Question

Identify which redox reactions occur spontaneously in the forward direction. Check all that apply.

a.Fe(s)+Mn2+(aq)→Fe2+(aq)+Mn(s)
b. 2Ag+(aq)+Fe(s)→2Ag(s)+Fe2+(aq)
c. Mg2+(aq)+Zn(s)→Mg(s)+Zn2+(aq)
d. 2Al(s)+3Pb2+(aq)→2Al3+(aq)+3Pb(s)

Answer 1

For a: The reaction is not spontaneous.

For b: The reaction is spontaneous.

For c: The reaction is not spontaneous.

For d: The reaction is spontaneous.

Step-by-step explanation:

For the reaction to be spontaneous, the Gibbs free energy of the reaction must come out to be negative.

Relationship between standard Gibbs free energy and standard electrode potential follows:

`\Delta G^o=-nFE^o_(cell)`

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)

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

• For a:

The chemical reaction follows:

`Fe(s)+Mn^(2+)(aq.)\rightarrow Fe^(2+)(aq.)+Mn(s)`

We know that:

`E^o_(Fe^(2+)/Fe)=-0.44V\\E^o_(Mn^(2+)/Mn)=-1.18V`

Calculating the
`E^o_(cell)` using equation 1, we get:

`E^o_(cell)=-1.18-(-0.44)=-0.74V`

As, the standard electrode potential is coming out to be negative. So, the reaction is not spontaneous.

• For b:

The chemical reaction follows:

`Fe(s)+2Ag^(+)(aq.)\rightarrow Fe^(2+)(aq.)+2Ag(s)`

We know that:

`E^o_(Fe^(2+)/Fe)=-0.44V\\E^o_(Ag^(+)/Ag)=0.80V`

Calculating the
`E^o_(cell)` using equation 1, we get:

`E^o_(cell)=0.80-(-0.44)=1.24V`

As, the standard electrode potential is coming out to be positive. So, the reaction is spontaneous.

• For c:

The chemical reaction follows:

`Zn(s)+Mg^(2+)(aq.)\rightarrow Zn^(2+)(aq.)+Mg(s)`

We know that:

`E^o_(Zn^(2+)/Zn)=-0.76V\\E^o_(Mg^(2+)/Mg)=-2.37V`

Calculating the
`E^o_(cell)` using equation 1, we get:

`E^o_(cell)=-2.37-(-0.76)=-1.61V`

As, the standard electrode potential is coming out to be negative. So, the reaction is not spontaneous.

• For d:

The chemical reaction follows:

`2Al(s)+3Pb^(2+)(aq.)\rightarrow 2Al^(3+)(aq.)+3Pb(s)`

We know that:

`E^o_(Al^(3+)/Al)=-1.66V\\E^o_(Pb^(2+)/Pb)=-0.13V`

Calculating the
`E^o_(cell)` using equation 1, we get:

`E^o_(cell)=-0.13-(-1.66)=1.53V`

As, the standard electrode potential is coming out to be positive. So, the reaction is spontaneous.