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3 years ago

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

Physics

1 answer:

polet [3.4K]3 years ago

7 0

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

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

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

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:
2K+ (aq) + 2e- ---> 2K (s) 
Hence, through reduction, solid potassium is deposited on the plate.

Half reactions:

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

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

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9. A 50 kg physics i student trudges from the 1st floor to the 3rd floor, going up a total height of 13 m.

SVETLANKA909090 [29]

Answer:

6370 J

Explanation:

By the law of energy conservation, the work done by the student would be the change in potential enegy from 1st floor to 3rd floor, or a change of 13 m

$W = E_p = mgh$

where m = 50kg is the mass of the student, g = 9.8 m/s2 is the gravitational constant and h = 13 m is the height difference

$W = 50*9.8*13 = 6370 J$

3 0

3 years ago

Chapter 21, Problem 009 Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of 0.12

PilotLPTM [1.2K]

Answer:

a) -1.325 μC

b) 4.17μC

Explanation:

First, you need to know that charge is conserved. So, the adition of the charges, as there is no lost in charge, should always be the same. Also, after the wire is removed, both spheres will have the same charge, trying to find equilibrium. In summary:

$q_1 + q_2 = constant\\q_1_f = q_2_f |Then\\q_1_f + q_2_f = 2q_1_f = q_1_o+q_2_o\\q_1_f = q_2_f = \frac{q_1_o+q_2_o}{2}$

We know both q1f and q2f must be positive, because the negative charge at the beginning was the the smaller.

The electrostatic force is equal to:

$F_e = k\frac{q_1q_2}{r^2}$

K is the Coulomb constant, equal to 9*10^9 Nm^2/C^2

Now, we are told that the electrostatic force after the wire is equal to 0.0443 N:

$F_e_f = k \frac{q_1_fq_2_f}{r^2} = k\frac{\frac{q_1_o+q_2_o}{2}\frac{q_1_o+q_2_o}{2}}{r^2} = k\frac{(q_1_o+q_2_o)^2}{4r^2} \\(q_1_o+q_2_o) = \sqrt{\frac{F_e_f*4r^2}{k}} = \sqrt{\frac{0.0443N *4(0.641m)^2}{9*10^9Nm^2/C^2} } = 2.844 *10^{-6}C \\ q_1_o = 2.844*10^{-6}C - q_2_o$

Originally, the force is negative because it was an attraction force, therefore, its direction was opposite to the direction of the repulsive force after the wire:

$F_e_o = k\frac{q_1_oq_2_o}{r^2}\\ q_1_oq_2_o = \frac{F_e_o*r^2}{k} = \frac{-0.121N(0.641m)^2}{9*10^9Nm^2/C^2} = -5.524*10^{-12}$

$(2.844*10^{-6}C - q_2_o)q_2_o = -5.524*10^{-12}\\0 = q_2_o^2 - 2.844*10^{-6}q_2_o - 5.524*10^{-12}$

Solving the quadratic equation:

$q_2_o = 4.17*10^{-6}C | -1.325 * 10^{-6}C$

for this values q_1 wil be:

$q_1_o = -1.325 *10^{-6}C | 4.17*10^{-6}C$

So as you can see, the negative charge will always be -1.325 μC and the positive 4.17μC

5 0

3 years ago

You have been hired by a "storm chaser" as an assistant. This individual loves to find locations at which tornadoes and violent

Studentka2010 [4]

Answer:

58.44 C

Explanation:

Electric field is found by

$E=\frac {\sigma}{\epsilon_o}=\frac {Q}{A\epsilon_o}$

Therefore, the charge is

$Q=EA\epsilon_o
$

$Q= 4.00 ✕ 10^{6} N/C*1.65 *10^{6} m^{2}*8.854*10^{-12}= 58.4364 C
$

Therefore, required charge is 58.44 C

5 0

3 years ago

If μs is greater than some critical value, the woman cannot start the crate moving no matter how hard she pushes. calculate this

fiasKO [112]

weight = mg acts downwards 
normal force = N acts upwards.
and force F acts at an angle θ below the horizontal.
(Let us assume that the woman pushes from the left, so F is acted towards the right, which is below the horizontal)
so that, Frictional force, f=us*N acts towards the left

Now we balance the forces along x and y directions:
y direction: N = mg + F sinΘ
x direction: us * N = F cosΘ

We let the value of µs be equal to a value such that any F will not be able to move the crate. Then, if we increase F by an amount F', then the force pushing the crate towards the right also increases by F' cosΘ. Additionally, the frictional force f must raise by exactly this amount.
Since f can’t exceed us*N, so the normal force must increase by F' cosΘ/us.
Also, from the y direction equation, the normal force exceeds by F' sin Θ.

These two values must be the same, therefore:
us = cot θ

4 0

3 years ago

Please help!!!!!! please help!!!!!! please help!!!!!!

natulia [17]

Answer:

The answer is: wavelength

Explain:

Or crest

3 0

3 years ago