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Arada [10]
2 years ago
6

A concentration cell is constructed by placing identical iron electrodes in two Fe2 solutions. The potential of this cell is obs

erved to be 0.047 V. If the more concentrated Fe2 solution is 0.10 M, what is the concentration of the other Fe2 solution
Chemistry
1 answer:
Stolb23 [73]2 years ago
8 0

<u>Answer:</u> The concentration of other (diluted) Fe^{2+} solution is 0.0026 M

<u>Explanation:</u>

The half reactions for the cell is:

<u>Oxidation half reaction (anode):</u>  Fe(s)\rightarrow Fe^{2+}+2e^-

<u>Reduction half reaction (cathode):</u>  Fe^{2+}+2e^-\rightarrow Fe(s)

In this case, the cathode and anode both are same. So, E^o_{cell} will be equal to zero.

To calculate cell potential of the cell, we use the equation given by Nernst, which is:

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Fe^{2+}_{(diluted)}]}{[Fe^{2+}_{(concentrated)}]}

where,

n = number of electrons in oxidation-reduction reaction = 2

E_{cell} = 0.047 V

[Fe^{2+}_{(diluted)}] = ? M

[Fe^{2+}_{(concentrated)}] = 0.10 M

Putting values in above equation, we get:

0.047=0-\frac{0.0592}{2}\log \frac{[Fe^{2+}_{(diluted)}]}{0.10M}

[Fe^{2+}_{(diluted)}]=0.0026M

Hence, the concentration of other (diluted) Fe^{2+} solution is 0.0026 M

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How many valence electrons are in chlorodifluoromethane
Elis [28]
<span>Well... first, let's recognize that the chemical formula for chlorodifluoromethane is CHClF2. Count out how many valence electrons there are. C = 4, H = 1, Cl = 7, F (X2) = 14. Total is 26. Let's put C as the central atom, and put the other elements surrounding it. Draw a pair of electrons beach each element and the central atom. Then fill the halogen elements with 3 pairs of electrons each to fill their octets. Count out how many dots you have. There should be 26, making this the correct lewis structure! Remember, hydrogen doesn't have a full octet, only a maximum of two electrons.</span>
7 0
2 years ago
Standard Hydrogen Electrode consists of platinum wire fused in a glass tube and a platinum plate coated with finely divided plat
iris [78.8K]

Answer:

To increase surface area of the platinum electrode which results in superior quality and action of the electrodes as opposed to normal platinum electrodes.

Explanation:

Platinization of Platinum is the process of covering platinum electrode with a layer of platinum black. Platinum black is a finally divided form of platinum, optimized for catalysing the addition of hydrogen to unsaturated organic compound. This increases the surface area of the platinum electrodes and therefore exhibits action superior to that of normal electrodes.

5 0
2 years ago
Consider the complex ion [CO(NH3)6]3+. Which response contains all of the following statements that are true, and no false state
Alja [10]

Answer:

III, IV, and V

Explanation:

The complex [CO(NH3)6]3+ is a diamagnetic complex. It a low spin d^6 complex. Most d^6 complexes are low spin due to the higher crystal field stabilization energy of the low spin over the high spin arrangement.

d^6 metal complexes are known to be octahedral (a coordination number of 6 leads to octahedral geometry). Octahedral complexes does not have geometric isomers rather, may exist as the fac or her stereo isomers.

8 0
2 years ago
4p + 5O2 -&gt; P4O10 ; the percent yield of PO4O10 when 6.20 g of phosphorus burns into excess oxygen is 67.0%. What is the yiel
denis23 [38]

Answer:    36.9 g

Explanation:

P4 + 5O2 = P4O10  Balanced equation

moles P4 present = 23.9 g x 1 mole/123.88 g = 0.193 moles

moles O2 present = 20.8 g x 1 mol/32 g = 0.65 moles O2

From balanced equation, mole ratio O2 : P4 is 5:1.  Is 0.65 moles O2 5x 0.193 moles?  NO.  You don't have enough O2.

O2 is limiting in this reaction.

 

theoretical moles of P4O10 = 0.65 moles O2 x 1 mole P4O10/5 moles O2 = 0.13 moles P4O10

mass of P4O10 produced = 0.13 moles x 283.9 g = 36.9 g

3 0
1 year ago
Calculate the ph of a solution formed by mixing 150.0 ml of 0.10 m hc7h5o2 with 100.0 ml of 0.30 m nac7h5o2. The ka for hc7h5o2
m_a_m_a [10]

Answer:

4.49

Explanation:

pKa = - log 6.5 x 10⁻⁵

pKa  =4.19

Given that :

Volume = 150 mL = 0.150 L

For solutions:

number of moles of acid = volume × concentration

number of moles of  acid = 0.150 L × 0.10 M = 0.0150

number of moles  of salt = 0.100 L × 0.30 M = 0.0300

total volume = 150 + 100

= 250 mL

= 0.250 L

Concentration = number of moles/ Volume

∴

For [salt] = 0.0300/ 0.250

= 0.12 M

For [acid] = 0.0150/ 0.250

=0.06 M

pH = 4.19 + log 0.12/0.06

=4.49

7 0
2 years ago
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