Electrochemistry numericals — Nernst equation, Faraday's law problem selection

Question

The standard electrode potential of a Daniell cell is 1.10 V. If the concentration of Zn $^{2+}$ is 0.1 M and Cu $^{2+}$ is 1.0 M, calculate the EMF of the cell at 298 K.

(JEE Main / CBSE 12 — Electrochemistry numerical)

Electrochemistry Problem Type Decision Tree

flowchart TD
    A["Electrochemistry Problem"] --> B{What is asked?}
    B -->|EMF at non-standard conditions| C["Nernst Equation"]
    B -->|Mass deposited / gas liberated| D["Faraday's Laws"]
    B -->|Conductivity / molar conductivity| E["Kohlrausch's Law"]
    C --> C1["E = E° - (RT/nF) ln Q"]
    C --> C2["At 298K: E = E° - (0.0591/n) log Q"]
    D --> D1["m = (M × I × t) / (n × F)"]
    D --> D2["F = 96500 C/mol"]
    E --> E1["Lambda_m = k / c"]

Solution — Step by Step

Daniell cell: $\text{Zn}(s) | \text{Zn}^{2+}(aq) || \text{Cu}^{2+}(aq) | \text{Cu}(s)$

Cell reaction: $\text{Zn}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu}(s)$

Here, $n = 2$ (two electrons transferred per Zn atom).

At 298 K:

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

The reaction quotient $Q = \dfrac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} = \dfrac{0.1}{1.0} = 0.1$

E_{\text{cell}} = 1.10 - \frac{0.0591}{2}\log(0.1)

E_{\text{cell}} = 1.10 - \frac{0.0591}{2} \times (-1)

E_{\text{cell}} = 1.10 + 0.02955

\boxed{E_{\text{cell}} = 1.13 \text{ V}}

The EMF increased from 1.10 V to 1.13 V. This makes sense: we decreased the product concentration (Zn $^{2+}$ = 0.1 M) and kept the reactant concentration high (Cu $^{2+}$ = 1.0 M). By Le Chatelier’s principle, the reaction is driven forward, increasing the EMF.

Why This Works

The Nernst equation accounts for the effect of concentration on cell potential. At standard conditions (all concentrations = 1 M), $E = E°$ . Deviating from standard conditions shifts the equilibrium, changing the driving force (EMF). The $\log Q$ term quantifies this shift.

The factor $0.0591/n$ at 298 K comes from $RT/F = 8.314 \times 298 / 96500 = 0.02569$ V, and converting $\ln$ to $\log_{10}$ multiplies by 2.303, giving 0.0591 V.

Alternative Method — Using the General Form

The general Nernst equation (at any temperature):

E = E° - \frac{RT}{nF}\ln Q = E° - \frac{2.303RT}{nF}\log Q

At 298 K: $\dfrac{2.303 \times 8.314 \times 298}{n \times 96500} = \dfrac{0.0591}{n}$

For temperatures other than 298 K, substitute the actual temperature. JEE sometimes gives T = 300 K or 350 K to check if you use the formula rather than the shortcut.

For Faraday’s law problems: $m = \dfrac{M \times I \times t}{n \times F}$ where $M$ = molar mass, $I$ = current in amperes, $t$ = time in seconds, $n$ = number of electrons, $F$ = 96500 C/mol. Always convert time to seconds and check whether $n$ is per ion or per mole.

Common Mistake

The most frequent error in Nernst equation problems: getting the reaction quotient $Q$ wrong. $Q$ is always products over reactants — same as the equilibrium expression. For the Daniell cell, $Q = [\text{Zn}^{2+}]/[\text{Cu}^{2+}]$ because Zn $^{2+}$ is the product and Cu $^{2+}$ is the reactant. Pure solids (Zn and Cu) are not included in $Q$ .

Another common slip: using $n = 1$ instead of $n = 2$ . Always count the electrons in the balanced half-reactions.