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Use oxidation states to identify the oxidizing agent and the reducing agent in the redox

Introductory Chemistry | 5th Edition | ISBN: 9780321910295 | Authors: Nivaldo J Tro ISBN: 9780321910295 34

Solution for problem 59P Chapter 16

Introductory Chemistry | 5th Edition

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Introductory Chemistry | 5th Edition | ISBN: 9780321910295 | Authors: Nivaldo J Tro

Introductory Chemistry | 5th Edition

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Problem 59P

Use oxidation states to identify the oxidizing agent and the reducing agent in the redox reaction.

2 Na(s) + 2 H2O(l) → 2 NaOH(aq) + H2(g)

Step-by-Step Solution:
Step 1 of 3

Answer :

2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g).

 Na is oxidized to Na+ and H in H2O is reduced to H2.

 The definitions of "oxidation" and "reduction" and you can assign oxidation numbers to the elements, and see how they change.

Any element, in the elemental state, like Na, has an oxidation number of zero.

 In NaOH, Na has an oxidation number of +1. Therefore, it is oxidized.

In H2O, hydrogen has an oxidation number of +1. In H2 it is zero. Therefore, H is the element reduced

 A Sodium ion (Na) has a oxidation number of +1. It gains an electron by reacting with OH, therefore it is the one being reduced.

Step 2 of 3

Chapter 16, Problem 59P is Solved
Step 3 of 3

Textbook: Introductory Chemistry
Edition: 5
Author: Nivaldo J Tro
ISBN: 9780321910295

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Use oxidation states to identify the oxidizing agent and the reducing agent in the redox