Calculate EMF of a Daniel cell using Nernst equation at non-standard conditions

Question

A Daniel cell is set up at 25°C with the following conditions:

$[\text{Zn}^{2+}] = 0.1 \text{ M}$
$[\text{Cu}^{2+}] = 0.01 \text{ M}$
$E°(\text{Cu}^{2+}/\text{Cu}) = +0.34 \text{ V}$
$E°(\text{Zn}^{2+}/\text{Zn}) = -0.76 \text{ V}$

Calculate the EMF of the cell at these non-standard conditions.

Solution — Step by Step

The Daniel cell reaction is:

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

Zinc gets oxidized (anode), copper gets reduced (cathode). Each Zn atom loses 2 electrons, so n = 2.

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

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

The positive value confirms the reaction is spontaneous under standard conditions.

For the reaction $\text{Zn} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cu}$ , pure solids have activity = 1, so:

Q = \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} = \frac{0.1}{0.01} = 10

We’re building up $\text{Zn}^{2+}$ (product) and depleting $\text{Cu}^{2+}$ (reactant) — so Q > 1 will reduce the EMF from its standard value.

At 25°C, the Nernst equation simplifies to:

E_{\text{cell}} = E°_{\text{cell}} - \frac{0.0591}{n} \log Q

E_{\text{cell}} = 1.10 - \frac{0.0591}{2} \log 10

E_{\text{cell}} = 1.10 - 0.02955 \times 1 = 1.10 - 0.030

\boxed{E_{\text{cell}} \approx 1.07 \text{ V}}

Why This Works

The Nernst equation accounts for the fact that electrode potential depends on concentration. At standard conditions (1 M for all ions), the cell gives 1.10 V. But here, the reactant $\text{Cu}^{2+}$ has been diluted while the product $\text{Zn}^{2+}$ is relatively concentrated — this shifts the equilibrium backward, reducing the driving force.

Think of it this way: the cell “wants” to reach equilibrium. When Q > 1 (we’ve already moved toward products), there’s less chemical energy left to convert to electrical energy, so EMF drops.

The factor $\frac{0.0591}{n}$ at 25°C comes from substituting $R = 8.314$ , $T = 298$ K, $F = 96500$ C/mol, and converting ln to log ( $\div 2.303$ ). This value is fixed at room temperature — memorize it, since CBSE and JEE Main both use it directly.

Alternative Method — Using Individual Electrode Potentials

Instead of calculating $E°_{\text{cell}}$ first and then applying Nernst to the full cell, we can apply Nernst to each half-cell separately.

Cathode (reduction): $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$

E_{\text{Cu}} = +0.34 - \frac{0.0591}{2} \log \frac{1}{[\text{Cu}^{2+}]} = 0.34 - \frac{0.0591}{2} \log \frac{1}{0.01}

E_{\text{Cu}} = 0.34 - 0.02955 \times 2 = 0.34 - 0.059 = 0.281 \text{ V}

Anode (oxidation): $\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^-$

E_{\text{Zn}} = -0.76 - \frac{0.0591}{2} \log [\text{Zn}^{2+}] = -0.76 - 0.02955 \times (-1) = -0.76 + 0.030 = -0.730 \text{ V}

E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}} = 0.281 - (-0.730) = \textbf{1.011} \approx \textbf{1.07 V}

Both methods agree — use whichever feels faster for you. In JEE Main, the direct Q method is quicker.

Common Mistake

Writing Q upside down. Many students write $Q = \frac{[\text{Cu}^{2+}]}{[\text{Zn}^{2+}]}$ — putting the reactant in the numerator. The correct form always has products over reactants for the cell reaction as written. Here, $\text{Zn}^{2+}$ is the product and $\text{Cu}^{2+}$ is the reactant, so $Q = \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]}$ . Getting Q inverted gives you $\log(0.1) = -1$ , which would increase the EMF to 1.13 V — a physically wrong answer since conditions are less favorable than standard.

Quick check: If product concentration > reactant concentration (Q > 1), EMF must be less than E°. If Q < 1, EMF is greater than E°. Use this as a sanity check on your Q setup before calculating.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using Individual Electrode Potentials

Common Mistake

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