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True [87]
3 years ago
8

Magnesium, the element, is produced commercially by electrolysis from a molten salt (the "electrolyte") using a cell similar to

the one. Recall that in an electrolytic cell the anode is given the + sign and the cathode is given the - sign, which is the opposite of what we see in batteries. What half-reaction occurs at the anode in this electrolytic cell?
Chemistry
1 answer:
kirill115 [55]3 years ago
7 0

Answer:

<u>The half-reaction that occurs on the anode of an electrolytic cell is called oxidation. </u>

This can easily be depicted by a chemical reaction as such <u>"Zn(s) = Zn2+ (aq) + (2e-)" </u>

While, the Zinc in this reaction loses two elections. Hence copper gains those electrons and becomes solid.

<em>These reactions occurring on the anode are called positive reactions. </em>

While the reaction that occurs on the cathode is called reduction. This reaction can be represented by this equation "<u>Cu2+ (aq) + 2e- = Cu(s)</u>".

<u>These non-spontaneous reactions occurring in an electrolytic cell together are called redox reactions. These generate electrical energy.</u>

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SO^3 = Sulfer Trioxide

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20 x 5 vials = 100mg

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Which of the following is TRUE about ideal gas particles?
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The answer is A.

Explanation:

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Why is a bowling ball harder to lift than a basket ball? Is it bc a bowling ball is made of metal?
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Learn how changes in binding free energy affect binding and the ratio of unbound and bound molecules.
Delvig [45]

C)[D]/[ED] = 5.20

D)[D]/[ED] = 5.20

E)[D']_T = 1.495* 10 ^-7 M

F)[D'] / [ED']  = 0.0579

Explanation:

E = 250 nM =2.5* 10 ^-7 mol/L , T=298.15 K

Dissociation constant of K_D = 1.3 μM (1.3 *10 ^-6 M)

E + D ⇄ ED → K_a = [ED] / [D][E]   (association constant)

ED ⇄ E + D → K_D = [E][D] / [ED]  (dissociation constant)

C)

[E] =2.5*10^-7 mol/L

K_D = 1.3* 10^-6 M

K_D = [E][D] / [ED] → [D]/[ED] = K_D / [E]

= [D]/[ED] = 1.3* 10 ^-6 / 2.5 *10^-7

= 13/25 * 10

=130/25 = 5.20

[D]/[ED] = 5.20

D)

ΔG =RTln Kd

ΔG_2 for E and D = 1.987 * 298.15 * ln 1.3*10^-6

ΔG_2 592.454 * [ln 1.3 +ln 10^-6]

ΔG_1 = 592.424 [0.2623 - 13.8155]

ΔG_2 = -592.424 * 13.553

ΔG_1 = -8184.633 cal/ mol

ΔG_1 = -8184.633  * 4.18 J/mol = -34244.508 J?mol

ΔG_1 = -34.245 KL/mol

so, ΔG_2 = ΔG_1 - 10.5 KJ/mol

ΔG_2 = -34.245 - 10.5

ΔG_2 = -44.745KJ / mol

ΔG_2 =RT ln K_D

-44.745 *10^3

=8.314 *298.15 lnK_D

lnK_D' = - 44745 / 2478.81 g

ln K_D' = -18.051

K_D' = -18.051

K_D' = e^-18.051

[D]/[ED] = 5.20

E)

[E] = 2.5* 10 ^-7 mol/ L = a

K_D' = [E][D] / [ED']                                  E +D' → ED'

K_D' = a/2(x-(a/2) / (a/2)

KD' = x - a/2

=2.447 *10^-8 = (2.5/2) * 10^-7

x=2.447 * 10^-8 + 1.25 * 10^-7

x = 2.447 *10^-8 + 1.25 * 10 ^-7

x= 10^-7 [1.25 + 0.2447]

x = 1.4947 * 10^-7

[D']_T = 1.495* 10 ^-7 M

F)

K_D' = [E][D'] / [ED']

[D'] / [ED'] = KD' / [E]

[D'] / [ED'] = 1.447 *10^-8 / 2.5* 10^-7

[D'] / [ED'] = 0.5788 * 10^-1

[D'] / [ED']  = 0.0579

5 0
3 years ago
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