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s344n2d4d5 [400]
3 years ago
12

In salt, what is the nature of the bond between sodium and chlorine? in salt, what is the nature of the bond between sodium and

chlorine? ionic nonpolar covalent hydrogen hydrophobic polar covalent
Chemistry
1 answer:
kifflom [539]3 years ago
4 0

A chemical bond that involves the transfer of electrons from a metal to a nonmetal and thus forms ions and the atoms in the molecule are attracted towards each other through electrostatic force of attraction are said be ionic bond. In short, the bond formed between the two oppositely charged ions.

In salt, the compound formed between sodium and chlorine is sodium chloride having molecular formula NaCl where sodium is a metal and chlorine is the non-metal. The formation of the compound takes place by transfer of an electron from sodium thus forming a cation of formula Na^{+} to chlorine which gains electron and thus results in an anion formation having formula Cl^{-}. The force of attraction between these oppositely charged ions that helds them together and results in formation of NaCl compound is electrostatic force of attraction.

Hence in salt, the nature of bond between sodium and chlorine is ionic.

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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.

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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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