How to predict if a reaction is spontaneous using E° values.

Standard Electrode Potential — Electrochemical Series

Question

The standard electrode potential of Zn²⁺/Zn is −0.76 V and that of Cu²⁺/Cu is +0.34 V.

(a) Write the cell representation for the galvanic cell formed using these two electrodes.
(b) Calculate the standard EMF of the cell.
(c) State whether the cell reaction is spontaneous.

(CBSE 2024 Board Exam)

Solution — Step by Step

The electrode with lower (more negative) E° value acts as the anode (oxidation). The one with higher E° value acts as the cathode (reduction).

Zn²⁺/Zn has E° = −0.76 V → anode
Cu²⁺/Cu has E° = +0.34 V → cathode

This is why zinc dissolves and copper deposits — zinc is a stronger reducing agent.

By IUPAC convention: anode on the left, cathode on the right, double line (||) for salt bridge.

\text{Zn} \mid \text{Zn}^{2+}(1\text{ M}) \parallel \text{Cu}^{2+}(1\text{ M}) \mid \text{Cu}

Single vertical line (|) = phase boundary. Double line (||) = salt bridge.

E°_{\text{cell}} = E°_{\text{cathode}} - E°_{\text{anode}}

E°_{\text{cell}} = (+0.34) - (-0.76) = +1.10 \text{ V}

Note the double negative: subtracting a negative number adds it. Many students lose a mark right here.

A cell reaction is spontaneous if $E°_{\text{cell}} > 0$ .

Since $E°_{\text{cell}} = +1.10\text{ V} > 0$ , the reaction is spontaneous.

You can also verify using $\Delta G° = -nFE°$ . Positive E° → negative ΔG° → spontaneous. Both ways give the same answer.

Why This Works

Every metal has a tendency to either lose electrons (get oxidized) or gain electrons (get reduced). The standard electrode potential quantifies this tendency — measured against the Standard Hydrogen Electrode (SHE), which is assigned E° = 0 V.

When we connect two half-cells, the one that is more eager to get reduced (higher E°) forces the other to get oxidized. The cell EMF is the difference between these two tendencies. A higher difference means more “push” for the reaction, so more voltage.

The electrochemical series is essentially all metals ranked by E° values. Any metal higher in the series (more negative E°) will displace metals below it from their salt solutions. This is exactly why zinc displaces copper from CuSO₄ — this reaction is the basis of the Daniel cell, one of the first practical batteries.

Alternative Method

Instead of memorising the formula, think in terms of half-reactions:

Oxidation at anode:

\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^- \quad E°_{\text{ox}} = +0.76\text{ V}

Reduction at cathode:

\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \quad E°_{\text{red}} = +0.34\text{ V}

Add them:

E°_{\text{cell}} = E°_{\text{ox (anode)}} + E°_{\text{red (cathode)}} = 0.76 + 0.34 = +1.10\text{ V}

Both methods give identical answers. The first method (cathode minus anode, using reduction potentials throughout) is safer in board exams because it avoids the sign flip on the anode value. Pick one method and stick to it consistently.

Common Mistake

Wrong formula: E°cell = E°anode − E°cathode

Students often subtract in the wrong order and get −1.10 V, then incorrectly conclude the reaction is non-spontaneous. The correct formula is always:

E°_{\text{cell}} = E°_{\text{cathode}} - E°_{\text{anode}}

Remember: C minus A (Cathode minus Anode). If your answer comes out negative, check whether you’ve swapped the two.

A negative E°cell means the reaction runs in the reverse direction spontaneously — so if your answer is negative, you’ve likely assigned the anode and cathode incorrectly.

Final Answers:

(a) $\text{Zn} \mid \text{Zn}^{2+}(1\text{ M}) \parallel \text{Cu}^{2+}(1\text{ M}) \mid \text{Cu}$

(b) $E°_{\text{cell}} = \mathbf{+1.10 \text{ V}}$

(c) The reaction is spontaneous (E°cell > 0, ΔG° < 0)

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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