°cell = −0.62 V, so the reaction as assigned is not spontaneous under standard conditions; the reverse cell direction has +0.62 V.
Look for these in your work
- Formula applied
- Sign correct
- Direction interpreted
A student doubles a balanced half-equation and doubles its E° value. Correct the calculation method.
Give me a hint
Electrode potential is not an amount that scales with reaction coefficients.
Compare my reasoning
- Balance electrons in the overall equation by scaling half-equations.
- Leave each tabulated E° unchanged.
- Calculate E°cell from the original reduction potentials.
Multiplying a half-equation does not change its E° value. Balance the reaction coefficients separately and use E°cell = E°cathode − E°anode.
Look for these in your work
- Coefficient scaling separated
- E° invariant stated
- Correct formula retained
Common mistakes
- Subtracting in anode-minus-cathode order
- Reversing a reduction value and then subtracting it again
- Multiplying E° by a half-equation coefficient
- Failing to identify which potential is more positive
What can you explain now?
Which tabulated half-cell is the cathode reduction in a spontaneous cell?
Compare with the explanation
The one with the more positive E°
It has the greater tendency to undergo reduction.
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Make it stick
Why does a more positive reduction potential identify the cathode and how does that give E°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.