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riadik2000 [5.3K]
3 years ago
9

If the reduction reaction has a reduction potential of 0.1 V, and the oxidation reaction has a reduction potential of -0.4V, and

2 electrons were transfered, what is the value of delta G expressed in terms of F?
Chemistry
1 answer:
aleksley [76]3 years ago
5 0

Answer : The value of ΔG expressed in terms of F is, -1 F

Explanation :

First we have to calculate the standard electrode potential of the cell.

E^o=E^o_{cathode}-E^o_{anode}

or,

E^o=E^o_{reduction}-E^o_{oxidation}

E^o=(0.1V)-(-0.4V)=+0.5V

Now we have to calculate the standard cell potential.

Formula used :

\Delta G^o=-nFE^o

where,

\Delta G^o = Gibbs free energy = ?

n = number of electrons = 2

F = Faraday constant

E^o = standard e.m.f of cell = +0.5 V

Now put all the given values in this formula, we get the Gibbs free energy.

\Delta G^o=-(2\times F\times 0.5)

\Delta G^o=-1F

Therefore, the value of ΔG expressed in terms of F is, -1 F

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Explanation:

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7 0
3 years ago
How many particles are there in 5.0 moles​
grigory [225]

Answer:

3.01 × 10^24 particles

Explanation:

According to Avagadro, in one mole of a substance, there are 6.02 × 10^23 atoms or particles.

Using the formula: N = n × NA

Where;

N= number of particles or atoms

n = number of moles

NA = Avagadro's constant or number

This means that for 5 moles of a substance, there will be:

5 × 6.02 × 10^23

= 30.1 × 10^23

= 3.01 × 10^24 particles

8 0
3 years ago
Examine the unbalanced electrolytic reaction.
romanna [79]
<span>2Li⁺(aq) + Zn⁰(s) → 2Li⁰(s) + Zn²⁺(aq) 

</span>2Li⁺(aq) + 2e⁻ → 2Li⁰(s) 
Zn⁰(s) → Zn²⁺(aq)  +2e⁻

2 electrons are transferred from atom of  Zn⁰ to 2 ions of Li⁺.
3 0
3 years ago
Which best describes the effect of j.j. thompsons theory
dangina [55]
 The accepted model of the atom was changed.
8 0
3 years ago
Antimony has two naturally occuring isotopes, 121 Sb and 123 Sb . 121 Sb has an atomic mass of 120.9038 u , and 123 Sb has an at
valina [46]

Answer:

Percentage abundance of 121 Sb is = 57.2 %

Percentage abundance of 123 Sb is = 42.8 %

Explanation:

The formula for the calculation of the average atomic mass is:

Average\ atomic\ mass=(\frac {\%\ of\ the\ first\ isotope}{100}\times {Mass\ of\ the\ first\ isotope})+(\frac {\%\ of\ the\ second\ isotope}{100}\times {Mass\ of\ the\ second\ isotope})

Given that:

Since the element has only 2 isotopes, so the let the percentage of first be x and the second is 100 -x.

For first isotope, 121 Sb :

% = x %

Mass = 120.9038 u

For second isotope, 123 Sb:

% = 100  - x  

Mass = 122.9042 u

Given, Average Mass = 121.7601 u

Thus,  

121.7601=\frac{x}{100}\times 120.9038+\frac{100-x}{100}\times 122.9042

120.9038x+122.9042\left(100-x\right)=12176.01

Solving for x, we get that:

x = 57.2 %

<u>Thus, percentage abundance of 121 Sb is = 57.2 % </u>

<u>percentage abundance of 123 Sb is = 100 - 57.2 %  = 42.8 %</u>

6 0
3 years ago
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