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Electrochemical Series

Divide the fo...

Question

Divide the following redox reactions into oxidation and reduction half reactions.

(v) 2Cu $^{+}$ $\to$ Cu + Cu $^{2 +}$

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Solution

The redox reactions into oxidation and reduction half reactions:-

Oxidation half reaction

Cu $^{+}$ $\to$ Cu $^{2 +}$ + e $^{-}$ (Loss of one electron)

Reduction half reaction

Cu $^{+}$ + e $^{-}$ $\to$ Cu (Gain of one electron)

Additional information:-This is an example of a disproportionation reaction as the same element copper is undergoing oxidation as well as reduction.

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Similar questions

Divide the following redox reactions into oxidation and reduction half reactions.

(i) $Z n + P b^{2} \to Z n^{2} + P b$

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

(iii) $C l_{2} + 2 B r^{-} \to B r_{2} + 2 C l^{-}$

P b^{2} +

Z n^{2}

^{+}

E^{⊖} f o r C u^{2 +} / C u^{\oplus}, C u^{\oplus} / C u, C u^{2 +} / C u,

2 C u^{+}, \to C u^{2 +} + C u

E^{⊖}

Δ G^{⊖}

{C u}^{+} (a q)

2 {C u}^{+} (a q) ⇌ {C u}^{2 +} (a q) + C u (s)

E

E_{{C u}^{2 +} / C u} = 0.34 V

E_{{c u}^{2 +} / {C u}^{+}} = 0.15 V

C u \to C u^{2 +} + 2 e^{-}

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