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vladimir2022 [97]
3 years ago
13

Oxygen has an electronegativity of 3.5, and carbon has an electronegativity of 2.5. How is charge distributed on an oxygen atom

when it bonds with a
carbon atom?
The oxygen atom has a slightly negative charge.
The oxygen atom has a charge of 1-
The oxygen atom has a charge of 1+.
The oxygen atom has a slightly positive charge.
Chemistry
1 answer:
Aleksandr [31]3 years ago
4 0

Answer:

The oxygen atom has a slightly negative charge.

Explanation:

3.5 - 2.5 = 1

For ionic bond electronegativity should be more than 1.6.

So, O and C do not have whole charge. Electronegativity of O more than C, so electrons slightly moved to the oxygen side.

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<u>Answer:</u> The correct answer is 3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.);E^o_{cell}=+1.53V

<u>Explanation:</u>

We are given:

Cr^{3+}(aq.)+3e^-\rightarrow Cr(s);E^o=-0.73V\\\\Ag^+(aq.)+e^-\rightarrow Ag(s);E^o=+0.80V

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

Chromium will undergo oxidation reaction and will get oxidized.

The half reactions for the above cell is:

Oxidation half reaction: Cr(s)\rightarrow Cr^{3+}+3e^-;E^o_{Cr^{3+}/Cr}=-0.73V

Reduction half reaction: Ag^{+}+e^-\rightarrow Ag(s);E^o_{Ag^{+}/Ag}=0.80V       ( × 3)

Net equation:  3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.)

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}=0.80-(-0.73)=1.53V

Hence, the correct answer is 3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.);E^o_{cell}=+1.53V

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