Question

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

Answer 1

Correct Answer: 983

To find the Gibbs free energy change for the given reaction, we use the Nernst equation and the formula for Gibbs free energy. The reaction is:

\(Cu_{(s)} + Sn^{2+} \rightarrow Cu^{2+} + Sn_{(s)}\)

Step 1: Calculate the standard cell potential, \(E^{\circ}_{cell}\)

The standard electrode potentials are given by:

• \(E^{\circ}_{Cu^{2+} / Cu} = 0.34\,V\)
• \(E^{\circ}_{Sn^{2+} / Sn} = -0.14\,V\)

Since the reaction involves oxidation of Sn and reduction of Cu, the standard cell potential:

\(E^{\circ}_{cell} = E^{\circ}_{Cu^{2+} / Cu} - E^{\circ}_{Sn^{2+} / Sn}\)

\(= 0.34 - (-0.14) = 0.48\,V\)

Step 2: Calculate the actual cell potential, \(E_{cell}\)

The Nernst equation is given by:

\(E_{cell} = E^{\circ}_{cell} - \dfrac{RT}{nF} \ln Q\)

Given that \(T = 298\,K\), \(R = 8.314\,J\,mol^{-1}\,K^{-1}\), and \(F = 96500\,C\,mol^{-1}\), and for this reaction \(n = 2\).

The reaction quotient, \(Q\), is:

\(Q = \dfrac{[Cu^{2+}]}{[Sn^{2+}]}\)

\(Q = \dfrac{0.01}{0.001} = 10\)

Using the Nernst equation:

\(E_{cell} = 0.48 - \dfrac{(8.314)(298)}{(2)(96500)} \ln(10)\)

\( \approx 0.48 - \dfrac{0.0257}{2} \times 2.302 \approx 0.48 - 0.0296 \approx 0.4504\,V\)

Step 3: Calculate Gibbs free energy change, \(\Delta G\)

The Gibbs free energy change is given by:

\(\Delta G = -nFE_{cell}\)

\(= -(2)(96500)(0.4504)\)

\(-87077.2\,J\,mol^{-1}\)

Convert to kJ/mol:

\(\Delta G \approx -87.1\,kJ\,mol^{-1}\)

According to the question, this is x × 10^-1 kJ/mol. Therefore,

\(x = 871\)

The value of x is 871, which fits within the provided range of 983,983.