The Gibbs energy change (in J ) for the given reaction at [ Cu ^ 2 + ] = [ Sn ^ 2 +] = 1M and 298K is : Cu(s) + Sn ^ 2 + (aq .) → Cu ^ 2 + (aq .) + Sn(s) ; ( E _ Sn ^ 2 + |Sn\ ^ 0 = - 0 .16 V, E _ Cu ^ 2 + |Cu ^ 0 = 0

Correct answer: 96500 array ll E _ cell ^ 0 = E _ Sn ^ 2 + / Sn ^ 0 - E _ Cu ^ 2 + / Cu ^ 0 array array ll = - 0 . 16 - 0 . 34 array array ll = - 0 . 50V array array ll Δ G ^ 0 = - nFE _ cell ^ 0 array array ll = - 2 × 96500 × ( - 0 . 5 ) array = 96500J = 96 . 5KJ = 96500J

JEE Main 20202 SeptemberMorning Shift

Exam: JEE Main
Year: 2020
Date: 2 September
Shift: Morning Shift
Subject: Chemistry
Chapter: Thermodynamics (C)

JEE Main 2020 · 2 September · Morning Shift · Chemistry · Thermodynamics (C)

The Gibbs energy change (in J ) for the given reaction at [ Cu ^ 2...

The Gibbs energy change (in $J$ ) for the given reaction at \left[{Cu}^{2 +}\right] = \left[{Sn}^{2} +\right] = 1\text{M} and 298K $is :</p><p>$ Cu\left(s\right) + {Sn}^{2 +}\left(aq .\right) \to {Cu}^{2 +}\left(aq .\right) + Sn\left(s\right) $; </p><p>\left({E}_{{Sn}^{2 +}\left|Sn\right\}^{0} = - 0 .16 V, {E}_{{Cu}^{2 +}|Cu}^{0} = 0 .34 V,\right)Take \leftF = 96500C{mol}^{- 1})$

Correct answer

96500

Step-by-step solution

$\begin{array}{ll}{E}_{\text{cell}}^{0} = {E}_{{Sn}^{2 +} / Sn}^{0} - {E}_{{Cu}^{2 +} / Cu}^{0}\end{array}$

$\begin{array}{ll}= - 0 . 16 - 0 . 34\end{array}$

$\begin{array}{ll}= - 0 . 50V\end{array}$

$\begin{array}{ll}\Delta{G}^{0} = - {nFE}_{cell}^{0}\end{array}$

$\begin{array}{ll}= - 2 \times 96500 \times ( - 0 . 5 )\end{array}$

$= 96500J$

$= 96 . 5KJ = 96500J$

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