An aluminum-air cell operates in an alkaline medium according to the reaction: 4 A l ( s ) + 3 O 2 ( g ) + 6 H 2 O ( l ) + 4 O H − ( a q ) ⟶ 4 [ A l ( O H ) 4 ] − ( a q ) The standard cell poten

Question

An aluminum-air cell operates in an alkaline medium according to the reaction:

$4 A l (s) + 3 O_{2} (g) + 6 H_{2} O (l) + 4 O H^{-} (a q) ⟶ 4 [A l (O H)_{4}]^{-} (a q)$

The standard cell potential is $E_{ce ll}^{\circ} = 2.73 V$ .

Given: $Δ G_{f}^{\circ} (O H^{-}, a q) = - 157 k J m o l^{- 1}, Δ G_{f}^{\circ} (H_{2} O, l) = - 237 k J m o l^{- 1},$

$F = 96500 C m o l^{- 1}$

If 2 moles of Al are completely consumed under standard conditions and the standard Gibbs energy of formation of $[A l (O H)_{4}]^{-}$ is $x k J m o l^{- 1}$ , then the value of $(\frac{W}{1000} + \frac{∣ x ∣}{260})$ , where $W$ is the total electrical energy obtained (in kJ), is (Answer to the nearest integer).

TestHub · Accepted Answer

For full reaction ( 4 mol Al ): n = 12 Δ G o = − nFE o = − 12 ( 96500 ) ( 2.73 ) = − 3160 kJ Using formation energies: − 3160 = 4 x − [ 6 ( − 237 ) + 4 ( − 157 )] − 3160 = 4 x + 2050 x ≈ − 1300 260 ∣ x ∣ ​ = 5 For 2 mol Al : 1 mol Al ⟶ 3 e − 2 mol ⟶ n = 6 W = nFE o = 6 ( 96500 ) ( 2.73 ) = 1580 kJ 1000 W ​ = 1.58 ≈ 2 Final answer: 2 + 5 = 7

Chemistry - Electrochemistry Question with Solution | TestHub

Doubts & Discussion