Question

Q1.calculate the Gibbs free energy of the reaction Mg/Mg+2//Cu+2/Cu given Ecell =1.1V​.

Answer 1

Explanation:

GIVEN :–

$\begin{gathered} \\ \longrightarrow \bf Mg | Mg^{ + 2} || Cu^{ + 2}|Cu \\ \end{gathered} </p><p>⟶Mg∣Mg </p><p>+2</p><p> ∣∣Cu </p><p>+2</p><p> ∣Cu</p><p>$

$\begin{gathered} \\ \longrightarrow \bf E_{cell}=1.1v = 1.1\:J/C\\ \end{gathered} </p><p>⟶E </p><p>cell</p><p> </p><p> =1.1v=1.1J/C</p><p>$

TO FIND :–

• Gibbs free energy = ?

SOLUTION :–

$\begin{gathered} \\ \implies\bf Mg +Cu^{ + 2}\longrightarrow Mg^{ + 2} + Cu \\ \end{gathered} </p><p>⟹Mg+Cu </p><p>+2</p><p> ⟶Mg </p><p>+2</p><p> +Cu</p><p>$

• Total number of transferred electron (n) = 2 mol

$\begin{gathered} \\ \implies\bf F = 96500\:C/mol\\ \end{gathered} </p><p>⟹F=96500C/mol</p><p>$

• We know that Gibbs free energy –

$\begin{gathered} \\ \implies \large{ \boxed{\bf\triangle G = - nFE_{cell}}}\\ \end{gathered} </p><p>⟹ </p><p>△G=−nFE </p><p>cell</p><p> </p><p>$

• Now put the values –

$\begin{gathered} \\ \implies\bf\triangle G = - (2)(96500)(1.1)\\ \end{gathered} </p><p>⟹△G=−(2)(96500)(1.1)</p><p>$

$\begin{gathered} \\ \implies\bf\triangle G = -(96500)(1.1 \times 2)\\ \end{gathered} </p><p>⟹△G=−(96500)(1.1×2)$

$\begin{gathered} \\ \implies\bf\triangle G = -96500 \times 2.2\\ \end{gathered} </p><p>⟹△G=−96500×2.2</p><p>$

$\begin{gathered} \\ \implies \bf\triangle G = -212300 \: \: J\\ \end{gathered} </p><p>⟹△G=−212300J$

$\begin{gathered} \\ \implies \large \red{\boxed{\bf\triangle G = -212.3\: \: KJ}}\\ \end{gathered} </p><p>⟹ </p><p>△G=−212.3KJ</p><p> </p><p>$