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MAXImum [283]
3 years ago
9

Which combination of an element and an ion will react? View Available Hint(s) Which combination of an element and an ion will re

act? Sn(s) and Mn2+(aq) will react. Fe(s) and Ca2+(aq) will react. Ni(s) and Pt2+(aq) will react. H2(g) and Na+(aq) will react.
Chemistry
1 answer:
Aleks04 [339]3 years ago
8 0

<u>Answer:</u> The combination of element ad an ion that will react is Ni(s)\text{ and }Pt^{2+}(aq.)

<u>Explanation:</u>

Oxidation reaction is defined as the reaction in which an atom looses its electrons. The oxidation number of the atom gets increased during this reaction.

X\rightarrow X^{n+}+ne^-

Reduction reaction is defined as the reaction in which an atom gains electrons. The oxidation number of the atom gets reduced during this reaction.

X^{n+}+ne^-\rightarrow X

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

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

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

E^o_{cell}=E^o_{cathode}-E^o_{anode}       ......(1)

For the given options:

  • <u>Option 1:</u>  Sn(s)\text{ and }Mn^{2+}(aq.)

Here, tin must undergo oxidation reaction and manganese undergo reduction reaction.

<u>Oxidation half reaction:</u>  Sn(s)\rightarrow Sn^{2+}(aq.)+2e^-;E^o_{Sn^{2+}/Sn}=-0.14V

<u>Reduction half reaction:</u>  Mn^{2+}(aq.)+2e^-\rightarrow Mn(s);E^o_{Mn^{2+}/Mn}=-1.18V

Putting values in equation 1, we get:

E^o_{cell}=-1.18-(-0.14)=-1.04V

As, the standard potential is coming out to be negative, the given reaction will not take place.

  • <u>Option 2:</u>  Fe(s)\text{ and }Ca^{2+}(aq.)

Here, iron must undergo oxidation reaction and calcium undergo reduction reaction.

<u>Oxidation half reaction:</u>  Fe(s)\rightarrow Fe^{2+}(aq.)+2e^-;E^o_{Fe^{2+}/Fe}=-0.44V

<u>Reduction half reaction:</u>  Ca^{2+}(aq.)+2e^-\rightarrow Ca(s);E^o_{Ca^{2+}/Ca}=-2.87V

Putting values in equation 1, we get:

E^o_{cell}=-2.87-(-0.44)=-2.43V

As, the standard potential is coming out to be negative, the given reaction will not take place.

  • <u>Option 3:</u>  Ni(s)\text{ and }Pt^{2+}(aq.)

Here, nickel must undergo oxidation reaction and platinum undergo reduction reaction.

<u>Oxidation half reaction:</u>  Ni(s)\rightarrow Ni^{2+}(aq.)+2e^-;E^o_{Ni^{2+}/Ni}=-0.25V

<u>Reduction half reaction:</u>  Pt^{2+}(aq.)+2e^-\rightarrow Pt(s);E^o_{Pt^{2+}/Pt}=1.2V

Putting values in equation 1, we get:

E^o_{cell}=1.2-(-0.25)=1.45V

As, the standard potential is coming out to be positive, the given reaction will take place.

  • <u>Option 4:</u>  H_2(g)\text{ and }Na^{+}(aq.)

Here, hydrogen must undergo oxidation reaction and sodium undergo reduction reaction.

<u>Oxidation half reaction:</u>  H_2(g)\rightarrow 2H^{+}(aq.)+2e^-;E^o_{2H^{+}/H_2}=0V

<u>Reduction half reaction:</u>  Na^{+}(aq.)+e^-\rightarrow Na(s);E^o_{Na^{+}/Na}=-0.27V

Putting values in equation 1, we get:

E^o_{cell}=-0.27-(-0)=-0.27V

As, the standard potential is coming out to be negative, the given reaction will not take place.

Hence, the combination of element ad an ion that will react is Ni(s)\text{ and }Pt^{2+}(aq.)

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