Question

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)
(b) Br2(l) + 2 Cl-(aq) ➝ 2 Br-(aq) + Cl2( g)
(c) MnO2(s) + 4 H+(aq) + Cu(s) ➝ Mn2+(aq) + 2 H2O(l) + Cu2+(aq)

Answer 1

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

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

$Mg(s)\rightarrow Mg^2^+(aq)+2e^-E^0=2.37V$

The reduction half equation is:

$Pb^2^+(aq)+2e^-\rightarrow Pb(s)E^0=-0.13V$

$E^0_c_e_l_l=E^0_r_e_d_u_c_t_i_o_n+E^0_o_x_i_d_a_t_i_o_n$

$E^0_c_e_l_l=-0.13V+2.37V$

$E^0_c_e_l_l=2.24V$

$\Delta G^0=-nFE^0_c_e_l_l$

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.

$\Delta G^0=-2mol*96485\frac{C}{mol}*2.44V$

$\Delta G^0=-432252.8J$

or $\Delta G^0=-432.25kJ$

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

$2Cl^-(aq)\rightarrow Cl_2(g)+2e^-E^0=-1.36V$

Reduction half equation is:

$Br_2(l)+2e^-\rightarrow 2Br^-E^0=1.07V$

$E^0_c_e_l_l=1.07V+(-1.36V)$

$E^0_c_e_l_l=1.07V-1.36V$

$E^0_c_e_l_l=-0.29V$

Now, we can calculate the $\Delta G^0$ same as we did for part a.

$\Delta G^0=-2mol*96485\frac{C}{mol}*(-0.29V)$

$\Delta G^0=55961.3J$

or $\Delta G^0=55.96kJ$

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

$Cu(s)\rightarrow Cu^2^+(aq)+2e^-E^0=-0.34V$

reduction half equation is:

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

$E^0_c_e_l_l=1.23V+(-0.34V)$

$E^0_c_e_l_l=0.89V$

$\Delta G^0=-2mol*96485\frac{C}{mol}*0.89V$

$\Delta G^0=-171743.3J$

or $\Delta G^0=-171.74kJ$