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Gelneren [198K]
3 years ago
15

What is the cell potential for the reaction mg(s+fe2+(aq?mg2+(aq+fe(s at 77 ?c when [fe2+]= 3.40 m and [mg2+]= 0.210 m . express

your answer to three significant figures and include the appropriate units?
Chemistry
1 answer:
Galina-37 [17]3 years ago
8 0
First, you need to calculate the standard cell potential using standard reduction potential from a textbook or online. Since Mg becomes Mg+2, magnesium is being oxidized because it is losing electrons, you need to flip its potential

Fe+2 + 2e- --> Fe                  potential= -0.44
Mg+2 + 2e- --> Mg                potential= -2.37


Cell potential= (-0.44) + (+2.37)= 1.93 V

Now, you need to use Nernst formula to get the answer. I have attached a PDF with the work.
Download pdf
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This is an incomplete question, here is a complete question.

The Henry's law constant for oxygen dissolved in water is 4.34 × 10⁹ g/L.Pa at 25⁰C.If the partial pressure of oxygen in air is 0.2 atm, under atmospheric conditions, calculate the molar concentration of oxygen in air-saturated and oxygen saturated water.

Answer : The molar concentration of oxygen is, 2.67\times 10^2mol/L

Explanation :

As we know that,

C_{O_2}=k_H\times p_{O_2}

where,

C_{O_2} = molar solubility of O_2 = ?

p_{O_2} = partial pressure of O_2 = 0.2 atm  = 1.97×10⁻⁶ Pa

k_H = Henry's law constant  = 4.34 × 10⁹ g/L.Pa

Now put all the given values in the above formula, we get:

C_{O_2}=(4.34\times 10^9g/L.Pa)\times (1.97\times 10^{-6}Pa)

C_{O_2}=8.55\times 10^3g/L

Now we have to molar concentration of oxygen.

Molar concentration of oxygen = \frac{8.55\times 10^3g/L}{32g/mol}=2.67\times 10^2mol/L

Therefore, the molar concentration of oxygen is, 2.67\times 10^2mol/L

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