Electrode potential

When a metal electrode sits in a solution of its own ions, an equilibrium forms right at the surface between the metal losing electrons and the ions regaining them:

The charge separation this equilibrium sets up across the metal–solution boundary is the electrode potential. It cannot be read off with a single wire and a voltmeter, because a voltmeter itself needs a complete circuit — a second electrode. So chemists agreed on a universal yardstick.

The standard hydrogen electrode (SHE). A platinum foil coated with finely divided platinum black is dipped into $1\ M$ $\text{H}^+$ solution, with pure $\text{H}_2$ gas bubbled over it at $1$ atm and $298\ \text{K}$. The platinum black catalyses the equilibrium $2\text{H}^+(aq) + 2e^- \rightleftharpoons \text{H}_2(g)$, and by international convention this half-cell is assigned $E^\circ = 0.00\ \text{V}$ at every temperature. It is not that hydrogen has no tendency to gain or lose electrons — the zero is an arbitrary but universally shared reference point, the way sea level is an arbitrary zero for altitude.

• Why the SHE and not something else: any lab in the world can (in principle) build one from the same recipe, so every reported $E^\circ$ value becomes comparable across the globe.
• Practical limitation: the SHE is finicky — it needs perfectly pure $\text{H}_2$ at exactly $1$ atm, scrupulously clean platinised platinum (easily poisoned by trace impurities), and is not portable. Routine lab work instead uses a secondary reference such as the calomel electrode, itself calibrated once against the SHE.

Sign convention. By agreement, a tabulated standard electrode potential $E^\circ$ always describes the reduction half-reaction, at $298\ \text{K}$ with every species at unit activity ($1\ M$ solutions, $1$ atm gases, pure solids/liquids). A more positive $E^\circ$ signals a stronger pull toward gaining electrons (easier to reduce); a more negative $E^\circ$ signals a stronger push to lose electrons (easier to oxidise). Ranking every half-reaction by $E^\circ$ produces the electrochemical series, and pairing any two half-cells lets you predict a cell's voltage:

where: $E^\circ_{cell}$ is the standard cell EMF, $E^\circ_{cathode}$ is the standard reduction potential of the electrode where reduction actually occurs, and $E^\circ_{anode}$ is the standard reduction potential of the electrode where oxidation actually occurs (its own $E^\circ$ is still looked up as a reduction value; the subtraction handles the sign flip automatically).

What shifts an electrode potential. Three qualitative factors matter, each a preview of the Nernst equation you will meet next: concentration (raising the $M^{n+}$ concentration pushes the equilibrium toward $M$, making the reduction potential more positive, by Le Chatelier's principle), temperature (the equilibrium position — and hence $E^\circ$ — is temperature-dependent, which is why standard values are always quoted at $298\ \text{K}$), and the nature of the electrode itself (each metal's ionisation energy, atomic size, and lattice/hydration energies give it a characteristic $E^\circ$, which is why zinc and copper behave so differently).

Problem. A cell is built as $\text{Zn(s)} \mid \text{Zn}^{2+}(1M) \parallel \text{H}^+(1M) \mid \text{H}_2(g,\ 1\ \text{atm}) \mid \text{Pt(s)}$ and gives a measured EMF of $0.76\ \text{V}$ with zinc as the negative electrode. A second cell, $\text{Pt(s)} \mid \text{H}_2(g,\ 1\ \text{atm}) \mid \text{H}^+(1M) \parallel \text{Cu}^{2+}(1M) \mid \text{Cu(s)}$, gives $+0.34\ \text{V}$ with copper as the positive electrode. Find $E^\circ_{\text{Zn}^{2+}/\text{Zn}}$ and $E^\circ_{\text{Cu}^{2+}/\text{Cu}}$, then use both to predict the EMF of a direct $\text{Zn}$–$\text{Cu}$ cell (the Daniell cell), without building it.

Step 1. Zinc cell: zinc is negative, so it is the anode ($\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^-$) and the SHE is the cathode. $E^\circ_{cell} = E^\circ_{\text{SHE}} - E^\circ_{\text{Zn}^{2+}/\text{Zn}} = 0 - E^\circ_{\text{Zn}^{2+}/\text{Zn}}$, and with $E^\circ_{cell} = 0.76\ \text{V}$ this gives $E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.76\ \text{V}$.

Step 2. Copper cell: copper is positive, so it is the cathode ($\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$) and the SHE is the anode. $E^\circ_{cell} = E^\circ_{\text{Cu}^{2+}/\text{Cu}} - E^\circ_{\text{SHE}} = E^\circ_{\text{Cu}^{2+}/\text{Cu}} - 0$, giving directly $E^\circ_{\text{Cu}^{2+}/\text{Cu}} = +0.34\ \text{V}$.

Step 3. Combine, without a new experiment. In a direct Zn–Cu cell, the more positive potential wins the cathode role: copper ($+0.34\ \text{V}$) is reduced, zinc ($-0.76\ \text{V}$) is oxidised. $E^\circ_{cell} = E^\circ_{\text{Cu}^{2+}/\text{Cu}} - E^\circ_{\text{Zn}^{2+}/\text{Zn}} = 0.34 - (-0.76) = 1.10\ \text{V}$.

Answer: $E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.76\ \text{V}$, $E^\circ_{\text{Cu}^{2+}/\text{Cu}} = +0.34\ \text{V}$, and the predicted Daniell-cell EMF is $\mathbf{1.10\ \text{V}}$ — matching the classic textbook value, obtained purely by combining two separate SHE measurements.

• Reporting a tabulated $E^\circ$ as an oxidation potential without flipping the sign — by convention every listed $E^\circ$ is a reduction potential; you only flip the sign when you explicitly write the reverse (oxidation) half-reaction.
• Thinking the SHE reads zero because "nothing happens" there — it is an arbitrary, universally agreed zero, not a claim that hydrogen has no tendency to gain or lose electrons.
• Dropping the strict standard conditions ($1\ M$, $1$ atm, $298\ \text{K}$); a cell measured at other concentrations needs the Nernst equation before you can call the result "standard."
• Assuming the more-negative electrode can never be reduced under any circumstance — that ranking only holds at standard conditions; extreme concentration changes can shift which electrode acts as cathode (a Nernst-equation effect, not covered here).

Ask the tutor: if you would like to work through more SHE-based EMF calculations or see how these same ideas lead into the Nernst equation, just ask.