$\mathrm{Cu}(\mathrm{s})+\mathrm{Sn}^{2+}(0.001 \mathrm{M}) \rightarrow \mathrm{Cu}^{2+}(0.01 \mathrm{M})+\mathrm{Sn}(\mathrm{s})$ The Gibbs free energy change for the above reaction at 298 K is $\mathrm{x} \times 10^{-1} \mathrm{~kJ} \mathrm{~mol}^{-1}$; The value of $x$ is $\_\_\_\_$. [nearest integer] $[$ Given : $\mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\ominus}=0.34 \mathrm{~V}$; $\mathrm{E}_{\mathrm{Sn}^{2+} / \mathrm{Sn}}^{\ominus}=-0.14 \mathrm{~V}$; $\mathrm{F}=96500 \mathrm{C} \mathrm{~mol}^{-1}]$

JEE Main · 2022 · 26 Jun 2022 · Shift 2 · Redox Reactions and Electrochemistry

Cu (s) + Sn^{2 +} (0.001 M) \to Cu^{2 +} (0.01 M) + Sn (s)

The Gibbs free energy change for the above reaction at 298 K is

x \times 1 0^{- 1} kJ mol^{- 1}

;

The value of

x

____

. [nearest integer]

[

Given :

E_{Cu^{2 +} / Cu}^{⊖} = 0.34 V

;

E_{Sn^{2 +} / Sn}^{⊖} = - 0.14 V

;

F = 96500 C mol^{- 1}]

Answer: 983

First, calculate

E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ} = - 0.14 - 0.34 = - 0.48

V.

Using the Nernst equation:

E_{cell} = E_{cell}^{\circ} - \frac{0.059}{2} lo g \frac{[ Cu ^{2 +} ]}{[ Sn ^{2 +} ]} = - 0.48 - \frac{0.059}{2} lo g (10) = - 0.48 - 0.0295 = - 0.5095

Δ G = - n F E_{cell} = - 2 \times 96500 \times (- 0.5095) = 98333.5

J/mol

= 98.335

kJ/mol

= 983.35 \times 1 0^{- 1}

kJ/mol.

Nearest integer:

x = 983

JEE Main · 2022 · 26 Jun 2022 · Shift 2 · Integer answer