For the cell..........| Electrochemistry

JEE MAIN 2019

11-01-2019 S1

Question

For the cell $\mathrm{Zn}(\mathrm{s})\left|\mathrm{Zn}^{2+}(\mathrm{ag}) \| \mathrm{M}^{x+}(\mathrm{ag})\right| \mathrm{M}(\mathrm{s})$, different half cells and their standard electrode potentials are given below

If $\mathrm{E}_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{\circ}=-0.76 \mathrm{~V}$, which cathode will give a maximum value of $\mathrm{E}_{\text {meal }}^{\circ}$ per electron transferred?

Select the correct option:

$\mathrm{Fe}^{2+} / \mathrm{Fe}$

$\mathrm{Ag}^{+} / \mathrm{Ag}$

$\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}$

$\mathrm{Au}^{3+} / \mathrm{Au}$

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

$$ E_{\text {coll }}=\left(E_{\mathrm{R} . \mathrm{P} .}^{\circ}\right)_{\text {Cathode }}-\left(E_{\mathrm{R} . \mathrm{P} .}^{\circ}\right)_{\text {Anode }} $$ All electrodes act as cathode wr r. Zn so the ion which has highest reduction potential will give maximum value of $\mathrm{E}_{\text {तcal }}^{\circ}$ so $\mathrm{Au}^{3+} / \mathrm{Au}$ produce highest $\mathrm{E}_{\text {call }}^{\circ}$

Question Tags

JEE Main

Chemistry

Easy

Start Preparing for JEE with Competishun

Report Issue