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3 years ago

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It turns out that the van der Waals constant bb equals four times the total volume actually occupied by the molecules of a mole

of gas. Calculate the fraction of the volume in a container actually occupied by ArAr atoms (bb

1 answer:

liberstina [14]3 years ago

7 0

Answer:

Explanation:

nommidealksjdo

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

Aleks04 [339]

<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 $Ni(s)\text{ and }Pt^{2+}(aq.)$

8 0

3 years ago