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

Answer:

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.

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.