The standard Gibbs energy for the reaction is -212.27 kJ mol−1 .Zn(s)…

The correct answer is Option (1) → 1.1 V

The electrical work done by a galvanic cell equals the decrease in Gibbs energy of its reaction. For the standard state, the Gibbs energy and the standard cell potential are related by:

    \( \displaystyle \Delta_rG^{\circ}=-nFE^{\circ}_{cell} \)

Here n is the number of electrons transferred in the reaction and F = 96487 C mol⁻¹.

In the Daniell cell, zinc is oxidised at the anode and \( \ce{Cu^2+} \) is reduced at the cathode:

    Anode: \( \ce{Zn(s) -> Zn^2+(aq) + 2e^-} \)
    Cathode: \( \ce{Cu^2+(aq) + 2e^- -> Cu(s)} \)
    Cell reaction:\( \ce{Zn(s) + Cu^2+(aq) -> Zn^2+(aq) + Cu(s)} \)

So n = 2. Convert the Gibbs energy to joules, because F is in coulombs and 1 J = 1 C × 1 V:

    \( \displaystyle \Delta_rG^{\circ}=-212.27\ \mathrm{kJ\,mol^{-1}}=-212270\ \mathrm{J\,mol^{-1}} \)

Rearranging the equation for E°_cell and substituting:

    \( \displaystyle E^{\circ}_{cell}=-\frac{\Delta_rG^{\circ}}{nF} \)

    \( \displaystyle E^{\circ}_{cell}=\frac{212270\ \mathrm{J\,mol^{-1}}}{2\times96487\ \mathrm{C\,mol^{-1}}} \)

    \( \displaystyle E^{\circ}_{cell}=\frac{212270}{192974}\ \mathrm{V}=1.10\ \mathrm{V} \)

So the standard electrode potential of the Daniell cell is 1.1 V. The positive value goes with the negative Δ_rG°: the cell reaction is spontaneous.