Calculating E°cell

7 min read Updated Mar 26, 2026

Calculating the standard cell potential is the most common quantitative task in MCAT electrochemistry. The formula is straightforward, and every calculation follows the same pattern.

The Master Equation

Important: Both E°cathode and E°anode are the standard reduction potentials from the table. You do not need to reverse any signs before plugging in. The subtraction does the work.

Step-by-Step Process

Identify the cathode and anode. The species with the higher (more positive) E° is reduced at the cathode. The species with the lower E° is oxidized at the anode.
Look up both reduction potentials from the table.
Subtract: E°cell = E°cathode - E°anode.
Interpret the sign. Positive = spontaneous (galvanic). Negative = nonspontaneous (requires external energy).

Worked Example 1: Daniell Cell

Half-reactions from the table:

Cu2+ + 2e- -> Cu E° = +0.34 V
Zn2+ + 2e- -> Zn E° = -0.76 V

Copper has the higher E°, so it is the cathode (reduction). Zinc has the lower E°, so it is the anode (oxidation).

E°cell = E°cathode - E°anode = (+0.34) - (-0.76) = +1.10 V

The positive value confirms this is a spontaneous reaction.

Worked Example 2: Iron-Silver Cell

Half-reactions:

Ag+ + e- -> Ag E° = +0.80 V
Fe2+ + 2e- -> Fe E° = -0.44 V

E°cell = (+0.80) - (-0.44) = +1.24 V

Silver is reduced, iron is oxidized. Spontaneous.

Worked Example 3: Testing a Nonspontaneous Direction

What if someone proposes: Cu + Zn2+ -> Cu2+ + Zn?

That would make zinc the cathode and copper the anode:

E°cell = E°cathode - E°anode = (-0.76) - (+0.34) = -1.10 V

Negative E°cell means this direction is nonspontaneous. You would need to supply at least 1.10 V of external energy to force this reaction (making it an electrolytic cell).

Critical Rule: Do NOT Multiply E° by Coefficients

When balancing a cell reaction, you may need to multiply one half-reaction to equalize electrons. Never multiply the E° value.

Example: Building a cell from Al and Cu:

Cu2+ + 2e- -> Cu E° = +0.34 V
Al3+ + 3e- -> Al E° = -1.66 V

To balance electrons, you multiply the copper half-reaction by 3 and the aluminum half-reaction by 2:

3Cu2+ + 6e- -> 3Cu E° is still +0.34 V (not 3 x 0.34)
2Al3+ + 6e- -> 2Al E° is still -1.66 V (not 2 x -1.66)

E°cell = (+0.34) - (-1.66) = +2.00 V

Calculate E°cell for a cell with Ni (E° = -0.26 V) and Ag (E° = +0.80 V). Which metal is the anode?

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E°cell = +1.06 V. Nickel is the anode. Silver has the higher reduction potential (+0.80 V), so it is reduced at the cathode. Nickel has the lower reduction potential (-0.26 V), so it is oxidized at the anode. E°cell = 0.80 - (-0.26) = +1.06 V.

If E°cell for a proposed reaction is -0.50 V, what does this tell you?

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The reaction is nonspontaneous as written. A negative E°cell means the reaction will not proceed in the forward direction without external energy. To make it happen, you would need to supply at least 0.50 V from an external power source (electrolytic cell). The reverse reaction, however, would be spontaneous with E°cell = +0.50 V.