For the reaction, CuSO$_4$ (aq) + Zn (s) → ZnSO$_4$ (aq) + Cu (s), the value of \( \Delta G^\circ \) (in kJ mol$^{-1}$) is .......... (Round off to the nearest integer) (Reduction potential: Cu$^{2+}$(aq)/Cu(s) = +0.34 V; Zn$^{2+}$(aq)/Zn(s) = -0.76 V))
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To calculate the Gibbs free energy for a redox reaction, use the relationship between the standard cell potential and the number of moles of electrons transferred.
Step 1: Using the Formula for Gibbs Free Energy.
The Gibbs free energy change for a redox reaction can be calculated using the equation:
\[
\Delta G^\circ = -nF \Delta E^\circ
\]
Where:
\( n \) is the number of moles of electrons transferred,
\( F \) is the Faraday constant (96485 C mol$^{-1}$),
\( \Delta E^\circ \) is the standard cell potential.
Step 2: Calculating the Standard Cell Potential.
The standard cell potential is calculated as:
\[
\Delta E^\circ = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}
\]
Substitute the given values for the reduction potentials of Cu$^{2+}$/Cu and Zn$^{2+}$/Zn, and calculate \( \Delta G^\circ \).
Step 3: Conclusion.
The value of \( \Delta G^\circ \) for the reaction is -99 kJ mol$^{-1}$.
