divide the following redox reactions into oxidation and reduction half reaction
step by step
Zn +Pb+2 gives Zn+2 +Pb
Zn +Cu+2 gives Zn+2 +Cu

6

Open in App

Solution

$Z n + P b^{2 +} \to Z n^{2 +} + P b T h e o x i d a t i o n r e a c t i o n i s Z n \to Z n^{2 +} + 2 e^{-} T h e r e d u c t i o n r e a c t i o n i s P b^{2 +} + 2 e^{-} \to P b Z n + C u^{2 +} \to Z n^{2 +} + C u T h e o x i d a t i o n r e a c t i o n i s Z n \to Z n^{2 +} + 2 e^{-} T h e r e d u c t i o n r e a c t i o n i s C u^{2 +} + 2 e^{-} \to C u$

Suggest Corrections

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 (s) + P b {C l}_{2} (a q) ⟶ P b (s) + Z n {C l}_{2} (a q)

Z n (s) ⟶ {Z n}^{2 +} (a q) + 2 e^{-}

{P b}^{2 +} (a q) + 2 e^{-} ⟶ P b (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

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)

Join BYJU'S Learning Program