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mihalych1998 [28]
2 years ago
14

What is the half-reaction that occurs at the cathode during the electrolysis of molten potassium bromide?

Physics
1 answer:
polet [3.4K]2 years ago
7 0

The complete ionization of KBr into its constituents is:<span>
<span>KBr (s)  --->  K+ (aq)  +  Br- (aq)</span></span>

<span>
During electrolysis, oxidation takes place at the anode electrode. This means that an ion is stripped off its electron hence becoming more positive:
<span>2 Br- (aq)  --->  Br2 (g) + 2e- </span></span>

We can see that Bromine gas Br2 is evolved at the anode. 

<span>
<span>Meanwhile at the cathode, the reduction reaction occurs. Which means that the electron from the anode electrode is used to make an ion more negative:
<span>2K+ (aq)  +  2e-  --->  2K (s) </span></span>
Hence, through reduction, solid potassium is deposited on the plate.</span>

 

 

Half reactions:

<span>Anode: 2 Br- (aq)  --->  Br2 (g) + 2e- </span>                       

<span>Cathode: 2K+ (aq)  +  2e-  --->  2K (s) </span>

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Suppose two point charges, Q1 = -6.25 x 10-9 C and Q2 = -6.25 x 10-9 C, are separated by a distance d = 0.617 m.
n200080 [17]

Answer:

The answer to your question is the letter A) F =  9.23 x 10⁻⁷ N

Explanation:

Data

q₁ = -6.25 x 10⁻⁹ C

q₂ = -6.25 x 10⁻⁹ C

d = 0.617 m

k = 9 x 10⁹ Nm²/C²

F = ?

Formula

              F = k q₁q₂ /r²

-Substitution

              F = (9 x 10⁹)(-6.25 x 10⁻⁹)(-6.25 x 10⁻⁹) / (0.617)²

-Simplification

              F = 3.512 x 10⁻⁷ / 0.381

-Result

              F = 9.227 x 10⁻⁷ N  ≈ 9.23 x 10⁻⁷ N

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I hope it will be helpful for you

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Note: the actual force is 55.89 Newtons.
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3 years ago
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