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dem82 [27]
3 years ago
11

Consider the group 1A element sodium (atomic number 11), the group 3A element aluminum (atomic number 13), and the group 7A elem

ent chlorine (atomic number 17). These elements are in period 3. How are the first ionization energies of these elements related?
Chemistry
1 answer:
hoa [83]3 years ago
3 0
Ionization energy is the energy needed to remove a valence electrón from a gaseous atom.

The energy needed to remove the first electron is the first ionization energy.

The stronger the atom atracts its electrons the higher the ionization energy.

A low ionization energy means that the atom may loose one electron easily to form a positive ion (cation)

In the period 3, sodium has the lowest ionization energy (it forms sodium ion easily), aluminium has a higher ionization energy (meaning that it is more difficult to form a positive ion) and Chlorine will never lose an electron to form a positive ion (on the contrary Chlorine is willing to accept one electron from other atom to form a negative ion, i.e. an anion).

The growing of the ionization energy inside a period is due to the size of the growing positive charge (number of protons) in the nucleous, which implies a stronger atraction to the electrons of valence.
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A chemist designs a galvanic cell that uses these two half-reactions:
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Answer :

(a) Reaction at anode (oxidation) : 4Fe^{2+}\rightarrow 4Fe^{3+}+4e^-  

(b) Reaction at cathode (reduction) : O_2+4H^++4e^-\rightarrow 2H_2O  

(c) O_2+4H^++4Fe^{2+}\rightarrow 2H_2O+4Fe^{3+}

(d) Yes, we have have enough information to calculate the cell voltage under standard conditions.

Explanation :

The half reaction will be:

Reaction at anode (oxidation) : Fe^{2+}\rightarrow Fe^{3+}+e^-     E^0_{anode}=+0.771V

Reaction at cathode (reduction) : O_2+4H^++4e^-\rightarrow 2H_2O     E^0_{cathode}=+1.23V

To balance the electrons we are multiplying oxidation reaction by 4 and then adding both the reaction, we get:

Part (a):

Reaction at anode (oxidation) : 4Fe^{2+}\rightarrow 4Fe^{3+}+4e^-     E^0_{anode}=+0.771V

Part (b):

Reaction at cathode (reduction) : O_2+4H^++4e^-\rightarrow 2H_2O     E^0_{cathode}=+1.23V

Part (c):

The balanced cell reaction will be,

O_2+4H^++4Fe^{2+}\rightarrow 2H_2O+4Fe^{3+}

Part (d):

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

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

E^o=(1.23V)-(0.771V)=+0.459V

For a reaction to be spontaneous, the standard electrode potential must be positive.

So, we have have enough information to calculate the cell voltage under standard conditions.

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