Answered

Welcome to Westonci.ca, where you can find answers to all your questions from a community of experienced professionals. Our Q&A platform provides quick and trustworthy answers to your questions from experienced professionals in different areas of expertise. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.

Which half-reaction equation represents the
reduction of a potassium ion?
(1) K+ + e-- ==>K (3) K+ ==>K + e--
(2) K + e-- ==> K+ (4) K==> K+ + e--

Sagot :

lulu1991
The reduction reaction is the gain of electrons while oxidation reaction is the loss of electrons. For potassium ion(K+), the reaction should be K+ + e- ==> K. So the answer is (1).

Answer : The correct option is, (1) [tex]K^++e^-\rightarrow K[/tex]

Explanation :

Oxidation-reduction reaction : It is a type of reaction where the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is a reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

As per question, the half-reaction equation of reduction of a potassium ion will be :

[tex]K^++e^-\rightarrow K[/tex]

In this reaction, the oxidation state of potassium changes from (+1) to (0) that means the oxidation state decreases. So, this reaction shows the half-reaction equation of reduction of a potassium ion.

Therefore, the correct option is, (1) [tex]K^++e^-\rightarrow K[/tex]

We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Thank you for visiting Westonci.ca. Stay informed by coming back for more detailed answers.