Electrode potential — how to predict which metal reduces which ion

Question

Given the standard reduction potentials:

$E^\circ(\text{Cu}^{2+}/\text{Cu}) = +0.34$ V
$E^\circ(\text{Zn}^{2+}/\text{Zn}) = -0.76$ V
$E^\circ(\text{Fe}^{2+}/\text{Fe}) = -0.44$ V

Can zinc metal reduce Cu²⁺ ions? Can copper metal reduce Fe²⁺ ions? Predict and explain.

(CBSE Class 12 / JEE Main pattern)

Solution — Step by Step

A higher (more positive) $E^\circ$ means the species is a better oxidising agent — it wants to gain electrons. A lower (more negative) $E^\circ$ means the metal is a better reducing agent — it easily loses electrons.

So the metal with lower $E^\circ$ can reduce the ion of a metal with higher $E^\circ$ .

flowchart TD
    A["Given: Metal A and Ion B²⁺"] --> B{"Compare E° values"}
    B -->|"E°(A²⁺/A) < E°(B²⁺/B)"| C["A can reduce B²⁺\n(Reaction is spontaneous)"]
    B -->|"E°(A²⁺/A) > E°(B²⁺/B)"| D["A cannot reduce B²⁺\n(Reaction is non-spontaneous)"]
    C --> E["E°cell = E°cathode - E°anode > 0"]
    D --> F["E°cell would be negative"]

Rule: Metal with lower $E^\circ$ reduces the ion of metal with higher $E^\circ$ .

$E^\circ(\text{Zn}^{2+}/\text{Zn}) = -0.76$ V (lower) vs $E^\circ(\text{Cu}^{2+}/\text{Cu}) = +0.34$ V (higher).

Zn has lower $E^\circ$ , so yes, Zn can reduce Cu²⁺.

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

Positive $E^\circ_{\text{cell}}$ confirms spontaneous reaction.

$E^\circ(\text{Cu}^{2+}/\text{Cu}) = +0.34$ V (higher) vs $E^\circ(\text{Fe}^{2+}/\text{Fe}) = -0.44$ V (lower).

Cu has higher $E^\circ$ , so no, Cu cannot reduce Fe²⁺.

E^\circ_{\text{cell}} = -0.44 - 0.34 = -0.78 \text{ V (negative)}

Negative cell potential = non-spontaneous. In fact, Fe can reduce Cu²⁺, not the other way around.

Why This Works

The electrochemical series is essentially a ranking of how desperately each species wants electrons. Species at the top (more negative $E^\circ$ ) are strong reducing agents — they readily give up electrons. Species at the bottom (more positive $E^\circ$ ) are strong oxidising agents — they grab electrons.

A spontaneous redox reaction occurs when the reducing agent is higher in the activity series (lower $E^\circ$ ) than the oxidising agent. The $\Delta G^\circ = -nFE^\circ_{\text{cell}}$ relationship ties it all together: positive cell potential means negative free energy, which means spontaneous.

Alternative Method — Activity Series Shortcut

For quick prediction without calculating $E^\circ_{\text{cell}}$ , just remember the activity series order: K, Na, Ca, Mg, Al, Zn, Fe, Ni, Sn, Pb, H, Cu, Ag, Au. Any metal higher in this list can displace ions of metals lower in the list from their salt solutions.

JEE and NEET love asking: “Which of the following metals can displace hydrogen from dilute acid?” The answer is any metal above hydrogen in the activity series (i.e., $E^\circ < 0$ V). Cu, Ag, Au cannot — they sit below hydrogen.

Common Mistake

The classic blunder: confusing which is cathode and which is anode when calculating $E^\circ_{\text{cell}}$ . Remember — the species getting reduced (higher $E^\circ$ ) is at the cathode. The species getting oxidised (lower $E^\circ$ ) is at the anode. The formula is $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$ , and the result must be positive for a spontaneous reaction.

Never subtract randomly — always identify cathode and anode first.

Question

Solution — Step by Step

Why This Works

Alternative Method — Activity Series Shortcut

Common Mistake

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