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Westkost [7]
2 years ago
10

N2+3H2 forms 2NH3

Chemistry
1 answer:
Nookie1986 [14]2 years ago
4 0

ok ther only one solution n2 is ÷ 2h×2 is ok but the problem is that the standard form of the square is not yet

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The standard reduction potentials of the following half-reactions are given in Appendix E in the textbook:
german

<u>Answer:</u>

<u>For 1:</u> The largest positive cell potential is of cell having 1st and 4th half reactions.

<u>For 2:</u> The standard electrode potential of the cell is 1.539 V

<u>For 3:</u> The smallest positive cell potential is of cell having 3rd and 4th half reactions. The standard electrode potential of the cell is 0.46 V

<u>Explanation:</u>

The substance having highest positive E^o potential will always get reduced and will undergo reduction reaction.

We are given:

Ag^++(aq.)+e^-\rightarrow Ag(s);E^o_{Ag^+/Ag}=0.799V\\\\Cu^{2+}+(aq.)+2e^-\rightarrow Cu(s);E^o_{Cu^{2+}/Cu}=0.337V\\\\Ni^{2+}(aq.)+2e^-\rightarrow Ni(s);E^o_{Ni^{2+}/Ni}=-0.28V\\\\Cr^{3+}(aq.)+3e^-\rightarrow Cr(s);E^o_{Cr^{3+}/Cr}=-0.74V

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

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

  • <u>Cell having 1st and 2nd half reactions:</u>

Silver has higher electrode potential. So, this will undergo reduction reaction and act as anode. Copper will undergo oxidation reaction and act as cathode.

E^o_{cell}=0.799-0.337=0.462V

  • <u>Cell having 1st and 3rd half reactions:</u>

Silver has higher electrode potential. So, this will undergo reduction reaction and act as anode. Nickel will undergo oxidation reaction and act as cathode.

E^o_{cell}=0.799-(-0.28)=1.079V

  • <u>Cell having 1st and 4th half reactions:</u>

Silver has higher electrode potential. So, this will undergo reduction reaction and act as anode. Chromium will undergo oxidation reaction and act as cathode.

E^o_{cell}=0.799-(-0.74)=1.539V

  • <u>Cell having 2nd and 3rd half reactions:</u>

Copper has higher electrode potential. So, this will undergo reduction reaction and act as anode. Nickel will undergo oxidation reaction and act as cathode.

E^o_{cell}=0.337-(-0.28)=0.617V

  • <u>Cell having 3rd and 4th half reactions:</u>

Nickel has higher electrode potential. So, this will undergo reduction reaction and act as anode. Chromium will undergo oxidation reaction and act as cathode.

E^o_{cell}=-0.28-(-0.74)=0.46V

Hence,

<u>For 1:</u> The largest positive cell potential is of cell having 1st and 4th half reactions.

<u>For 2:</u> The standard electrode potential of the cell is 1.539 V

<u>For 3:</u> The smallest positive cell potential is of cell having 3rd and 4th half reactions. The standard electrode potential of the cell is 0.46 V

8 0
4 years ago
What is the term for a push or a pull on an object
Katyanochek1 [597]
The term for a push or pull is called Force.
4 0
3 years ago
Which is an example of a chemical reaction
valkas [14]

Answer:iron and oxygen combining to make rust. vinegar and baking soda combining to make sodium acetate, carbon dioxide and water

Explanation:

7 0
3 years ago
Determine the number of grams in each of the quantities<br><br> 1.39.0 x 1024 molecules Cl2
STatiana [176]

Mass of Cl₂ : 164.01 g

<h3>Further explanation</h3>

A mole is a number of particles(atoms, molecules, ions)  in a substance

This refers to the atomic total of the 12 gr C-12  which is equal to 6.02.10²³, so 1 mole = 6.02.10²³ particles  

Can be formulated :

N = n x No

N = number of particles

n = mol

No = 6.02.10²³ = Avogadro's number

mol Cl₂ :

\tt n=\dfrac{N}{No}\\\\n=\dfrac{1.39.10^{24}}{6.02.10^{23}}\\\\n=2.31

mass Cl₂(MW=71 g/mol) :

\tt mass=mol\times MW\\\\mass=2.31\times 71=164.01

8 0
3 years ago
What is the pH of a 3.5 x 10 to the negative 11th M H+ solution
Ghella [55]

Answer:

Explanation:

-log(3.5 * 10^-11)

= 10.4559

Be careful how you put this into your calculator. I had to use Exp to get it to work properly.

-log

(3.5 * 10 exp -11)

=

4 0
3 years ago
Read 2 more answers
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