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AveGali [126]
2 years ago
5

A certain half-reaction has a standard reduction potential +0.80 V . An engineer proposes using this half-reaction at the anode

of a galvanic cell that must provide at least 0.9 V of electrical power. The cell will operate under standard conditions. Note for advanced students: assume the engineer requires this half-reaction to happen at the anode of the cell.
a. Is there a minimum standard reduction potential that the half-reaction used at the cathode of this cell can have?
b. Is there a maximum standard reduction potential that the half-reaction used at the cathode of this cell can have?
Chemistry
1 answer:
Strike441 [17]2 years ago
8 0

Answer:

a. Minimum 1.70 V

b. There is no maximum.

Explanation:

We can solve this question by remembering that the cell potential is given by the formula

ε⁰ cell = ε⁰ reduction -  ε⁰  oxidation

Now the problem states the cell must provide at least 0.9 V and that the reduction potential of the  oxidized species  0.80 V, thus

ε⁰ reduction -  ε⁰  oxidation ≥  ε⁰ cell

Since ε⁰  oxidation is by definition the negative of ε⁰ reduction , we have

ε⁰ reduction - ( 0.80 V )  ≥  0.90 V

⇒ ε⁰ reduction  ≥ 1.70 V

Therefore,

(a) The minimum standard reduction potential is 1.70 V

(b) There is no maximum standard reduction potential since it is stated in the question that we want to have a cell that provides at leat 0.9 V

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Therefore M₁ = 4÷0.238² = 70.61

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Answer:

The correct answer is <em>C)  Two atoms of silver are needed to complete the reaction.</em>

Explanation:

The Law of Conservation of Matter postulates that "the mass is not created or destroyed, only transformed." This means that the reagents interact with each other and form new products with physical and chemical properties different from those of the reagents because the atoms of the substances are ordered differently. But the amount of matter or mass before and after a transformation (chemical reaction) is always the same, that is, the quantities of the masses involved in a given reaction must be constant at all times, not changing in their proportions when the reaction ends.

Then, taking into account the Law of Conservation of Matter, as an atom cannot be created or destroyed in a chemical reaction, the number of atoms that are present in the reagents has to be equal to the number of atoms present in the products.

For this, the chemical equation must be balanced. For that, you must first look at the subscripts next to each atom to find the number of atoms in each compound in the equation. If the same atom appears in more than one molecule, you must add its quantities. On the other hand, the coefficients located in front of each molecule indicate the amount of each molecule for the reaction. This coefficient can be modified to balance the equation, just as you should never alter the subscripts. By multiplying the coefficient mentioned by the subscript, you get the amount of each element present in the reaction.

In this case:

Left side: 2 silver  (Ag), 2 hydrogen (H) and 1 sulfur (S)

Right side: 2 silver  (Ag), 2 hydrogen (H) and 1 sulfur (S)

In this case the equation is balanced because you have the same amount of all the elements on each side of the reaction. And <u><em>the 2 in front of 2Ag indicates that,since silver is a reagent, two atoms of silver are needed to complete the reaction. (option C).</em></u>

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Answer:

Explanation:

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2 years ago
What are the mole fraction and the mass percent of a solution made by dissolving 0.21 g KBr in 0.355 L water? (d = 1.00 g/mL.) m
IRISSAK [1]

Answer:

Mol fraction H2O = 0.99991

Mol fraction KBr = 0.00009

mass % KBr = 0.059 %

mass % H2O = 99.941 %

Explanation:

Step 1: Data given

Mass of KBr = 0.21 grams

Molar mass KBr = 119 g/mol

Volume of water = 355 mL

Density of water = 1.00 g/mL

Molar mass water = 18.02 g/mol

Step 2: Calculate mass water

Mass water = 355 mL * 1g /mL

Mass water = 355 grams

Step 3: Calculate moles water

Moles water = mass water / molar mass water

Moles water = 355 grams / 18.02 g/mol

Moles water = 19.7 moles

Step 4: Calculate moles KBr

Moles KBr = 0.21 grams / 119 g/mol

Moles KBr = 0.00176 moles

Step 5: Calculate total moles

Total moles = 19.7 moles + 0.00176 moles

Total moles = 19.70176 moles

Step 6: Calculate mol fraction

Mol fraction H2O = 19.7 moles / 19.70176 moles

Mol fraction H2O = 0.99991

Step 7: Calculate mol fraction KBr

Mol fraction KBr = 0.00176 / 19.70176

Mol fraction KBr = 0.00009

Step 6: Calculate mass %

mass % KBR = (0.21 grams / (0.21 + 355) grams) *100%

mass % KBr = 0.059 %

mass % H2O = (355 grams / 355.21 grams) *100%

mass % H2O = 99.941 %

8 0
3 years ago
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