Electrochemistry problem solving — Nernst, Faraday, conductance selection

medium CBSE JEE-MAIN NEET 3 min read

Question

An electrochemistry problem gives you cell EMF, concentration, and time of electrolysis. Which formula do you pick — Nernst equation, Faraday’s law, or conductance formula? Solve: Find the EMF of a Daniell cell when [Cu2+]=0.1 M[\text{Cu}^{2+}] = 0.1\text{ M} and [Zn2+]=1 M[\text{Zn}^{2+}] = 1\text{ M}. Ecell=1.1 VE^\circ_{\text{cell}} = 1.1\text{ V}.

(CBSE 12 + JEE Main + NEET pattern)


Solution — Step by Step

Electrochemistry has three major formula families:

  • Nernst equation: Use when concentrations are non-standard (not 1 M) and you need EMF
  • Faraday’s laws: Use when the problem involves electrolysis — mass deposited, time, current
  • Conductance/molar conductivity: Use when the problem involves solution conductivity, Λm\Lambda_m, or Kohlrausch’s law

The Nernst equation at 25 degrees C:

Ecell=Ecell0.0591nlogQE_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n}\log Q

For Daniell cell: ZnZn2+Cu2+Cu\text{Zn} | \text{Zn}^{2+} || \text{Cu}^{2+} | \text{Cu}

Here n=2n = 2 (two electrons transferred), and reaction quotient:

Q=[Zn2+][Cu2+]=10.1=10Q = \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} = \frac{1}{0.1} = 10
Ecell=1.10.05912log10=1.10.05912(1)=1.10.02955E_{\text{cell}} = 1.1 - \frac{0.0591}{2}\log 10 = 1.1 - \frac{0.0591}{2}(1) = 1.1 - 0.02955 Ecell=1.07 V\boxed{E_{\text{cell}} = 1.07\text{ V}}

The EMF decreases when the product concentration (Zn2+\text{Zn}^{2+}) is high and reactant concentration (Cu2+\text{Cu}^{2+}) is low — Le Chatelier’s principle applied to electrochemistry.


Electrochemistry Formula Selection Flowchart

flowchart TD
    A["Electrochemistry problem"] --> B{"What is given/asked?"}
    B -->|"EMF at non-standard conditions"| C["Nernst equation: E = E° - (0.0591/n) log Q"]
    B -->|"Mass deposited / time / current"| D["Faraday's law: m = (M × I × t) / (n × F)"]
    B -->|"Conductivity / molar conductivity"| E["Λm = κ × 1000 / C"]
    B -->|"Λm at infinite dilution"| F["Kohlrausch's law: Λ°m = λ°₊ + λ°₋"]
    C --> G{"Standard conditions?"}
    G -->|"Yes (1M, 1atm)"| H["E = E° directly"]
    G -->|"No"| I["Use full Nernst equation"]

Why This Works

Electrochemistry has three distinct problem domains: galvanic cells (EMF calculation), electrolytic cells (mass/charge calculation), and conductance (solution properties). Identifying which domain the problem belongs to is 80% of the battle.

The Nernst equation modifies the standard EMF based on actual concentrations. Faraday’s law connects charge flow to mass change. Kohlrausch’s law predicts molar conductivity of weak electrolytes at infinite dilution.


Common Mistake

In the Nernst equation, students frequently get QQ inverted. The reaction quotient Q=[products][reactants]Q = \frac{[\text{products}]}{[\text{reactants}]} for the cell reaction as written. For a Daniell cell, Zn is oxidised (anode) and Cu is reduced (cathode). So the cell reaction is Zn+Cu2+Zn2++Cu\text{Zn} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cu}, giving Q=[Zn2+][Cu2+]Q = \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]}. Swapping numerator and denominator flips the sign of the correction term.

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