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mart [117]
3 years ago
5

Balance the following oxidation-reduction reactions, which take place in acidic solution, by using the "half-reaction" method.

Chemistry
1 answer:
LuckyWell [14K]3 years ago
8 0
First identify which is being oxidized and reduced. In this case, the Mg is being oxidized and the Hg is being reduced. 

Mg --> Mg+2 
 
<span>Hg+2 --> Hg+1 
</span>
Then you have to balance each half reaction first with electrons before adding them together in one equation

Mg ⇒ Mg^{+2}  +  2e^{-1}

and

2Hg^{+2} + 2e^{-1} ⇒ Hg_{2} ^{+2}
 
and then combine them together to form 

Mg + 2 Hg^{+2} + 2e ⇒ Mg^{+2} + 2e ^{-} + 2 Hg^{+1} + 2 e^{-}

It isn't necessary to keep the electrons but its essential to know how many there are in order to know how many are in the equation in order to calculate the reaction energy. Note: A<span>dd H+ and H2O to balance the H's and O's in acidic solution if needed.</span>
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3. What mass of copper could be deposited from a copper(II) Sulphate solution using a current of 5.0 A over 100 seconds? ( F =96
arsen [322]

Mass of copper : 0.165 g

<h3>Further explanation</h3>

Given

5.0 A over 100 seconds

Required

Mass of copper

Solution

Faraday's law:

<em>The mass of the substance formed at each electrode is proportional to the electric current flowing in the electrolysis</em>

<em />\tt W=\dfrac{e.i.t}{96500}<em />

e = Ar / valence = eqivalent weight

i = current

t = time

W = weight

CuSO₄ ----> Cu²⁺ + SO₄²⁻

Cu ----> Cu²⁺ + 2e

e = Ar/2

= 63,5/2 = 31,75

\tt W=\dfrac{31.75\times 5\times 100}{96500}=0.165~g

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2 years ago
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