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erik [133]
3 years ago
15

Standard reduction potentials for nickel(II) and copper(II) The standard reduction potential for a substance indicates how readi

ly that substance gains electrons relative to other substances at standard conditions. Consider the following:
Ni2+(aq)+2e- -> Ni(s), E red = -0.23V Cu2+(aq)+2e- -> Cu(s), E red = +0.337V
What is the standard potential, E cell, for this galvanic cell? Express your answer to four significant figures and include the appropriate units.
Chemistry
1 answer:
AleksandrR [38]3 years ago
6 0

Answer:

The standard potential, E cell, for this galvanic cell is 0.5670V

Explanation:

Ni²⁺(aq)  +  2e⁻   →   Ni(s)           E red =  - 0.23V      ANODE

Cu²⁺(aq) +  2e-    →  Cu(s)          E red = + 0.337V    CATHODE

ΔE° = E cathode - E anode

ΔE° = 0.337V - (0.23V) = 0.5670 V

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A scientist has a container with a substance inside. At first, the molecules in the substance move away from each other. Later,
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Base on the properties the substances possess the substance change phase from gas to liquid.

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Read 2 more answers
Iron (III) oxide and hydrogen react to form iron and water, like this: Fe 03(s)+3H9)2Fe(s)+3HO) At a certain temperature, a chem
belka [17]

The question is incomplete, here is the complete question:

Iron (III) oxide and hydrogen react to form iron and water, like this:

Fe_2O_3(s)+3H_2(g)\rightarrow 2Fe(s)+3H_2O(g)

At a certain temperature, a chemist finds that a 8.9 L reaction vessel containing a mixture of iron(III) oxide, hydrogen, Iron, and water at equilibrium has the following composition.

Compound             Amount

  Fe₂O₃                     3.95 g

     H₂                        4.77 g

     Fe                        4.38 g

    H₂O                      2.00 g

Calculate the value of the equilibrium constant Kc for this reaction. Round your answer to 2 significant digits.

<u>Answer:</u> The value of equilibrium constant for given equation is 1.0\times 10^{-4}

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}

  • <u>For hydrogen gas:</u>

Given mass of hydrogen gas = 4.77 g

Molar mass of hydrogen gas = 2 g/mol

Volume of the solution = 8.9 L

Putting values in above expression, we get:

\text{Molarity of hydrogen gas}=\frac{4.77}{2\times 8.9}\\\\\text{Molarity of hydrogen gas}=0.268M

  • <u>For water:</u>

Given mass of water = 2.00 g

Molar mass of water = 18 g/mol

Volume of the solution = 8.9 L

Putting values in above expression, we get:

\text{Molarity of water}=\frac{2.00}{18\times 8.9}\\\\\text{Molarity of water}=0.0125M

For the given chemical equation:

Fe_2O_3(s)+3H_2(g)\rightarrow 2Fe(s)+3H_2O(g)

The expression of equilibrium constant for above equation follows:

K_{eq}=\frac{[H_2O]^3}{[H_2]^3}

Concentration of pure solids and pure liquids are taken as 1 in equilibrium constant expression.

Putting values in above expression, we get:

K_{c}=\frac{(0.0125)^3}{(0.268)^3}\\\\K_{c}=1.0\times 10^{-4}

Hence, the value of equilibrium constant for given equation is 1.0\times 10^{-4}

6 0
3 years ago
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