Galvanic cell electrode potentials — standard hydrogen electrode and EMF series

Question

How is the standard electrode potential measured using the Standard Hydrogen Electrode (SHE)? How do we use the electrochemical (EMF) series to predict which electrode is anode/cathode and calculate cell EMF?

(JEE Main, NEET, CBSE 12 — EMF series applications and cell potential calculations appear in every electrochemistry paper)

Solution — Step by Step

We cannot measure the absolute potential of a single electrode — we can only measure the potential DIFFERENCE between two electrodes. So we need a reference.

The SHE is assigned a potential of exactly 0.00 V at all temperatures. It consists of:

Pt electrode coated with Pt black
Dipped in 1 M $H^+$ solution
$H_2$ gas bubbled at 1 atm (1 bar)
Temperature: 298 K

Any other electrode is connected to SHE, and the measured cell potential IS the standard electrode potential ( $E^0$ ) of that electrode.

When we measure $E^0$ for many electrodes against SHE, we get the electrochemical series:

Electrode	$E^0$ (V)	Tendency
$Li^+/Li$	-3.04	Strongest reducing agent
$K^+/K$	-2.93
$Zn^{2+}/Zn$	-0.76
$Fe^{2+}/Fe$	-0.44
$H^+/H_2$	0.00	Reference
$Cu^{2+}/Cu$	+0.34
$Ag^+/Ag$	+0.80
$Au^{3+}/Au$	+1.50	Strongest oxidising agent

More negative $E^0$ = stronger reducing agent (loses electrons easily) More positive $E^0$ = stronger oxidising agent (gains electrons easily)

E^0_{cell} = E^0_{cathode} - E^0_{anode}

Rule: The electrode with the more negative (or less positive) $E^0$ becomes the anode (oxidation occurs). The electrode with the more positive $E^0$ becomes the cathode (reduction occurs).

Example — Daniel cell: Zn-Cu cell

$E^0_{Zn^{2+}/Zn} = -0.76$ V (more negative, so anode)
$E^0_{Cu^{2+}/Cu} = +0.34$ V (more positive, so cathode)
$E^0_{cell} = +0.34 - (-0.76) = +1.10$ V

A positive $E^0_{cell}$ means the reaction is spontaneous ( $\Delta G^0 \lt 0$ ).

Predicting spontaneity of redox reactions: A metal higher in the series (more negative $E^0$ ) can displace a metal lower in the series from its salt solution. Zn can displace Cu from $CuSO_4$ , but Cu cannot displace Zn.
Predicting reaction with acids: Metals above hydrogen in the series ( $E^0 \lt 0$ ) react with dilute acids to liberate $H_2$ . Metals below hydrogen (Cu, Ag, Au) do not react with dilute $HCl$ or $H_2SO_4$ .
Predicting corrosion: More reactive metals (higher in series) corrode faster. Iron rusts because $E^0_{Fe} \lt E^0_{O_2}$ .

flowchart TD
    A["Two half-cells given"] --> B["Look up E° values"]
    B --> C["More negative E° → Anode (oxidation)"]
    B --> D["More positive E° → Cathode (reduction)"]
    C --> E["E°cell = E°cathode - E°anode"]
    D --> E
    E --> F{"E°cell > 0?"}
    F -->|"Yes"| G["Reaction is spontaneous"]
    F -->|"No"| H["Reaction is non-spontaneous<br/>(needs external voltage)"]

Why This Works

The EMF series ranks elements by their tendency to lose or gain electrons — a fundamental thermodynamic property. The cell EMF directly relates to Gibbs free energy: $\Delta G^0 = -nFE^0_{cell}$ . A positive cell EMF means negative $\Delta G$ , which means spontaneous reaction. The entire series is internally consistent — if A can reduce B, and B can reduce C, then A can definitely reduce C.

Common Mistake

The formula trap: students sometimes write $E^0_{cell} = E^0_{anode} - E^0_{cathode}$ , which gives a negative value and confuses them. The correct formula is always cathode minus anode. Think of it as: $E^0_{cell} = E^0_{reduction} - E^0_{oxidation}$ . If your answer is negative, either your formula is flipped or you have assigned anode/cathode incorrectly.

Shortcut for “will metal A displace metal B?” — whichever metal has the more negative $E^0$ displaces the other. No need to calculate $E^0_{cell}$ explicitly. Zn ( $-0.76$ ) displaces Cu ( $+0.34$ ) because $-0.76 \lt +0.34$ . Done in 5 seconds.

Question

Solution — Step by Step

Why This Works

Common Mistake

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