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Kazeer [188]
3 years ago
5

Question 15 (1 point)

Chemistry
1 answer:
Digiron [165]3 years ago
6 0

Answer:

Atoms are electrically neutral because the number of protons, which carry a 1+ charge, in the nucleus of an atom is equal to the number of electrons, which carry a 1- charge, in the atom. The result is that the total positive charge of the protons cancels out the total negative charge of the electrons so that the net charge of the atom is zero. Most atoms, however, can either gain or lose electrons; when they do so, the number of electrons becomes different from the number of protons in the nucleus. The resulting charged species is called an ion.

Cations and anions

When a neutral atom loses one or more electrons, the total number of electrons decreases while the number of protons in the nucleus remains the same. The result is that the atom becomes a cation—an ion with a net positive charge.

The opposite process can also occur. When a neutral atom gains one or more electrons, the number of electrons increases while the number of protons in the nucleus remains the same. The result is that the atom becomes an anion—an ion with a net negative charge. We can illustrate this by examining some very simple cations and anions, those formed when a single hydrogen atom loses or gains an electron.

Note: Hydrogen is actually somewhat unusual in that it readily forms both cations and anions. Most elements much prefer to form only one or the other. In terms of its electron configuration, can you explain why hydrogen can form both cations and anions? Feel free to post in the comments at the end of the article!

A hydrogen cation, a hydrogen atom, and a hydrogen anion.

A hydrogen cation, a hydrogen atom, and a hydrogen anion.

Classification cation neutral atom anion

No. of protons 111 111 111

No. of electrons 000 111 222

Net charge 111++plus 000 111-−minus

If a neutral hydrogen atom ( \text{H}Hstart text, H, end text, center) loses an electron, it becomes a hydrogen cation ( \text{H}^+H  

+

start text, H, end text, start superscript, plus, end superscript, left). Conversely, if the neutral \text{H}Hstart text, H, end text atom gains an electron, it becomes a hydrogen anion ( \text{H}^-H  

−

start text, H, end text, start superscript, minus, end superscript, right), also known as a hydride ion. Image credit: adapted from Boundless Learning, CC BY-SA 4.0.

In the center column, we have a diagram of a single, neutral hydrogen atom. It contains one proton and one electron; thus, its net charge is zero. If hydrogen loses its electron, it forms the cation \text{H}^+H  

+

start text, H, end text, start superscript, plus, end superscript (left column). The \text{H}^+H  

+

start text, H, end text, start superscript, plus, end superscript cation has a net charge of 1+ from the one proton in the nucleus since there are no electrons to cancel out the positive charge. If neutral hydrogen gains an electron, it forms the anion \text{H}^-H  

−

start text, H, end text, start superscript, minus, end superscript (right column). The \text{H}^-H  

−

start text, H, end text, start superscript, minus, end superscript anion has a net charge of 1- because it has one extra electron compared to the total number of protons.

Explanation: Hopes this gives a better explanation on them!

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Thus, the anode and cathode will be E^o_{(Cu^{2+}/Cu)}

and E^o_{(Fe^{3+}/Fe^{2+})} respectively.

(b) Now we have to calculate the potential of a cell.

Oxidation half reaction (anode):  Cu\rightarrow Cu^{2+}+2e^-

Reduction half reaction (cathode):  Fe^{3+}+1e^-\rightarrow Fe^{2+}

In order to balance that electrons, we will multiple the reduction reaction by 2, we get:

Oxidation half reaction (anode):  Cu\rightarrow Cu^{2+}+2e^-

Reduction half reaction (cathode):  2Fe^{3+}+2e^-\rightarrow 2Fe^{2+}

The overall cell reaction will be,

2Fe^{3+}+Cu\rightarrow Cu^{2+}+2Fe^{2+}

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

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

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Using Nernest equation :

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

E_{cell} = emf of the cell = ?

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

E_{cell}=1.20-\frac{0.0592}{2}\log \frac{(0.20)^2(0.25)}{(0.0001)^2}

E_{cell}=1.022V

Therefore, the emf of cell potential is 1.022 V

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