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3 years ago

What is the reduction half equation of Fe(s)+ 2 HC2H3O2(aq) → Fe(C2H3O2)2(aq) + H2(g)

Chemistry

1 answer:

Sonbull [250]3 years ago

4 0

Answer:

2 H⁺ + 2e = H₂ ( reduction )

Explanation:

Fe( s ) + 2 CH₃COOH = Fe ( OOCCH₃ ) ₂ + H₂

Fe( s ) = Fe⁺² + 2e ( oxidation )

2 H⁺ + 2e = H₂ ( reduction )

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Identify the solute with the highest van't Hoff factor. And how do you determine which one is highest?

german

Answer : The correct option is, (A) $AlCl_3$

Explanation :

Van't Hoff factor : It is defined as the ratio of the experimental value of the colligative property to the calculated value of the colligative property.

We can determine the van't Hoff factor by the association and dissociation of the compound.

(A) $AlCl_3$

It is an electrolyte that dissociates to give aluminum ion and chloride ion.

The dissociation of $AlCl_3$ will be,

$AlCl_3\rightarrow Al^{3+}+3Cl^{-}$

So, Van't Hoff factor = Number of solute particles = $Al^{3+}+3Cl^{-}$ = 1 + 3 = 4

(B) $KI$

It is an electrolyte that dissociates to give potassium ion and iodide ion.

The dissociation of $KI$ will be,

$KI\rightarrow K^{+}+I^{-}$

So, Van't Hoff factor = Number of solute particles = $K^{+}+I^{-}$ = 1 + 1 = 2

(C) $CaCl_2$

It is an electrolyte that dissociates to give calcium ion and chloride ion.

The dissociation of $CaCl_2$ will be,

$CaCl_2\rightarrow Ca^{2+}+2Cl^{-}$

So, Van't Hoff factor = Number of solute particles = $Ca^{2+}+2Cl^{-}$ = 1 + 2 = 3

(D) $MgSO_4$

It is an electrolyte that dissociates to give magnesium ion and sulfate ion.

The dissociation of $MgSO_4$ will be,

$MgSO_4\rightarrow Mg^{2+}+SO_4^{2-}$

So, Van't Hoff factor = Number of solute particles = $Mg^{2+}+SO_4^{2-}$ = 1 + 1 = 2

(E) Non-electrolyte

The dissociation non-electrolyte is not possible. So, the Van't Hoff factor will always be, 1.

Hence, the highest van't Hoff factor of solute is, $AlCl_3$

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A mineral that breaks into irregular pieces is said to show which of the following?

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The original options for this question were cleavage, luster and hardness. The answer would be cleavage.

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Answer:

Explanation:

The element which is oxidized is the element that losses electrons and its oxidation state be more positive.

The element which is reduced is the element that gain electrons and its oxidation state be more negative.

<em> O goes from 0 to -2, so, it is the element that is reduced.</em>

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