JEE Main 2022 · 29 Jul · Shift 2 · Chemistry Q84

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ChemistryInteger+4 / -1Question 84 of 90

For a cell, Cu(s) | Cu $^{2+}$ (0.001 M) || Ag $^+$ (0.01 M) | Ag (s) the cell potential is found to be 0.43 V at 298 K. The magnitude of standard electrode potential for Cu $^{2+}$ /Cu is _______ $\times10^{-2}$ V.

[Given : $E^\ominus_{\mathrm{Ag^+/Ag}}$ = 0.80 V and $\dfrac{2.303RT}F$ = 0.06 V]

Your answer

Answer and worked solution

Official answer: 34

Nernst. $E=E^\circ-\dfrac{0.06}2\log\dfrac{[\mathrm{Cu^{2+}}]}{[\mathrm{Ag^+}]^2}=E^\circ-0.03\log10=E^\circ-0.03$ , so $E^\circ_{cell}=0.46$ V.

Copper $E^\circ=0.80-0.46=0.34$ V $=34\times10^{-2}$ .

Answer: 34

Topics:Nernst equation

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JEE Main 2022 · 29 Jul · Shift 2, Chemistry question 84