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AleksAgata [21]
2 years ago
8

A voltaic cell consists of a nickel electrode in a solution containing Ni2+ ions, and a copper electrode in a solution containin

g Cu2+ ions. Which is the cell potential? a. E subscript c e l l space end subscript superscript 0 space end superscript equals minus 0.59 V b. E subscript c e l l space end subscript superscript 0 space end superscript equals plus 0.59 V c. E subscript c e l l space end subscript superscript 0 space end superscript equals minus 0.09 V d. E subscript c e l l space end subscript superscript 0 space end superscript equals plus 0.09 V
Chemistry
1 answer:
Marysya12 [62]2 years ago
4 0
First, we write the half equations for the reduction of the chemical species present:

Cu⁺² + 2e → Cu; E° = 0.34 V
Ni⁺² + 2e → Ni; E° = - 0.23 V

In order to determine the potential of the cell, we find the difference between the two values. For this:

E(cell) = 0.34 - (-0.23)
E(cell) = 0.57 V

The second option is correct. (The difference in values is due to different values in literature, and it is negligible)
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<em><u>Neutrons</u></em>

<em>Although elements have a specific number of protons, atoms of the same element may have different numbers of neutrons and are termed isotopes. For example, hydrogen has three isotopes, each with a single proton. Protium is an isotope of hydrogen with zero neutrons, deuterium has one neutron, and tritium has two neutrons. Although the number of neutrons may differ between isotopes, the isotopes all behave in a chemically similar manner.</em>

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<u><em>Electrons</em></u>

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3 years ago
What element is necessary for a healthy eye
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For the following reaction, 38.3 grams of sulfuric acid are allowed to react with 33.5 grams of calcium hydroxide sulfuric acid(
Likurg_2 [28]

Answer:

What is the maximum amount of calcium sulfate that can be formed? 53.1 grams CaSO4

What is the FORMULA for the limiting reagent? H2SO4

What amount of the excess reagent remains after the reaction is complete? 4.59 grams of Ca(OH)2

Explanation:

Step 1: Data given

Mass of sulfuric acid = 38.3 grams

Molar mass of H2SO4 = 98.08 g/mol

Mass of calcium hydroxide = 33.5 grams

Molar mass of Ca(OH)2 = 74.09 g/mol

Step 2: The balanced equation

H2SO4 + Ca(OH)2 → CaSO4 + 2H2O

Step 3: Calculate moles of H2SO4

moles H2SO4 = mass H2SO4 / molar mass H2SO4

moles H2SO4 = 38.3 grams / 98.08 g/mol

moles H2SO4 = 0.390 moles

Step 4: Calculate moles of Ca(OH)2

moles Ca(OH)2 = 33.5 grams / 74.09 g/mol

moles Ca(OH)2 =0.452 moles

Step 5: Calculate limiting reactant

For 1 mol H2SO4, we need 1 mol of Ca(OH)2 to produce, 1 mol of CaSO4 and 2 mol of H2O

H2SO4 is the limiting reactant. It will completely be consumed (0.390 moles).

Ca(OH)2 is in excess. There will be consumed 0.390 moles

There will remain 0.452 - 0.390 = 0.062 moles

This is 0.062 * 74.09 g/mol = 4.59 grams

Step 6: Calculate moles of calcium sulfate

For 1 mol H2SO4, we need 1 mol of Ca(OH)2 to produce, 1 mol of CaSO4 and 2 mol of H2O

For 0.390 moles of H2SO4, there will be produced 0.390 moles of CaSO4

Step 7: Calculate mass of CaSO4

Mass CaSO4 = moles CaSO4 * molar mass CaSO4

Mass CaSO4 = 0.390 moles * 136.14 g/mol

Mass of CaSO4 = 53.1 grams

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