Calculate the standard cell potentials of galvanic cells in which the following reactions take place:
$(i)$ $2Cr_{(s)} + 3Cd^{2+}_{(aq)} \rightarrow 2Cr^{3+}_{(aq)} + 3Cd_{(s)}$
$(ii)$ $Fe^{2+}_{(aq)} + Ag^{+}_{(aq)} \rightarrow Fe^{3+}_{(aq)} + Ag_{(s)}$
Calculate the $\Delta_r G^\Theta$ and equilibrium constant of the reactions.
$(i)$ $2Cr_{(s)} + 3Cd^{2+}_{(aq)} \rightarrow 2Cr^{3+}_{(aq)} + 3Cd_{(s)}$
$(ii)$ $Fe^{2+}_{(aq)} + Ag^{+}_{(aq)} \rightarrow Fe^{3+}_{(aq)} + Ag_{(s)}$
Calculate the $\Delta_r G^\Theta$ and equilibrium constant of the reactions.
(N/A) $(i)$ $E_{Cr^{3+}/Cr}^\Theta = -0.74 \, V$
$E_{Cd^{2+}/Cd}^\Theta = -0.40 \, V$
The galvanic cell is: $Cr_{(s)} | Cr^{3+}_{(aq)} || Cd^{2+}_{(aq)} | Cd_{(s)}$
$E_{cell}^\Theta = E_R^\Theta - E_L^\Theta = -0.40 - (-0.74) = +0.34 \, V$
For $n = 6$,$\Delta_r G^\Theta = -nFE_{cell}^\Theta = -6 \times 96487 \times 0.34 = -196833.48 \, J \, mol^{-1} = -196.83 \, kJ \, mol^{-1}$
$\log K = -\frac{\Delta_r G^\Theta}{2.303 RT} = \frac{196833.48}{2.303 \times 8.314 \times 298} = 34.496$
$K = \text{antilog}(34.496) = 3.13 \times 10^{34}$
$(ii)$ $E_{Fe^{3+}/Fe^{2+}}^\Theta = 0.77 \, V$,$E_{Ag^{+}/Ag}^\Theta = 0.80 \, V$
The galvanic cell is: $Pt_{(s)} | Fe^{2+}_{(aq)}, Fe^{3+}_{(aq)} || Ag^{+}_{(aq)} | Ag_{(s)}$
$E_{cell}^\Theta = 0.80 - 0.77 = 0.03 \, V$
For $n = 1$,$\Delta_r G^\Theta = -1 \times 96487 \times 0.03 = -2894.61 \, J \, mol^{-1} = -2.89 \, kJ \, mol^{-1}$
$\log K = \frac{2894.61}{2.303 \times 8.314 \times 298} = 0.5073$
$K = \text{antilog}(0.5073) = 3.22$
$E_{Cd^{2+}/Cd}^\Theta = -0.40 \, V$
The galvanic cell is: $Cr_{(s)} | Cr^{3+}_{(aq)} || Cd^{2+}_{(aq)} | Cd_{(s)}$
$E_{cell}^\Theta = E_R^\Theta - E_L^\Theta = -0.40 - (-0.74) = +0.34 \, V$
For $n = 6$,$\Delta_r G^\Theta = -nFE_{cell}^\Theta = -6 \times 96487 \times 0.34 = -196833.48 \, J \, mol^{-1} = -196.83 \, kJ \, mol^{-1}$
$\log K = -\frac{\Delta_r G^\Theta}{2.303 RT} = \frac{196833.48}{2.303 \times 8.314 \times 298} = 34.496$
$K = \text{antilog}(34.496) = 3.13 \times 10^{34}$
$(ii)$ $E_{Fe^{3+}/Fe^{2+}}^\Theta = 0.77 \, V$,$E_{Ag^{+}/Ag}^\Theta = 0.80 \, V$
The galvanic cell is: $Pt_{(s)} | Fe^{2+}_{(aq)}, Fe^{3+}_{(aq)} || Ag^{+}_{(aq)} | Ag_{(s)}$
$E_{cell}^\Theta = 0.80 - 0.77 = 0.03 \, V$
For $n = 1$,$\Delta_r G^\Theta = -1 \times 96487 \times 0.03 = -2894.61 \, J \, mol^{-1} = -2.89 \, kJ \, mol^{-1}$
$\log K = \frac{2894.61}{2.303 \times 8.314 \times 298} = 0.5073$
$K = \text{antilog}(0.5073) = 3.22$
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