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Aleksandr [31]
2 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]2 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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the value of ksp for pbcl2 is 1.6. what is the lowest concentration of Cl- that would be needed to begin precipitation of PbCl2
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Answer:

The minimum concentration of Cl⁻ that produces precipitation is 12.6M

Explanation:

The Ksp of PbCl₂ is expressed as:

PbCl₂(s) → Pb²⁺(aq) + 2Cl⁻(aq)

The Ksp is:

Ksp = 1.6 = [Pb²⁺] [Cl⁻]²

When Ksp = [Pb²⁺] [Cl⁻]² the solution begind precipiration.

A 0.010M Pb(NO₃)₂ is 0.010M Pb²⁺, thus:

1.6 = [0.010M] [Cl⁻]²

160 = [Cl⁻]²

12.6M =  [Cl⁻]

<h3>The minimum concentration of Cl⁻ that produces precipitation is 12.6M</h3>
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How do you predict a reaction outcome for Ni(s) + H2O(L)?
ElenaW [278]

Answer:

Ni(s) + H₂O(l) -------------> [Ni(H₂O)₆]²⁺

Explanation:

Data Given:

Reactants:

Ni(s) + H₂O(l)

Product = ?

Solution:

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So.

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