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crimeas [40]
3 years ago
8

In this reaction, what is the substance oxidized? zn(s) + 2 hcl(aq) → zncl2(aq) + h2(g) zinc chloride chlorine hydrogen zn oxyge

n
Chemistry
2 answers:
Zarrin [17]3 years ago
7 0
Zn(s) + 2 HCl(aq) → ZnCl₂(aq) + H₂(g) 
Oxidation means lose of electrons and increase of positive charge so the part which oxidized in this equation is Zn(s) because it converted to Zn²⁺ (i.e. lost two electrons) 
AURORKA [14]3 years ago
5 0

Answer: Zinc (Zn)

Explanation:

Oxidation-reduction reaction or redox reaction is defined as the reaction in which oxidation and reduction reactions occur simultaneously.

Oxidation reaction is defined as the reaction in which a substance looses its electrons. The oxidation state of the substance increases.

Reduction reaction is defined as the reaction in which a substance gains electrons. The oxidation state of the substance gets reduced.

Zn(s)+2HCl(aq)\rightarrow ZnCl_2(aq)+H_2(g)

On reactant side:

Oxidation state of zinc = 0

Oxidation state of hydrogen = +1

On product side:

Oxidation state of zinc = +2

Oxidation state of hydrogen = 0

The oxidation state of zinc increases from 0 to +2, it is getting oxidized and it undergoes oxidation reaction.

The oxidation state of hydrogen decreases from +1 to 0. Thus, it is getting reduced and it undergoes reduction reaction.

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6 0
3 years ago
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3 0
3 years ago
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Cl2 + 2OH− → Cl− + ClO− + H2O
Nataly_w [17]
Cl2(g) -------> Cl-(aq) + ClO-(aq) 

2e- + Cl2(g) -------> 2Cl-(aq) [reduction] 

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______________________________________... 
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4 0
3 years ago
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3 years ago
What mass of copper is deposited when a current of 10.0a is passed through a solution of copper(ii) nitrate for 30.6 seconds?
asambeis [7]
Data Given:

Time = t = 30.6 s

Current = I = 10 A

Faradays Constant = F = 96500

Chemical equivalent = e = 63.54/2 = 31.77 g

Amount Deposited = W = ?

Solution:
             According to Faraday's Law,

                                          W  =  I t e / F

Putting Values,
 
                            W  =  (10 A × 30.6 s × 31.77 g) ÷ 96500

                            W  =  0.100 g

Result:
           0.100 g of Cu
²⁺ is deposited.
3 0
4 years ago
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