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

Question

Calculate the EMF of a Daniell cell at 298 K when the concentrations are $[\text{Zn}^{2+}] = 0.1\text{ M}$ and $[\text{Cu}^{2+}] = 0.01\text{ M}$ .

Given: $E^\circ_{\text{cell}} = +1.10\text{ V}$ , $R = 8.314\text{ J mol}^{-1}\text{K}^{-1}$ , $F = 96500\text{ C mol}^{-1}$

Solution — Step by Step

The Daniell cell involves:

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

Here, $n = 2$ (number of electrons transferred).

The Nernst equation relates cell EMF to concentration:

E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{RT}{nF} \ln Q

At 298 K, substituting $\frac{RT}{F} = \frac{8.314 \times 298}{96500} = 0.02569\text{ V}$ , and converting $\ln$ to $\log_{10}$ (multiply by 2.303):

E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0592}{n} \log Q

The factor 0.0592 V (sometimes written as 0.059 V) applies specifically at 298 K.

For the overall reaction $\text{Zn} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cu}$ :

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

Pure solids (Zn and Cu) are not included in Q — their activity is 1. Substituting:

Q = \frac{0.1}{0.01} = 10

E_{\text{cell}} = 1.10 - \frac{0.0592}{2} \log(10)

E_{\text{cell}} = 1.10 - \frac{0.0592}{2} \times 1

E_{\text{cell}} = 1.10 - 0.0296 = \mathbf{1.0704 \text{ V}}

Why This Works

The Nernst equation comes from thermodynamics. The relationship between Gibbs free energy and EMF is $\Delta G = -nFE$ . The general equation $\Delta G = \Delta G^\circ + RT \ln Q$ leads directly to the Nernst equation when we substitute $\Delta G = -nFE$ and $\Delta G^\circ = -nFE^\circ$ .

Physically, here’s the intuition: we’re making a product ( $\text{Zn}^{2+}$ ) in relatively high concentration (0.1 M) and consuming a reactant ( $\text{Cu}^{2+}$ ) in low concentration (0.01 M). By Le Chatelier’s principle, this disfavors the forward reaction — so the driving force (EMF) decreases below the standard value. That’s exactly what the calculation shows: $1.0704 < 1.10\text{ V}$ .

If instead $[\text{Cu}^{2+}] \gg [\text{Zn}^{2+}]$ , Q would be small (log Q negative), making $E > E^\circ$ .

Alternative Method

Using the natural log form directly:

E_{\text{cell}} = 1.10 - \frac{8.314 \times 298}{2 \times 96500} \times \ln(10)

= 1.10 - 0.01285 \times 2.303 = 1.10 - 0.0296 = 1.0704\text{ V}

Same answer. The $0.0592/n$ shortcut is derived from this and is faster for exams — but only works at exactly 298 K.

For JEE and CBSE 12 board exams, always use the $0.0592/n$ form at 298 K unless the question gives a different temperature. If the temperature is different (say 310 K), you must use the full $\frac{RT}{nF}\ln Q$ form. Most questions are at 298 K, so memorize: $E = E^\circ - \frac{0.0592}{n} \log Q$ .

Common Mistake

The most common error is inverting Q — writing $Q = [\text{Cu}^{2+}]/[\text{Zn}^{2+}]$ instead of the products over reactants form. For the Nernst equation, Q is always products/reactants for the overall cell reaction as written. Here, $\text{Zn}^{2+}$ is the product and $\text{Cu}^{2+}$ is the reactant, so $Q = [\text{Zn}^{2+}]/[\text{Cu}^{2+}]$ . Getting Q inverted means you’d calculate $Q = 0.01/0.1 = 0.1$ and get $E = 1.10 + 0.0296 = 1.1296\text{ V}$ — a positive correction instead of negative, which is physically wrong given our concentrations.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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