Verified Solution • 2025–26 Session moderate

QJEEM22Cha8ac: JEE Main 2022 Chemistry - Class 12 JEEM22Cha8ac Solution: The cell potential for is at . The reaction quotient for the…

The cell potential for Zn∣Zn2+(aq)∣∣Snx+∣Sn\mathrm{Zn}\left|\mathrm{Zn}^{2+}(\mathrm{aq})\right|\left|\mathrm{Sn}^{x+}\right| \mathrm{Sn} is 0.801 V0.801 \mathrm{~V} at 298 K298 \mathrm{~K} . The reaction quotient for the above reaction is 10−210^{-2} . The number of electrons involved in the given electrochemical cell reaction is ____________.

(\left(\right. Given :EZn2+∣Zno=−0.763 V,ESnx+∣Sno=+0.008 V: \mathrm{E}_{\mathrm{Zn}^{2+} \mid \mathrm{Zn}}^{\mathrm{o}}=-0.763 \mathrm{~V}, \mathrm{E}_{\mathrm{Sn}^{x+} \mid \mathrm{Sn}}^{\mathrm{o}}=+0.008 \mathrm{~V} and 2.303RTF=0.06 V)\left.\frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.06 \mathrm{~V}\right)

Step-by-Step Verified Solution

A:Zn→Zn2++2e−\mathrm{A}: \mathrm{Zn} \rightarrow \mathrm{Zn}^{2+}+2 e^{-}

C:Sn+x+xe−→Sn\mathrm{C}: \mathrm{Sn}^{+\mathrm{x}}+\mathrm{xe}^{-} \rightarrow \mathrm{Sn}

ECell∘=EZn∣Zn2+∘+ESn+x∣Sn∘\mathrm{E}_{\mathrm{Cell}}^{\circ}=\mathrm{E}_{\mathrm{Zn} \mid \mathrm{Zn}^{2+}}^{\circ}+\mathrm{E}_{\mathrm{Sn}^{+x} \mid \mathrm{Sn}}^{\circ}

⇒0.763+0.008=0.771 V\Rightarrow 0.763+0.008=0.771 \mathrm{~V}

From the Nernst equation,

ECell =ECell ∘−2.303 RTnFlog⁡Q\mathrm{E}_{\text {Cell }}=\mathrm{E}_{\text {Cell }}^{\circ} \frac{-2.303 \,\mathrm{RT}}{\mathrm{nF}} \log \mathrm{Q}

0.801=0.771−0.06nlog⁡10−20.801=0.771-\frac{0.06}{\mathrm{n}} \log 10^{-2}

0.03=0.06n×20.03=\frac{0.06}{n} \times 2

n=4\mathrm{n}=4

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring