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Electrochemical Series

Divide the fo...

Question

Divide the following redox reactions into oxidation and reduction half reactions.
Zn + $C u^{2} +$ → $Z n^{2} +$ + Cu

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Solution

The redox reactions into oxidation and reduction half reactions:-

Oxidation half reaction:

Zn → $Z n^{2} +$ + 2e- (Loss of two electrons)

Reduction half reaction:

$C u^{2} +$ + 2e-→ Cu (Gain of two electrons)

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

What is the oxidizing agent in the following redox reaction?

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

Z n + C u^{2 +} \to Z n^{2 +} + C u

M g + Z n^{2 +} \to M g^{2 +} + Z n

C u + 2 A g^{+} \to C u^{2 +} + 2 A g

For cell reaction, $Z n + C u^{2 +} \to Z n^{2 +} + C u,$ cell representation is

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)

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