Which statement about van der Waals forces is true?
a)When the forces are weaker, a substance will have higher volatility.
b)When the forces are stronger, a substance will have lower viscosity.
c)When the forces are weaker, the boiling point of a substance will be higher.
d)When the forces are stronger, the melting point of a substance will be lower.

Answer: (a) [tex]\Delta G^0=-432.25kJ[/tex] , (b) [tex]\Delta G^0=55.96kJ[/tex] and (c)  [tex]\Delta G^0=-171.74kJ[/tex]

Explanation: (a) Oxidation half reaction for the given equation is:

[tex]Mg(s)\rightarrow Mg^2^+(aq)+2e^-E^0=2.37V[/tex]

The reduction half equation is:

[tex]Pb^2^+(aq)+2e^-\rightarrow Pb(s)E^0=-0.13V[/tex]

[tex]E^0_c_e_l_l=E^0_r_e_d_u_c_t_i_o_n+E^0_o_x_i_d_a_t_i_o_n[/tex]

[tex]E^0_c_e_l_l=-0.13V+2.37V[/tex]

[tex]E^0_c_e_l_l=2.24V[/tex]

[tex]\Delta G^0=-nFE^0_c_e_l_l[/tex]

where n is the number of moles of electrons transferred and F is faraday constant.

2 moles of electrons are transferred in the cell reaction which is also clear from both the half equations.

[tex]\Delta G^0=-2mol*96485\frac{C}{mol}*2.44V[/tex]

[tex]\Delta G^0=-432252.8J[/tex]

or [tex]\Delta G^0=-432.25kJ[/tex]

(b) Oxidation half reaction for the given equation is:

[tex]2Cl^-(aq)\rightarrow Cl_2(g)+2e^-E^0=-1.36V[/tex]

Reduction half equation is:

[tex]Br_2(l)+2e^-\rightarrow 2Br^-E^0=1.07V[/tex]

[tex]E^0_c_e_l_l=1.07V+(-1.36V)[/tex]

[tex]E^0_c_e_l_l=1.07V-1.36V[/tex]

[tex]E^0_c_e_l_l=-0.29V[/tex]

Now, we can calculate the [tex]\Delta G^0[/tex] same as we did for part a.

[tex]\Delta G^0=-2mol*96485\frac{C}{mol}*(-0.29V)[/tex]

[tex]\Delta G^0=55961.3J[/tex]

or [tex]\Delta G^0=55.96kJ[/tex]

(c) Oxidation half reaction for the given equation is:

[tex]Cu(s)\rightarrow Cu^2^+(aq)+2e^-E^0=-0.34V[/tex]

reduction half equation is:

[tex]MnO_2(s)+4H^+(aq)+2e^-\rightarrow Mn^2^+(aq)+2H_2O(l)E^0=1.23V[/tex]

[tex]E^0_c_e_l_l=1.23V+(-0.34V)[/tex]

[tex]E^0_c_e_l_l=0.89V[/tex]

[tex]\Delta G^0=-2mol*96485\frac{C}{mol}*0.89V[/tex]

[tex]\Delta G^0=-171743.3J[/tex]

or [tex]\Delta G^0=-171.74kJ[/tex]

Final answer:

To calculate the standard Gibbs free energy change (∆G°) for each reaction at 25°C in kJ, we can use tabulated electrode potentials. By writing half-cell reactions and summing the electrode potentials, we can determine the overall reaction's standard potential (E°). Then, using the formula ∆G° = -nFE°, we can calculate the standard Gibbs free energy change.

Explanation:

Use tabulated electrode potentials to calculate ∆G° rxn for each reaction at 25 °C in kJ.

a) Pb2+(aq) + Mg(s) ➝ Pb(s) + Mg2+(aq)First, we need to write half-cell reactions for the given equation:Pb2+ + 2e- ➝ Pb (E° = -0.126 V)Mg2+ + 2e- ➝ Mg (E° = -2.37 V)Next, we can sum up the electrode potentials to calculate the overall reaction's standard potential:E° rxn = E°(Pb) - E°(Mg)E° rxn = -0.126 V - (-2.37 V) = 2.244 VFinally, we can use the formula ∆G° = -nFE° to calculate the standard Gibbs free energy change:∆G° = -2F(2.244 V)

b) Br2(l) + 2 Cl-(aq) ➝ 2 Br-(aq) + Cl2( g)First, we need to write half-cell reactions for the given equation:Br2 + 2e- ➝ 2Br- (E° = 1.07 V)Cl2 + 2e- ➝ 2Cl- (E° = 1.36 V)Next, we can sum up the electrode potentials to calculate the overall reaction's standard potential:E° rxn = E°(2Br-) - E°(2Cl-)E° rxn = 2(1.07 V) - 2(1.36 V) = -0.640 VFinally, we can use the formula ∆G° = -nFE° to calculate the standard Gibbs free energy change:∆G° = -2F(-0.640 V)

c) MnO2(s) + 4 H+(aq) + Cu(s) ➝ Mn2+(aq) + 2 H2O(l) + Cu2+(aq)First, we need to write half-cell reactions for the given equation:MnO2 + 4H+ + 2e- ➝ Mn2+ + 2H2O (E° = 1.23 V)Cu2+ + 2e- ➝ Cu (E° = 0.34 V)Next, we can sum up the electrode potentials to calculate the overall reaction's standard potential:E° rxn = E°(Mn2+) - E°(Cu)E° rxn = 1.23 V - 0.34 V = 0.89 VFinally, we can use the formula ∆G° = -nFE° to calculate the standard Gibbs free energy change:∆G° = -2F(0.89 V)

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