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  • The cell potential for <img alt="\mathrm{Zn}\left|\mathrm{Zn}^{2+}(\mathrm{aq}) \| \mathrm{Sn}^{x+}\right| \mathrm{Sn} \text{ is } 0.801 \mathrm{~V} \text{ at } 298 \mathrm{~K}" src="/latex-image/?

The cell potential for \mathrm{Zn}\left|\mathrm{Zn}^{2+}(\mathrm{aq}) \| \mathrm{Sn}^{x+}\right| \mathrm{Sn} \text{ is } 0.801 \mathrm{~V} \text{ at } 298 \mathrm{~K}. The reaction quotient for the above reaction is 10^{-2}. The number of electrons involved in the given electrochemical cell reaction is_____________.

\text { (Given: } \left.\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} \text { and } \frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.06 \mathrm{~V}\right)

Option: 1

4


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

We know,

\mathrm{E_{eu}=E^{o}-\frac{2.303 \; RT}{nF}log\; Q}.........(1)

and \mathrm{E^{o}=E^{o}_{R}-E_{L}^{o}=0.008-(0.763)}

       \mathrm{E^{o}=0.771\; V}

Given,

      \mathrm{E_{cell}=0.801\; V}

So, from ........(1)

    0.801=0.771 - \frac{2.303\text{ RT}}{\text{nxF}}\log(10^{-2})

    0.801=0.771 - \frac{0.06}{\text{n}}(-2)

    \text{n}=\frac{0.06 \times2}{0.03}=4

Answer is (4)

Posted by

Ritika Kankaria

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