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JEE Main 202225 JulyMorning Shift

Exam: JEE Main
Year: 2022
Date: 25 July
Shift: Morning Shift
Subject: Chemistry
Chapter: Electrochemistry

JEE Main 2022 · 25 July · Morning Shift · Chemistry · Electrochemistry

The cell potential for Zn| Zn ^ 2 + ( aq ) ‖ Sn ^ x + |Sn is 0 ....

The cell potential for

Zn\left|{Zn}^{2 +} ( aq ) ‖ {Sn}^{x +}\right|Sn

is

0 . 801V

at

298K

. The reaction quotient for the above reaction is

{10}^{- 2}

. The number of electrons involved in the given electrochemical cell reaction is _____ .
(Given

{E}_{{Zn}^{2 +} ∣ Zn}^{0} = - 0 . 763V, {E}_{{Sn}^{x +} | Sn}^{0} = + 0 . 008V

and

\dfrac{2 . 303RT}{F} = 0 . 06V

)

Correct answer

0

Step-by-step solution

$E = {E}^{0} - \dfrac{2 . 303RT}{nF}logQ$
Here,
$E = + 0 . 801V{E}^{0} = {E}_{Cathode}^{∘} - {E}_{Anode}^{∘}\left(BothareSRPvalues\right){E}^{0} = 0 . 008 - \left(- 0 . 763\right)$
$= + 0 . 771V$
$∴ 0 . 801 = + 0 . 771 - \dfrac{0 . 06}{n}log{10}^{- 2}$
$⇒ n = 4$

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