Respuesta :
Answer: The combination of element ad an ion that will react is [tex]Ni(s)\text{ and }Pt^{2+}(aq.)[/tex]
Explanation:
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.
[tex]X\rightarrow X^{n+}+ne^-[/tex]
Reduction reaction is defined as the reaction in which an atom gains electrons. The oxidation number of the atom gets reduced during this reaction.
[tex]X^{n+}+ne^-\rightarrow X[/tex]
The substance having highest positive [tex]E^o[/tex] 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 [tex]E^o_{cell}[/tex] of the reaction, we use the equation:
[tex]E^o_{cell}=E^o_{cathode}-E^o_{anode}[/tex] ......(1)
For the given options:
- Option 1: [tex]Sn(s)\text{ and }Mn^{2+}(aq.)[/tex]
Here, tin must undergo oxidation reaction and manganese undergo reduction reaction.
Oxidation half reaction: [tex]Sn(s)\rightarrow Sn^{2+}(aq.)+2e^-;E^o_{Sn^{2+}/Sn}=-0.14V[/tex]
Reduction half reaction: [tex]Mn^{2+}(aq.)+2e^-\rightarrow Mn(s);E^o_{Mn^{2+}/Mn}=-1.18V[/tex]
Putting values in equation 1, we get:
[tex]E^o_{cell}=-1.18-(-0.14)=-1.04V[/tex]
As, the standard potential is coming out to be negative, the given reaction will not take place.
- Option 2: [tex]Fe(s)\text{ and }Ca^{2+}(aq.)[/tex]
Here, iron must undergo oxidation reaction and calcium undergo reduction reaction.
Oxidation half reaction: [tex]Fe(s)\rightarrow Fe^{2+}(aq.)+2e^-;E^o_{Fe^{2+}/Fe}=-0.44V[/tex]
Reduction half reaction: [tex]Ca^{2+}(aq.)+2e^-\rightarrow Ca(s);E^o_{Ca^{2+}/Ca}=-2.87V[/tex]
Putting values in equation 1, we get:
[tex]E^o_{cell}=-2.87-(-0.44)=-2.43V[/tex]
As, the standard potential is coming out to be negative, the given reaction will not take place.
- Option 3: [tex]Ni(s)\text{ and }Pt^{2+}(aq.)[/tex]
Here, nickel must undergo oxidation reaction and platinum undergo reduction reaction.
Oxidation half reaction: [tex]Ni(s)\rightarrow Ni^{2+}(aq.)+2e^-;E^o_{Ni^{2+}/Ni}=-0.25V[/tex]
Reduction half reaction: [tex]Pt^{2+}(aq.)+2e^-\rightarrow Pt(s);E^o_{Pt^{2+}/Pt}=1.2V[/tex]
Putting values in equation 1, we get:
[tex]E^o_{cell}=1.2-(-0.25)=1.45V[/tex]
As, the standard potential is coming out to be positive, the given reaction will take place.
- Option 4: [tex]H_2(g)\text{ and }Na^{+}(aq.)[/tex]
Here, hydrogen must undergo oxidation reaction and sodium undergo reduction reaction.
Oxidation half reaction: [tex]H_2(g)\rightarrow 2H^{+}(aq.)+2e^-;E^o_{2H^{+}/H_2}=0V[/tex]
Reduction half reaction: [tex]Na^{+}(aq.)+e^-\rightarrow Na(s);E^o_{Na^{+}/Na}=-0.27V[/tex]
Putting values in equation 1, we get:
[tex]E^o_{cell}=-0.27-(-0)=-0.27V[/tex]
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 [tex]Ni(s)\text{ and }Pt^{2+}(aq.)[/tex]
