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3 years ago

What are the oxidation numbers of the atoms in this reaction?

Chemistry

2 answers:

SVEN [57.7K]3 years ago

6 0

Answer : The correct options are:

0 for Mg(s) and H in $H_2$

+1 for H in HCl

+2 for Mg in $MgCl_2$

Explanation :

The given chemical reaction is:

$Mg(s)+2HCl(aq)\rightarrow MgCl_2(aq)+H_2(g)$

Oxidation state of Mg(s) is, 0

Oxidation state of hydrogen in HCl is, (+1)

Oxidation state of chlorine in HCl is, (-1)

Oxidation state of Mg in $MgCl_2$ is, (+2)

Oxidation state of Cl in $MgCl_2$ is, (-1)

Oxidation state of hydrogen in $H_2$ is, 0

The given reaction is a single displacement reaction in which the most reactive metal (Mg) displaces the least reactive element (H) from its compound.

It is also a redox reaction because the oxidation state of Mg changes from 0 to (+2) and shows oxidation and the oxidation state of hydrogen changes from (+1) to 0 and shows reduction.

OLEGan [10]3 years ago

5 0

all these are correct:
0 for Mg(s) and H in H2

+1 for H in HCl

-1 for Cl in HCl

+2 for Mg in MgCl2

only the first statement is wrong

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Answer:

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Explanation:

Zn(OH)2 --> Zn+2 + 2OH- Ksp = 3X10^-15

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K = Ksp X Kf

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Zn(OH)₂ + 2OH⁻(aq) --> Zn(OH)₄²⁻(aq)

Initial: 0 0.3 0

Change: -2x +x

Equilibrium: 0.3 - 2x x

K = Zn(OH)₄²⁻/[OH⁻]²

6 = x/(0.3 - 2x)²

6 = x/(0.3 -2x)(0.3 -2x)

6(0.09 -1.2x + 4x²) = x

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24x² - 8.2x + 0.54 = 0

Upon solving as quadratic equation, we obtain;

x = 0.089

Therefore,

Concentration of (OH⁻) = 0.3 - 2x

= 0.3 -(2*0.089)

= 0.122

pOH = -log[OH⁻]

= -log 0.122

= 0.91

pH = 14-0.91

= 13.09

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