Verified Solution • 2025–26 Session moderate

QJEEM22Ch0c39: JEE Main 2022 Chemistry - Class 12 JEEM22Ch0c39 Solution: For a cell, the cell potential is found to be at . The…

For a cell, Cu(s)∣Cu2+(0.001 M)∥Ag+(0.01 M)∣Ag(s)\mathrm{Cu}(\mathrm{s})\left|\mathrm{Cu}^{2+}(0.001 \,\mathrm{M}) \| \mathrm{Ag}^{+}(0.01 \,\mathrm{M})\right| \mathrm{Ag}(\mathrm{s})

the cell potential is found to be 0.43 V0.43 \mathrm{~V} at 298 K298 \mathrm{~K} . The magnitude of standard electrode potential for Cu2+/Cu\mathrm{Cu}^{2+} / \mathrm{Cu} is _________ ×10−2 V\times 10^{-2} \mathrm{~V} .

[Given : EAg+/AgΘE_{A{g^ + }/Ag}^\Theta = 0.80 V and 2.303RTF{{2.303RT} \over F} = 0.06 V]

Step-by-Step Verified Solution

Anode : Cu(s)⟶Cu2+(aq)+2e−\mathrm{Cu}(\mathrm{s}) \longrightarrow \mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-}

Cathode : [Ag++e−⟶Ag(s)]2\left.[\mathrm{Ag}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(\mathrm{s})\right] 2


Cus(s) +2Ag+(aq)⟶Cu2+(aq)+2Ag(s)+2 \mathrm{Ag}^{+}(\mathrm{aq}) \longrightarrow \mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s})

amp;Ecell =Ecell 0−0.062log⁡[Cu2+][Ag+]2amp;0.43=Ecell 0−0.062log⁡(10−3(10−2)2)amp;0.43=Ecell 0−0.03log⁡10amp;Ecell 0=0.46 Vamp;Ecell 0=EAg+/Ag0−ECu2+/Cu0amp;ECu2+/Cu0=(0.80−0.46)=0.34 V=34×10−2\begin{aligned}& E_{\text {cell }}=E_{\text {cell }}^{0}-\frac{0.06}{2} \log \frac{\left[\mathrm{Cu}^{2+}\right]}{\left[\mathrm{Ag}^{+}\right]^{2}} \\& 0.43=\mathrm{E}_{\text {cell }}^{0}-\frac{0.06}{2} \log \left(\frac{10^{-3}}{\left(10^{-2}\right)^{2}}\right) \\& 0.43=E_{\text {cell }}^{0}-0.03 \log 10 \\& \mathrm{E}_{\text {cell }}^{0}=0.46 \mathrm{~V} \\& \mathrm{E}_{\text {cell }}^{0}=\mathrm{E}_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{0}-\mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{0} \\& \mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{0}=(0.80-0.46)=0.34 \mathrm{~V}=34 \times 10^{-2}\end{aligned}

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