Chemistry · Electrochemistry

JEE Main 2024 — 8 April, Shift 2 — Question 72

The emf of cell

T1∣T1+(0.01M) ∣∣Cu2+ (0.01M) ∣Cu\text{T}1\left| \underset{\left( 0.01\text{M} \right)}{\mathop{\text{T}{{1}^{+}}}}\, \right|\left| \text{C}{{\text{u}}^{2+}}\underset{\left( 0.01M \right)}{\mathop{~}}\, \right|\text{Cu}

is 0.83 V at 298 K . It could be increased by :

  1. Option A:

    increasing concentration of Tl+\mathrm{Tl}^{+}ions

  2. Option B:

    increasing concentration of both Tl+\mathrm{Tl}^{+}and Cu2+\mathrm{Cu}^{2+} ions

  3. Option C:

    decreasing concentration of both Tl+\mathrm{Tl}^{+}and Cu2+\mathrm{Cu}^{2+} ions

  4. Option D:

    increasing concentration of Cu2+\mathrm{Cu}^{2+} ions

    Correct

Answer: D

Step-by-step solution

Tl++cu→Tl+cu+2E=E0−0.05911log⁡[cu+2][Tl+]As↑[Tl+]⇒↓ses         [cu+2][Tl+]As  [cu+2]↑ses⇒E.M.F   will  be   increses\begin{aligned}& T{{l}^{+}}+cu\to Tl+c{{u}^{+2}} \\ & E={{E}^{0}}-\frac{0.0591}{1}\log \frac{\left[ c{{u}^{+2}} \right]}{\left[ T{{l}^{+}} \right]} \\ & As\uparrow \left[ T{{l}^{+}} \right]\Rightarrow \downarrow ses\,\,\,\,\,\,\,\,\,\frac{\left[ c{{u}^{+2}} \right]}{\left[ T{{l}^{+}} \right]} \\ & As\,\,\left[ c{{u}^{+2}} \right]\uparrow ses\Rightarrow E.M.F\,\,\,will\,\,be\,\,\,increses \\ \end{aligned}

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Subject
Chemistry
Chapter
Electrochemistry
Topic
Nernst Equation and Electrochemical Series