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fgiga [73]
2 years ago
15

The Lewis structure for a chlorate ion, ClO3-, should show ____ single bond(s), ____ double bond(s), and ____ lone pair(s).

Chemistry
2 answers:
Yuliya22 [10]2 years ago
7 0

<u>Answer:</u> A molecule of ClO_3^- has 1 single bond, 2 double bonds and 8 lone pairs.

<u>Explanation:</u>

Lewis-dot structure is defined as the structure which represents the number of valence electrons around the atoms. The electrons are represented as dots.

This structure helps in the determination of bonding electrons and non-bonding electrons.

For the given compound ClO_3^-

As we know that chlorine has '7' valence electrons and oxygen has '6' valence electron.  The negative charge gets add up in the total number of valence electrons.

Total number of valence electrons in ClO_3^- = 7 + 3(6) + 1 = 26

In ClO_3^-, two electrons of 2 oxygen atoms combine with 2  electrons of chlorine and 1 electron of one oxygen atom combines with 1 electron of chlorine atom.

A lone pair remains on the central atom which is chlorine.

Hence, a molecule of ClO_3^- has 1 single bond, 2 double bonds and 8 lone pairs.

eduard2 years ago
3 0
1 single bond, 2 double bond 2 lone pairs
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Answer:

1.18 V

Explanation:

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Al(s)/Al^{3+}(0.10M)||Fe^{2+}(0.020M)/Fe(s)

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Oxidation half reaction: Al(s)\rightarrow Al^{3+}(0.10M)+2e^-;E^o_{Al^{3+}/Al}=-1.66V

Reduction half reaction: Fe^{2+}(0.020M)+2e^-\rightarrow Fe(s);E^o_{Fe^{2+}/Fe}=-0.45V

Multiply Oxidation half reaction by 2 and Reduction half reaction by 3

Net reaction: 2Al(s)+3Fe^{2+}(0.020M)\rightarrow 2Al^{3+}(0.10M)+3Fe(s)

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

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

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

Putting values in above equation, we get:

E^o_{cell}=-0.45-(-1.66)=1.21V

To calculate the EMF of the cell, we use the Nernst equation, which is:

E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Al^{3+}]^2}{[Fe^{2+}]^3}

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E_{cell} = electrode potential of the cell = ?V

E^o_{cell} = standard electrode potential of the cell = +1.21 V

n = number of electrons exchanged = 6

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E_{cell}=1.21-\frac{0.059}{6}\times \log(\frac{0.10^2}{0.020^3})\\\\E_{cell}=1.18V

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