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Art [367]
3 years ago
11

Chlorine can form a maximum of bonds

Chemistry
1 answer:
stepladder [879]3 years ago
7 0

Answer: it can form a maximum of one bond

Explanation:

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1.What zodiac sign is October 19,2006?<br> A). Sagittarius<br> B). Libra<br> C). Leo<br> D). Pisces
MrMuchimi

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B. Libra

Explanation:

Libra's range from September 23rd to October 22nd

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An electrochemical cell at 25°C is composed of pure copper and pure lead solutions immersed in their respective ionis. For a 0.6
ExtremeBDS [4]

Answer :

(a) The concentration of Pb^{2+} is, 0.0337 M

(b) The concentration of Pb^{2+} is, 6.093\times 10^{32}M

Solution :

<u>(a) As per question, lead is oxidized and copper is reduced.</u>

The oxidation-reduction half cell reaction will be,

Oxidation half reaction:  Pb\rightarrow Pb^{2+}+2e^-

Reduction half reaction:  Cu^{2+}+2e^-\rightarrow Cu

The balanced cell reaction will be,  

Pb(s)+Cu^{2+}(aq)\rightarrow Pb^{2+}(aq)+Cu(s)

Here lead (Pb) undergoes oxidation by loss of electrons, thus act as anode. Copper (Cu) undergoes reduction by gain of electrons and thus act as cathode.

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

E^o_{[Pb^{2+}/Pb]}=-0.13V

E^o_{[Cu^{2+}/Cu]}=+0.34V

E^o=E^o_{[Cu^{2+}/Cu]}-E^o_{[Pb^{2+}/Pb]}

E^o=0.34V-(-0.13V)=0.47V

Now we have to calculate the concentration of Pb^{2+}.

Using Nernest equation :

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Pb^{2+}]}{[Cu^{2+}]}

where,

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

E_{cell} = 0.507 V

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

0.507=0.47-\frac{0.0592}{2}\log \frac{[Pb^{2+}]}{(0.6)}

[Pb^{2+}]=0.0337M

Therefore, the concentration of Pb^{2+} is, 0.0337 M

<u>(b) As per question, lead is reduced and copper is oxidized.</u>

The oxidation-reduction half cell reaction will be,

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

Reduction half reaction:  Pb^{2+}+2e^-\rightarrow Pb

The balanced cell reaction will be,  

Cu(s)+Pb^{2+}(aq)\rightarrow Cu^{2+}(aq)+Pb(s)

Here Copper (Cu) undergoes oxidation by loss of electrons, thus act as anode. Lead (Pb) undergoes reduction by gain of electrons and thus act as cathode.

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

E^o_{[Pb^{2+}/Pb]}=-0.13V

E^o_{[Cu^{2+}/Cu]}=+0.34V

E^o=E^o_{[Pb^{2+}/Pb]}-E^o_{[Cu^{2+}/Cu]}

E^o=-0.13V-(0.34V)=-0.47V

Now we have to calculate the concentration of Pb^{2+}.

Using Nernest equation :

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cu^{2+}]}{[Pb^{2+}]}

where,

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

E_{cell} = 0.507 V

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

0.507=-0.47-\frac{0.0592}{2}\log \frac{(0.6)}{[Pb^{2+}]}

[Pb^{2+}]=6.093\times 10^{32}M

Therefore, the concentration of Pb^{2+} is, 6.093\times 10^{32}M

6 0
3 years ago
What is the nature of the Ca-Cl bond in a molecule of calcium chloride (CaCl2) if the electronegativity value of calcium is 1.0
nevsk [136]
Ca-Cl is an ionic bond. We know this because the difference between their electronegativities is 2.16.The electronegativity tells us which atom will attract the electron more than the other. Hence, chlorine attracts the electron a lot more than the calcium.A difference of more than 1.6 (or 1.7 depending on the source) implies that the electrons are so unevenly shared, that the bond is ionic, rather than polar covalent.

3 0
3 years ago
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