Standard electrode potentials only hold at 1.00 mol dm−3. Dilute the ions and the potential moves, in a direction Le Chatelier predicts and by an amount the Nernst equation calculates.
For any half-cell written as a reduction, raising the concentration of the oxidised form pushes the equilibrium to the right, so E becomes more positive; raising the reduced form makes it less positive. The Nernst equation puts a number on this at 298 K: E = E⊖ + (0.059 ÷ z) log ([oxidised] ÷ [reduced]), where z is the number of electrons in the half-equation. Solids are left out, so for a metal E = E⊖ + (0.059 ÷ z) log [Mz+]. For Cu2+ at 0.010 mol dm−3: E = 0.34 + 0.0295 × log 0.010 = 0.34 − 0.059 = +0.281 V. A tenfold dilution changes E by only 0.059 ÷ z volts, which is why E⊖ predictions are reliable unless E⊖cell is small.
Treat each half-equation as an equilibrium and apply Le Chatelier. Anything that drives it to the right (towards the reduced form) makes the electrode more positive.
| Half-cell | Change | Effect on E | Consequence |
|---|---|---|---|
| Cu2+ + 2e− ⇌ Cu | lower [Cu2+] | less positive | Cu2+ a weaker oxidising agent |
| Zn2+ + 2e− ⇌ Zn | lower [Zn2+] | more negative | Zn a stronger reducing agent; Ecell of a Zn/Cu cell rises |
| Fe3+ + e− ⇌ Fe2+ | raise [Fe3+] : [Fe2+] | more positive | the ratio, not the individual values, matters |
| Cl2 + 2e− ⇌ 2Cl− | raise [Cl−] | less positive | Cl− easier to oxidise: chlorine from concentrated brine |
| MnO4− + 8H+ + 5e− ⇌ Mn2+ + 4H2O | raise [H+] | more positive | manganate(VII) is a stronger oxidant in acid |
Four quick checks. Each correct answer earns XP and records this skill on your profile.
What happens to E for the Cu2+/Cu half-cell when [Cu2+] is decreased below 1.00 mol dm−3?
In E = E⊖ + (0.059 ÷ z) log ([oxidised] ÷ [reduced]), what is z for the Ag+/Ag half-cell?
What is E for the Cu2+/Cu half-cell when [Cu2+] = 0.010 mol dm−3? (E⊖ = +0.34 V)
Why does increasing [Cl−] make chloride ions easier to oxidise?
Use E = E⊖ + (0.059 ÷ z) log ([oxidised] ÷ [reduced]). Identify z from the half-equation, put the oxidised form on top, and leave out solids.
| Half-cell and concentrations | Substitution | E / V |
|---|---|---|
| Cu2+/Cu, [Cu2+] = 0.010 mol dm−3 | 0.34 + (0.059 ÷ 2) log 0.010 | +0.281 |
| Ag+/Ag, [Ag+] = 0.100 mol dm−3 | 0.80 + 0.059 log 0.100 | +0.741 |
| Zn2+/Zn, [Zn2+] = 0.010 mol dm−3 | −0.76 + (0.059 ÷ 2) log 0.010 | −0.819 |
| Fe3+/Fe2+, [Fe3+] = 0.10, [Fe2+] = 1.0 mol dm−3 | 0.77 + 0.059 log (0.10 ÷ 1.0) | +0.711 |
| Cl2/Cl−, [Cl−] = 0.10 mol dm−3, Cl2 at 100 kPa | 1.36 + (0.059 ÷ 2) log (1 ÷ 0.102) | +1.42 |
Use log to base 10, not ln, with the 0.059 V form. Divide 0.059 by z, the electrons in the half-equation (2 for Cu2+, 1 for Ag+ and Fe3+). Put the oxidised species on the top of the fraction, and raise concentrations to their coefficients (2Cl− gives [Cl−]2). Diluting the oxidised form makes E less positive, never more.
Unlocks once the four checks above are done. Worth more XP, written to A Level Paper 4 standard.
What is E for the Zn2+/Zn half-cell when [Zn2+] = 0.010 mol dm−3? (E⊖ = −0.76 V)
A cell is Zn | Zn2+ (1.00 mol dm−3) || Cu2+ (0.010 mol dm−3) | Cu. What is Ecell?
For Fe3+ + e− ⇌ Fe2+ (E⊖ = +0.77 V), what is E when [Fe3+] = 1.0 and [Fe2+] = 0.10 mol dm−3?
Two Ag+/Ag half-cells contain 1.00 mol dm−3 and 0.0100 mol dm−3 Ag+. What is the cell e.m.f. and which electrode is positive?
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