Question: 59: Two half cell reactions are given below :
\mathrm{Co}^{3+}+e^{-} \rightarrow \mathrm{Co}^{2+}, \mathrm{E}_{\mathrm{Co}^{2+} / \mathrm{Co}^{3+}}^{\circ}=-1.81 \mathrm{~V} 2 \mathrm{Al}^{3+}+6 e^{-} \rightarrow 2 \mathrm{Al}(\mathrm{s}), \mathrm{E}_{\mathrm{Al} / \mathrm{Al}^{3+}}^{\circ}=+1.66 \mathrm{~V}The standard EMF of a cell with feasible redox reaction will be :
(1) +7.09 V
(2) +0.15 V
(3) +3.47 V
(4) -3.47 V
Answer: Option (3)
Explanation:
To calculate the standard EMF of the cell, we first identify which species will act as the anode (oxidation) and which as the cathode (reduction).
The more positive standard reduction potential will be reduced (cathode) and the more negative will be oxidized (anode).
Given half-reactions:
Cobalt: \mathrm{Co}^{3+} + e^{-} \rightarrow \mathrm{Co}^{2+}, \mathrm{E}^{\circ}=-1.81~\mathrm{V}
Aluminium: 2\mathrm{Al}^{3+} + 6e^{-} \rightarrow 2\mathrm{Al}, \mathrm{E}^{\circ}=+1.66~\mathrm{V}
For a feasible redox reaction, the species with lower (more negative) reduction potential gets oxidized.
So \mathrm{Co}^{2+} is oxidized to \mathrm{Co}^{3+}
and \mathrm{Al}^{3+} is reduced to \mathrm{Al}.
EMF of the cell is calculated using :
E_{\text{cell}}^{\circ} = E_{\text{cathode}}^{\circ} - E_{\text{anode}}^{\circ}Here:
E_{\text{cathode}}^{\circ} = +1.66~\mathrm{V} E_{\text{anode}}^{\circ} = -1.81~\mathrm{V}Thus,
E_{\text{cell}}^{\circ} = 1.66 - (-1.81) = 1.66 + 1.81 = 3.47~\mathrm{V}Therefore, the standard EMF of the feasible cell is +3.47~\mathrm{V}.