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Hydrogen Displacement by Alkalis

State wether ...

Question

State wether the following is oxidation reaction and which is reduction reaction. $C u^{2 +} + 2 e^{-} \to C u$

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Solution

Oxidation is the loss of electrons or an increase in oxidation state by a molecule, atom, or ion. Reduction is the gain of electrons or a decrease in oxidation state by a molecule, atom, or ion.
Here,
cupric ion gets two electrons and forms copper so its a reduction reaction.

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The oxidation and reduction half reactions of the reaction

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

S n^{4 +} + 2 e^{-} \to S n^{2 +}

Z n | Z n^{2 +} (a q) (1.0 M) | | C u^{2 +} (a q) (1.0 M) | | C u

0.350

C u^{2 +} (a q) + 2 e^{-} \to C u

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Z n^{2 +} (a q) + 2 e^{-} \to Z n

E^{\circ}

Z n \to Z n^{2 +} + 2 e^{-}; E^{\circ} = 0.76 V

C u \to C u^{2 +} + 2 e^{-}; E^{\circ} = 0.34 V

Z n (s) + C u^{2 +} \to Z n^{2 +} + C u (s)

Z n | Z n^{2 +} (a q) (1.0 M) | | C u^{2 +} (a q) (1.0 M) | | C u

C u^{2 +} (a q) + 2 e^{-} \to C u

Z n^{2 +} (a q) + 2 e^{-} \to Z n

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