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Hatshy [7]
3 years ago
5

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

Chemistry
1 answer:
lidiya [134]3 years ago
7 0

<u>Answer:</u>

<u>For a:</u> The reaction is not spontaneous.

<u>For b:</u> The reaction is spontaneous.

<u>For c:</u> The reaction is not spontaneous.

<u>For d:</u> The reaction is spontaneous.

<u>Explanation:</u>

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.

  • <u>For a:</u>

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.

  • <u>For b:</u>

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.

  • <u>For c:</u>

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.

  • <u>For d:</u>

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.

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0.55M=\frac{0.2216}{\text{Volume of solution}}\\\\\text{Volume of solution}=\frac{0.2216}{0.55}=0.4029L=402.9mL

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