Question

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

Answer 1

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

Step-by-step 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.

`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:

• Option 1:
  `Sn(s)\text{ and }Mn^(2+)(aq.)`

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

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

Reduction half reaction:
`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.

• Option 2:
  `Fe(s)\text{ and }Ca^(2+)(aq.)`

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

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

Reduction half reaction:
`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.

• Option 3:
  `Ni(s)\text{ and }Pt^(2+)(aq.)`

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

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

Reduction half reaction:
`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.

• Option 4:
  `H_2(g)\text{ and }Na^(+)(aq.)`

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

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

Reduction half reaction:
`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.)`