$A l^{3 +} (a q .) + 3 e^{-} \to A l (s); E^{\circ} = - 1.66 V$
$C u^{2 +} (a q .) + 2 e^{-} \to C u (s); E^{\circ} = + 0.34 V$

What voltage is produced under standard conditions by combining the half reactions with these standard electrode potentials?

1.32 V

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2.00 V

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2.30 V

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4.34 V

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Solution

The correct option is B $2.00 V$
The cell will be:
$A l (s) | A l^{3 +} (a q .) | | C u^{2 +} (a q .) | C u (s)$

$E_{c e l l}^{\circ} = E_{C a t h o d e}^{\circ} - E_{A n o d e}^{\circ}$

$= E_{C u^{2 +} / C u}^{\circ} - E_{A l^{3 +} / A l}^{\circ}$

$= + 0.34 - (- 1.66) = + 2.00 V$ .

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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

F e^{3 +} (a q) + e^{-} \to F e^{2 +} (a q); E^{\circ} = + 0.77 V

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

B r_{2} (a q) + 2 e^{-} \to 2 B r^{-}; E^{\circ} = + 1.09 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

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

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