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8090 [49]
3 years ago
11

The half-reaction occurring at the cathode in the balanced reaction shown below is __________?2Ag(s) + 1/2 O2(g) + 2H+(aq) → 2Ag

+(aq) + H2O (l)A. 2H2O (l) → O2 (g) + 4H+ (aq) + 4e-B. 2Ag (s) → 2Ag+ (aq) + 2e-C. 1/2 O2 (g) + 2H+ (aq) + 2e- → H2O (l)D. 1/2 O2 (g) + 2H+ (aq) + 2e- → H2O (l)
Chemistry
1 answer:
Shtirlitz [24]3 years ago
3 0

Answer:

Correct choice are C and D (they are both, the same).

Explanation:

Cathode is the positive(+) electrode where a reduction occurs.

Reduction is the chemical reaction where the oxidation state is reduced.

2Ag(s) + 1/2 O2(g) + 2H+(aq) → 2Ag+(aq) + H2O (l)

A. 2H2O (l) → O2 (g) + 4H+ (aq) + 4e-

B. 2Ag (s) → 2Ag+ (aq) + 2e-

C. 1/2 O2 (g) + 2H+ (aq) + 2e- → H2O (l)

D. 1/2 O2 (g) + 2H+ (aq) + 2e- → H2O (l)

C or D, are ok. They are the same equation.

Oxygen from ground state reduce the oxidation state from 0 to -2

You might be interested in
In addition to the two types of motion found in liquids, gases have <br> What kind of motion
makvit [3.9K]

Answer:

In liquids, particles are quite close together and move with random motion throughout the container. Particles move rapidly in all directions but collide with each other more frequently than in gases due to shorter distances between particles.

3 0
3 years ago
How many kJ of heat are released by the reaction of 25.0 g of Na2O2(s) in the following reaction? (M = 78.0 g/mol for Na2O2)
solong [7]

-20.16 KJ of heat are released by the reaction of 25.0 g of Na2O2.

Explanation:

Given:

mass of Na2O2 = 25 grams

atomic mass of Na2O2 = 78 gram/mole

number of mole = \frac{mass}{atomic mass of 1 mole}

                          = \frac{25}{78}

                          =0. 32 moles

The balanced equation for the reaction:

2 Na2O2(s) + 2 H2O(l) → 4 NaOH(aq) + O2(g) ∆Hο = −126 kJ

It can be seen that 126 KJ of energy is released when 2 moles of Na2O2 undergoes reaction.

similarly 0.3 moles of Na2O2 on reaction would give:

\frac{126}{2} = \frac{x}{0.32}

x = \frac{126 x 0.32}{2}

 = -20.16 KJ

Thus, - 20.16 KJ of energy will be released.

6 0
3 years ago
Express the van der Waals equation of state as a virial expansion in powers of 1/Vm and obtain expressions for B and C in terms
nirvana33 [79]

Answer:

PV_{m} = RT[1 + (b-\frac{a}{RT})\frac{1}{V_{m} } + \frac{b^{2} }{V^{2} _{m} } + ...]

B = b -a/RT

C = b^2

a = 1.263 atm*L^2/mol^2

b = 0.03464 L/mol

Explanation:

In the given question, we need to express the van der Waals equation of state as a virial expansion in powers of 1/Vm and obtain expressions for B and C in terms of the parameters a and b. Therefore:

Using the van deer Waals equation of state:

P = \frac{RT}{V_{m}-b } - \frac{a}{V_{m} ^{2} }

With further simplification, we have:

P = RT[\frac{1}{V_{m}-b } - \frac{a}{RTV_{m} ^{2} }]

Then, we have:

P = \frac{RT}{V_{m} } [\frac{1}{1-\frac{b}{V_{m} } } - \frac{a}{RTV_{m} }]

Therefore,

PV_{m} = RT[(1-\frac{b}{V_{m} }) ^{-1} - \frac{a}{RTV_{m} }]

Using the expansion:

(1-x)^{-1} = 1 + x + x^{2} + ....

Therefore,

PV_{m} = RT[1+\frac{b}{V_{m} }+\frac{b^{2} }{V_{m} ^{2} } + ... -\frac{a}{RTV_{m} }]

Thus:

PV_{m} = RT[1 + (b-\frac{a}{RT})\frac{1}{V_{m} } + \frac{b^{2} }{V^{2} _{m} } + ...]           equation (1)

Using the virial equation of state:

P = RT[\frac{1}{V_{m} }+ \frac{B}{V_{m} ^{2}}+\frac{C}{V_{m} ^{3} }+ ...]

Thus:

PV_{m} = RT[1+ \frac{B}{V_{m} }+ \frac{C}{V_{m} ^{2} } + ...]     equation (2)

Comparing equations (1) and (2), we have:

B = b -a/RT

C = b^2

Using the measurements on argon gave B = −21.7 cm3 mol−1 and C = 1200 cm6 mol−2 for the virial coefficients at 273 K.

b = \sqrt{C} = \sqrt{1200} = 34.64[tex]cm^{3}/mol[/tex] = 0.03464 L/mol

a = (b-B)*RT = (34.64+21.7)*(1L/1000cm^3)*(0.0821)*(273) = 1.263 atm*L^2/mol^2

3 0
2 years ago
If 75 grams of oxygen react, how many grams of aluminum are required?
german

Answer:

84.24 g

Explanation:

Given data:

Mass of oxygen = 75 g

Mass of Al required to react = ?

Solution:

Chemical equation:

4Al + 3O₂     →   2Al₂O₃

Number of moles of oxygen:

Number of moles = mass/ molar mass

Number of moles = 75 g/ 32 g/mol

Number of moles = 2.34 mol

Now we will compare the moles of oxygen with Al.

                          O₂         :          Al

                           3          :             4

                        2.34        :         4/3×2.34 = 3.12 mol

Mass of Al required:

Mass = number of moles × molar mass

Mass = 3.12 mol × 27 g/mol

Mass = 84.24 g

5 0
3 years ago
A gas with a volume of 4.0 L at a pressure of 2.02 atm is allowed to expand to a volume of 12.0
Maurinko [17]
<h3>Answer:</h3>

P₂ = 0.67 atm

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right  

Equality Properties

  • Multiplication Property of Equality
  • Division Property of Equality
  • Addition Property of Equality
  • Subtraction Property of Equality<u> </u>

<u>Chemistry</u>

<u>Gas Laws</u>

Boyle's Law: P₁V₁ = P₂V₂

  • P₁ is pressure 1
  • V₁ is volume 1
  • P₂ is pressure 2
  • V₂ is volume 2
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] P₁ = 2.02 atm

[Given] V₁ = 4.0 L

[Given] V₂ = 12.0 L

[Solve] P₂

<u>Step 2: Solve</u>

  1. Substitute in variables [Boyle's Law]:                                                              (2.02 atm)(4.0 L) = P₂(12.0 L)
  2. [Pressure] Multiply:                                                                                           8.08 atm · L = P₂(12.0 L)
  3. [Pressure] [Division Property of Equality] Isolate unknown:                          0.673333 atm = P₂
  4. [Pressure] Rewrite:                                                                                           P₂ = 0.673333 atm

<u>Step 3: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs as our smallest.</em>

0.673333 atm ≈ 0.67 atm

4 0
3 years ago
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