Correct a standard electrode potential for real conditions: enter E°, the number of electrons n and the reaction quotient Q to get E = E° − (0.0592/n)·log Q at 298 K.
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From the article Nernst Equation Made Simple — Cell Potential at Any Concentration.
Find the potential of a copper electrode dipped in 1.0 × 10⁻³ M Cu²⁺ at 298 K. E°(Cu²⁺/Cu) = +0.34 V.
Half reaction: Cu²⁺ + 2e⁻ → Cu, so n = 2.
Q = 1 ÷ [Cu²⁺] = 1 ÷ 0.0010 = 1000, so log Q = 3.
E = 0.34 − (0.0592 ÷ 2) × 3 = 0.34 − 0.0296 × 3 = 0.34 − 0.0888
E = +0.251 V ≈ +0.25 V
Diluting the Cu²⁺ made the electrode a weaker oxidising agent, so the reduction potential dropped. That direction should always feel reasonable before you accept a number.
Worked in full in Nernst Equation Made Simple — Cell Potential at Any Concentration.
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