Electrochemical series

The electrochemical (activity) series is nothing more than the list of standard reduction potentials $E^\circ$ for electrode couples $M^{n+}/M$, measured at 298 K against the standard hydrogen electrode and arranged from most negative to most positive:

where: $E^\circ$ is the standard reduction potential in volts for the half-reaction $M^{n+} + ne^- \rightarrow M$, measured relative to the hydrogen electrode taken as $0.00\ \text{V}$ by convention. Two readings of the same list matter for exam problems.

Position tells you reducing/oxidising strength. A couple higher in the series (more negative $E^\circ$) has a metal that loses electrons readily — a strong reducing agent. A couple lower in the series (more positive $E^\circ$) has an ion (or, for fluorine, the element itself) that grabs electrons readily — a strong oxidising agent. Lithium's metallic form is a ferocious electron donor; fluorine's molecular form is a ferocious electron acceptor. Everything between them is a matter of degree.

Position tells you feasibility, via a subtraction. For any proposed redox reaction, identify which species is reduced (this becomes the cathode) and which is oxidised (the anode), then compute the cell potential:

where: $E^\circ_{\text{cathode}}$ is the standard reduction potential of the species being reduced and $E^\circ_{\text{anode}}$ is the standard reduction potential of the species being oxidised, both read straight off the series table. The reaction is spontaneous only when $E^\circ_{cell} > 0$. Equivalently, a metal higher in the series will displace the ion of a metal lower in the series from solution — the textbook basis of simple displacement reactions such as iron nails darkening in copper sulphate. The figure below shows the series as a number line, so the "higher up, more reducing" rule is visible at a glance rather than memorised as a sentence.

Other uses of the same ranking show up repeatedly across the electrochemistry unit: any metal above hydrogen in the series liberates $\text{H}_2$ gas from dilute acids, while metals below hydrogen do not; the series ranks metal reactivity and hence corrosion tendency; and it guides the choice of a sacrificial anode — a more reactive metal (higher up, more negative $E^\circ$) is deliberately connected to a structure so that it oxidises instead of the metal being protected.

Problem: Using $E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.76\ \text{V}$ and $E^\circ_{\text{Cu}^{2+}/\text{Cu}} = +0.34\ \text{V}$, predict whether zinc metal displaces copper from copper sulphate solution, and find $E^\circ_{cell}$.

Step 1. Compare the two tabulated potentials. Copper's is more positive ($+0.34\ \text{V}$) than zinc's ($-0.76\ \text{V}$), so copper ion has the greater tendency to be reduced; zinc has the greater tendency to be oxidised.

Step 2. Assign roles accordingly: zinc is oxidised at the anode, $\text{Cu}^{2+}$ is reduced at the cathode, giving the overall equation $\text{Zn(s)} + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu(s)}$.

Step 3. Subtract: $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.34 - (-0.76) = +1.10\ \text{V}$.

Step 4. Since $E^\circ_{cell} > 0$, the reaction is feasible. Zinc metal does displace copper from $\text{CuSO}_4$ solution, $E^\circ_{cell} = +1.10\ \text{V}$, matching the everyday observation of red-brown copper depositing as the blue solution fades.

Competitive stretch

Now check the reverse pairing that students often get wrong: will silver displace zinc from $\text{ZnSO}_4$ solution? Using $E^\circ_{\text{Ag}^+/\text{Ag}} = +0.80\ \text{V}$ and $E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.76\ \text{V}$, the proposed direction has Ag oxidised (anode) and $\text{Zn}^{2+}$ reduced (cathode), so $E^\circ_{cell} = E^\circ_{\text{Zn}^{2+}/\text{Zn}} - E^\circ_{\text{Ag}^+/\text{Ag}} = -0.76 - 0.80 = -1.56\ \text{V}$. Negative, so silver does not displace zinc; the reverse reaction (zinc displacing $\text{Ag}^+$) is the spontaneous one instead — consistent with silver sitting far below zinc in the series.

• Reading "more positive $E^\circ$" as "more reactive as a metal" — it is the opposite: a more negative $E^\circ$, higher up the series, marks the more reactive, more easily oxidised metal.
• Comparing two $E^\circ$ values without first fixing which species is being reduced and which oxidised in the specific proposed direction — always compute $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$ for that direction, not just eyeball the ranking.
• Treating the series as predicting reaction rate — it predicts only thermodynamic feasibility (the sign of $E^\circ_{cell}$, hence $\Delta G$); a spontaneous reaction can still be kinetically very slow.
• Forgetting the series describes standard-state (1 M, 298 K) conditions only; a small positive $E^\circ_{cell}$ can flip once concentrations move away from standard, a refinement handled by the Nernst equation.

Ask the tutor: want to practice computing $E^\circ_{cell}$ for a few more metal or halogen pairs, or see how this connects forward to the Nernst equation for non-standard conditions?