Question

Based on the reduction potentials listed in the textbook appendix, which of the following redox reactions do you expect to occur spontaneously?  

W. 2Al(s)+3Pb2+ (aq) → 2Al3+ (aq)+3Pb(s)  
X. Fe(s)+Cr3+ (aq) → Fe3+ (aq)+Cr(s)  
Y. Ca2+ (aq)+Zn(s) → Ca(s)+Zn2+(aq)  
Z. 2Cu+(aq)+Co(s) → 2Cu(s)+Co2+ (s)

a. W only
b. X, Y and Z
c. Y only
d. X and Z
e. Z only
f. X and Y
g. W, X and Z
h. X and Y
i.W and Z

Answer 1

Answer: The redox reactions that occur spontaneously are Reaction W and Reaction Z.

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 reaction W:

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.

• For reaction X:

The chemical reaction follows:

`Fe(s)+Cr^(3+)(aq.)\rightarrow Fe^(3+)(aq.)+Cr(s)`

We know that:

`E^o_(Fe^(3+)/Fe)=0.77V\\E^o_(Cr^(3+)/Cr)=-0.74V`

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

`E^o_(cell)=-0.74-(0.77)=-1.51V`

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

• For reaction Y:

The chemical reaction follows:

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

We know that:

`E^o_(Ca^(2+)/Ca)=-2.87V\\E^o_(Zn^(2+)/Zn)=-0.76V`

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

`E^o_(cell)=-2.87-(-0.76)=-2.11V`

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

• For reaction Z:

The chemical reaction follows:

`Co(s)+2Cu^(+)(aq.)\rightarrow Co^(2+)(aq.)+2Cu(s)`

We know that:

`E^o_(Cu^(+)/Cu)=0.34V\\E^o_(Co^(2+)/Co)=-0.28V`

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

`E^o_(cell)=0.34-(-0.28)=0.62V`

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

Hence, the redox reactions that occur spontaneously are Reaction W and Reaction Z.