CameraIcon

SearchIcon

MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Assigning Oxidation Numbers

Divide the fo...

Question

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

Open in App

Solution

Oxidation: A process in which the oxidation state of an atom or ion increases by losing electrons.
The Zn atom loses two electrons to form Zn2+ ion.

Oxidation half: Zn → Zn2+ + 2e-

Reduction: A process in which the oxidation state of an atom or ion decreases by gaining electrons.
The Cu2+ ion gains two electrons to form a Cu atom.

Reduction half: Cu2+ + 2e- → Cu

Suggest Corrections

thumbs-up

0

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

Q. 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

What is the oxidizing agent in the following redox reaction?

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

Q. Based on the following reactions, arrange the metals in increasing order of their reduction potentials.

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

Q. The standard reduction potential

E^{\circ}

, for the half reaction are:-

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

The emf for the cell reaction,

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

is_____________.

View More

Join BYJU'S Learning Program

explore_more_icon

Explore more

Assigning Oxidation Numbers

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon