Galvanic cell electrode potentials — standard hydrogen electrode and EMF series

Question

A galvanic cell is set up with zinc and copper electrodes. The standard reduction potentials are $E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.76$ V and $E^\circ_{\text{Cu}^{2+}/\text{Cu}} = +0.34$ V. Identify the anode and cathode, write the cell notation, and calculate the standard EMF. How does the EMF series help predict spontaneity?

(JEE Main 2023 pattern)

Solution — Step by Step

The electrode with the lower (more negative) standard reduction potential gets oxidised — it becomes the anode. The one with the higher reduction potential gets reduced — it becomes the cathode.

Here: Zn ( $-0.76$ V) is the anode, Cu ( $+0.34$ V) is the cathode.

Why? Because Zn has a greater tendency to lose electrons than Cu.

Cell notation follows the convention: anode on the left, cathode on the right, with a salt bridge (double vertical line) in between:

\text{Zn(s)} \mid \text{Zn}^{2+}\text{(aq)} \| \text{Cu}^{2+}\text{(aq)} \mid \text{Cu(s)}

Single vertical lines represent phase boundaries.

E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}

E^\circ_{\text{cell}} = (+0.34) - (-0.76) = +1.10 \text{ V}

A positive $E^\circ_{\text{cell}}$ confirms the reaction is spontaneous under standard conditions ( $\Delta G^\circ = -nFE^\circ < 0$ ).

When concentrations differ from 1 M:

E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{RT}{nF}\ln Q = E^\circ_{\text{cell}} - \frac{0.0591}{n}\log Q \text{ (at 25°C)}

For this cell, $n = 2$ (two electrons transferred per Zn atom oxidised).

Why This Works

The EMF series ranks metals by their tendency to lose electrons. A metal higher in the activity series (more negative $E^\circ$ ) is a stronger reducing agent. In a galvanic cell, electrons flow spontaneously from the stronger reducing agent (anode) to the weaker one (cathode).

The Standard Hydrogen Electrode (SHE) is the reference point — assigned exactly $0.00$ V. Every other electrode potential is measured relative to SHE. This gives us a universal scale for comparing reactivity.

graph TD
    A["Given two electrodes"] --> B{"Compare E° values"}
    B -->|"More negative E°"| C["Anode (oxidation)"]
    B -->|"More positive E°"| D["Cathode (reduction)"]
    C --> E["Write oxidation half-reaction"]
    D --> F["Write reduction half-reaction"]
    E --> G["Cell EMF = E°cathode - E°anode"]
    F --> G
    G --> H{"E°cell > 0?"}
    H -->|"Yes"| I["Spontaneous reaction"]
    H -->|"No"| J["Non-spontaneous (electrolytic)"]

Alternative Method — Using Oxidation Potentials

Some older textbooks use oxidation potentials (just the negative of reduction potentials). In that system:

E^\circ_{\text{cell}} = E^\circ_{\text{oxidation, anode}} + E^\circ_{\text{reduction, cathode}}

= (+0.76) + (+0.34) = +1.10 \text{ V}

Same answer. But stick with the IUPAC convention (reduction potentials) — that’s what CBSE, JEE, and NEET use.

Quick shortcut for MCQs: if they give two reduction potentials, just subtract the smaller from the larger. The answer is always positive for a working galvanic cell. The electrode with the smaller value is always the anode.

Common Mistake

The most frequent blunder: students reverse the formula and calculate $E^\circ_{\text{anode}} - E^\circ_{\text{cathode}}$ , getting $-1.10$ V. They then panic and flip the sign manually. The correct formula is always cathode minus anode for standard reduction potentials. Write “CRA” on your rough sheet — Cathode Reduction, Anode (subtract). This appeared as a trap option in JEE Main 2022 Shift 2.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using Oxidation Potentials

Common Mistake

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