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Aleksandr [31]
3 years ago
7

Able 1 Cell Type Operating Cell Potential for Commercial Batteries, E (V) Lithium-iodine Zinc-mercury +2.80 +1.35 Table 2 Standa

rd Reduction Potential, E' (V) -1.20 Half-Reaction [Zn(OH)212 +2e → Zn + 4 OH Zn(OH)2 +2e → Zn +20H- HgO + H2O +2e → Hg+20H O2 + 2H20 +40 →40H -1.25 +0.10 +0.40 ctronic devices that help regulate the heart rate. Currently, lithium-iodine cells are commonly used to power pacemakers and have replaced zinc-me otential, E, for each cell. Table 2 provides the standard reduction potentials for several half-reactions related to zinc-mercury and zinc-air cells. dation given, which of the following is a major difference between the zinc-mercury cell and the lithium-iodine cell? 0, +2H2O + 4e +40H +0.40 Pacemakers are electronic devices that help regulate the heart rate Currently, lithium-iodine cells are commonly used to power pacemakers and have replaced zinc-mercury cells. Table 1 provides the operating cell potential, E, for each cell. Table 2 provides the standard reduction potentials for several half-reactions related to zinc-mercury and zinc air cells. Based on the information given, which of the following is a major difference between the zinc mercury cell and the lithium-iodine cell?
A. During the initial cell operation, each reaction is thermodynamically favorable, but the larger operating potential of the lithium-iodine cell indicates that its cell reaction is less thermodynamically favorable.
B. During the initial cell operation, each reaction is thermodynamically favorable, but the larger operating potential of the lithium-iodine cell indicates that its cell reaction is more thermodynamically favorable.
C. During the initial cell operation, the oxidation of iodine is thermodynamically favorable but the oxidation of mercury is not.
D. During the initial cell operation, the oxidation of mercury is thermodynamically favorable but not the oxidation of iodine is not.
Chemistry
1 answer:
IrinaK [193]3 years ago
8 0

Answer:

During the initial cell operation, each reaction is thermodynamically favorable, but the larger operating potential of the lithium-iodine cell indicates that its cell reaction is more thermodynamically favorable.  ( B )

During the initial cell operation, the oxidation of iodine is  thermodynamically favorable but the oxidation of mercury is not. ( C )

Explanation:

<u>The major Differences between The Zinc mercury cell and Lithium-iodine cell are :</u>

During the initial cell operation, each reaction is thermodynamically favorable, but the larger operating potential of the lithium-iodine cell indicates that its cell reaction is more thermodynamically favorable.   and

During the initial cell operation, the oxidation of iodine is thermodynamically favorable but the oxidation of mercury is not.

Given the relationship below,

Δ G = -nFE

E = emf of cell ,  G = free energy.

This relationship shows that if E is positive the reaction will be thermodynamically favorable also if E is large it will increase the negativity of free energy  also From the question we can see that with the reduction of mercury the value of E is more positive and this shows that Mercury is thermodynamically unfavorable

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Select the statement that is FLASE about this experiment. Group of answer choices Changing metal solution will affect the redox
Radda [10]

Well arranged question is;

Select the statement that is FALSE about this experiment. Group of answer choices;

A) Changing metal solution will affect the redox potential.

B) To ensure consistent data, collect the redox potential values at least three times.

C) Changing temperature will affect the redox potential.

D) To ensure stable reading, insert the conductivity meter in the solution for at least 60 seconds.

E) Changing concentration will affect the redox potential.

Answer:

D) To ensure stable reading, insert the conductivity meter in the solution for at least 60 seconds.

Explanation:

Usually, redox potential is the measure of the ability of chemical elements to gain/lose electrons from/to an electrode respectively and after which they undergo reduction or oxidation respectively.

From online sources on experiments to determine redox potential, and taking into account the nernst equation which is; E_cell = E_0 – (RT/nF)•ln Q

Where;

E_cell is cell potential

E_0 is potential of the cell under standard conditions

R is universal gas constant

T is temperature

n is the number of electrons that are transferred in the redox reaction

F is Faraday's constant

Q is reaction quotient

Now, with all that in mind and looking at the options, the one that is false is option D.

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Answer:- Cl_2(g)+2e^-\rightarrow 2Cl^-(aq)

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Cl_2(g)+2e^-\rightarrow 2Cl^-(aq)

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When production first began some eighty years ago, ammonia production relied upon the direct reaction between gaseous hydrogen a
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Answer:

  • <u>Option 2. </u><u><em>Produce more ammonia.</em></u>

Explanation:

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This is the chemical reaction:

  • 3 H₂ (g) + N₂(g) ⇄ 2 NH₃(g) ∆H = −92.2 kJ

The information about the enthalpy of the reaction, ∆H = − 92.2 kJ,  indicates that energy (heat) has been released to the surroundings (the products of the forward reaction have less energy than the reactants), which is defined as an exothermic reaction.

Then, you can rewrite the equaition in the form:

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This is, the heat can be seen as a product of the direct reaction (or a reactant of the reverse reaction).

Now, it is quite straight to apply  Le Chatelier's principle:

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b) Then, the equilibrium must shift in a way that this lack of heat is compensated. Then, the reaction will shift to the right to produce more heat.

As conclusion, you can tell that in exothermic reactions, a decrase in temperature will cause the equilibrium to shift to the right.

This shift, of course, means the production of more ammonia.

The other choices are discarded following this brief reasoning:

1. increase the velocity of the gas molecules: the average velocity of the particles increases when the average kinetic energy increases, and the average kinetic energy will decrease if the temperature decreases. So, this statement is false.

3. increase the kinetic energy of the gas molecules: no, the average kinetic energy is proportional to the temperature, then reducing the temperature decreasese the average kinetic energy.

4. produce less ammonia: it was shown that reducing the temperature will produce more ammonia.

5. have no effect: no, it does have effect, as shown.

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