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Spontaneity

Al3+ aq + 3e-...

Question

$A l^{3 +} (a q) + 3 e^{-} \to A l (s); E^{\circ} = - 1.66 V$
$C u^{2 +} (a q) + 2 e^{-} \to C u (s); E^{\circ} = + 0.34 V$
What voltage is produced under standard conditions to give a spontaneous reactions by combination of these two-half cells?

1.32 V

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- 1.32 v

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2.00 V

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- 2.00 V

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Solution

The correct option is A $1.32 V$
${A l}^{3 +} + 3 e^{-} \to A l E^{0} = - 1.66$
${C u}^{2 +} + {2 e}^{-} \to C u (s); E^{0} = + 0.34 V$
$\Rightarrow$ When $1^{s t}$ reaction is reversed $\Rightarrow$ $A l ⟶ {A l}^{3 +} + 3 e^{-} \Rightarrow E^{0} = 1.66 V$
$\Rightarrow$ $A l$ gets reduced and ${C u}^{2 +}$ gets oxidized.
So, $E_{1} - E_{2} = 1.66 V - 0.34 V = 1.32 V$

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Similar questions

A l^{3 +} (a q .) + 3 e^{-} \to A l (s); E^{\circ} = - 1.66 V

C u^{2 +} (a q .) + 2 e^{-} \to C u (s); E^{\circ} = + 0.34 V

The oxidation and reduction half reactions of the reaction

2Al_(s) + 3Cu²⁺_(aq) → 2Al³⁺_(aq) + 3Cu_(s) are

A u B r_{4}^{-} (a q) + 3 e^{-} \to A u (s) + 4 B r^{-} (a q); E^{\circ} = - 0.86 V

E u^{3 +} (a q) + e^{-} \to E u^{2 +} (a q); E^{\circ} = - 0.43 V

S n^{2 +} (a q) + 2 e^{-} \to S n (s); E^{\circ} = - 0.14 V

I O^{-} (a q) + H_{2} O (l) + 2 e^{-} \to I^{-} (a q) + 2 O H^{-} (a q); E^{\circ} = + 0.49 V

F e

A g^{+} (a q) + e^{-} - > A g (s) E^{\circ} = + 0.80 V

C u^{2 +} (a q) + 2 e^{-} - - > C u (s) E^{\circ} = + 0.34 V

P b^{2 +} (a q) + 2 e^{-} - - > P b (s) E^{\circ} = - 0.13 V

F e^{2 +} (a q) + 2 e^{-} - - > F e (s) E^{\circ} = - 0.44 V

A l^{3 +} (a q) + 3 e^{-} - - > A l (s) E^{\circ} = - 1.66 V

F e^{2 +} (a q) + 2 e^{-} \to F e (s)

E^{⊖} = - 0.44 V

A l^{3 +} + 3 e^{-} \to A l (s)

E^{⊖} = - 1.66 V

B r^{2 +} 2 e^{-} \to 2 B r^{⊖} (a q)

E^{⊖} = - 1.08 V

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