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3 years ago

Balance this half-reaction: Re → ReO2 (acidic solution)

1 answer:

7 0

This redox reaction occurs in acidic solution, meaning that we need to add H+ and H2O to help balance it out (for comparison, balancing in basic solution would involve OH- and H2O instead).

Notice that the right side of the half reaction already has oxygen added to it, so we need to add the water to the other side. This way, the left side of the reaction would get enough oxygen to match the right side. Add enough water to the left to match the number of oxygens on the right.

2H2O + Re — ReO2

We now have two oxygens on each side, but the left has four more hydrogens than the right. Fix this by adding 4H+ to the right side of the equation.

2H2O + Re — ReO2 + 4H+

The number of each element matches the number of each on the other side; the only thing left is to balance the charges. The left side of the equation has a charge of 0, but the right has a charge of +4. We can fix this by adding four electrons to the right to cancel out the +4 charge.

2H2O + Re — ReO2 + 4H+ + 4e-

That’s the balanced equation.

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Car manufacturers are developing engines that use H₂ as fuel. In Iceland, Sweden, and other parts of Scandinavia, where hydroele

weeeeeb [17]

If the coulombs are supplied at 1.44 V, then 4.766 × 10¹¹ joules are produced.

(a) The reaction is

2H₂O + 2e⁻ → H₂ + 2OH⁻

According to the reaction, 2 moles of electrons flow per mole of reaction,

(b) Given

T = 25°C

Change Celsius into Kelvin

T = 25°C

= (25 + 273 ) K

= 298 K

<h3>What is Ideal Gas Law ?</h3>

It is expressed as

PV = nRT

Hence ,

n = $\frac{PV}{RT}$

= $\frac{(12)(3.5.10^{6} )}{(0.082057)(298)}$

= 1.7176 ×10⁶ mole

Now, From the given reaction we can say that ,1 mole of H₂ is produced by 2 moles of electron

Hence, 1.7176 × 10⁶ mole of H₂ is produced by

= 2 × 1.7176 × 10⁶ moles of electron

= 3.435 ×10⁶ moles of electron

We know that

charge = moles of electron × F

= (3.435 ×10⁶ × 96500) C

= 3.31 × 10¹¹ C

Voltage = $\frac{energy}{charge}$ = $\frac{J}{C}$

1.44 V = $\frac{energy}{3.31 . 10^{11} }$

Energy = 1.44 V × 3.31 × 10¹¹ C

= 4.766 × 10¹¹ Joules

Thus from the above conclusion we can say that, coulombs are supplied at 1.44 V, then 4.766 × 10¹¹ joules are produced.

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Disclaimer : The question given was incomplete on portal, Here is the complete question.

Question : Car manufacturers are developing engines that use H₂ as fuel. In Iceland, Sweden, and other parts of Scandinavia countries, where hydroelectric plants produce inexpensive electric power, the H₂ can be made industrially by the electrolysis of water.

a) How many moles of electrons flow per mole of reaction ?

b) How many coulombs are needed to produce 3.53X10⁶ L of H₂ gas at 12.0 atm and 25°C.

c) If the coulombs are supplied at 1.44 V, how many joules are produced?

8 0

1 year ago

Which molecule has a nonpolar covalent bond?

kondaur [170]

Nonpolar covalent bonds are chemical bonds where two atoms share a pair of electrons with each other and the electronegativities of the two atoms are equal. An example is methane. It has four carbon-hydrogen single covalent bonds. These bonds are nonpolar because the electrons are shared equally.

8 0

3 years ago

Read 2 more answers

Identify the number of moles in 369 grams of calcium hydroxide. Use the periodic table and the polyatomic ion resource.

topjm [15]

Answer : The number of moles in 369 grams of calcium hydroxide is, 4.98 moles

Explanation : Given,

Mass of calcium hydroxide = 369 g

Molar mass of calcium hydroxide = 74.093 g/mole

Formula used :

$\text{Moles of calcium hydroxide}=\frac{\text{Mass of calcium hydroxide}}{\text{Molar mass of calcium hydroxide}}$

Now put all the given values in this formula, we get the moles of calcium hydroxide.

$\text{Moles of calcium hydroxide}=\frac{369g}{74.093g/mole}=4.98mole$

Therefore, the number of moles in 369 grams of calcium hydroxide is, 4.98 moles

7 0

3 years ago

1pt Which statement best describes the relationship between the system and the surroundings?

Firlakuza [10]

Answer:

Hi how are you doing today Jasmine

4 0

3 years ago

Precipitation calculations with Ni2+ and Pb2+

Maurinko [17]

Precipitation calculations with Ni²⁺ and Pb²⁺ a. Use the solubility product for Ni(OH)₂ (s) . the pH at which Ni(OH)₂ begins to precipitate from a 0.18 M Ni²⁺ solution. (Ksp Ni(OH)₂ = 5.5x10⁻¹⁶) is 6.8.

When Ni(OH)₂ starts precipitate :

Ksp of Ni(OH)₂ = [ Ni²⁺ ] [ OH²⁻ ]

5.5x10⁻¹⁶ = [ 0.18 ] [ OH²⁻ ]

[ OH²⁻ ] = 5.5x10⁻¹⁶ / 0.18

[ OH⁻ ] = 5.5 × 10⁻⁸ M

pOH = 7.2

therefore , pH = 14 - 7.2

pH = 6.8

Thus, Precipitation calculations with Ni²⁺ and Pb²⁺ a. Use the solubility product for Ni(OH)₂ (s) . the pH at which Ni(OH)₂ begins to precipitate from a 0.18 M Ni²⁺ solution. (Ksp Ni(OH)₂ = 5.5x10⁻¹⁶) is 6.8.

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6 0

11 months ago