Question

Use appropriate data to calculate δg∘ for the reaction. 2mno−4(aq)+cd(s)→2mno2−4(aq)+cd2+(aq) express your answer to three significant figures and include the appropriate units.

Answer 1

∆G° = -185 kJ

Step-by-step explanation:

First, use the half-reaction method:

OX: Cd ⟶ Cd2+ + 2e- where E° = -0.40 V (anode)

RED: 2MnO-4 + e- ⟶ 2MnO2-4 where E° = 0.56 V (cathode)

Balance the chemical equation with the correct stoichiometric coefficients:

2 × (2MnO-4 + e- ⟶ 2MnO2-4) = 4MnO-4 + 2e- ⟶ 4MnO2-4

1 × (Cd ⟶ Cd2+ + 2e-) = Cd ⟶ Cd2+ + 2e-

Cancel out e- on both sides to get:

4MnO-4 + Cd ⟶ 4MnO2-4 + Cd2+

Using this balanced equation, we can determine:

number of moles of electrons exchanged in the cell reaction, n = 2

E°cell = E°cathode - E°anode = 0.56 - (-0.40) = 0.96 V

F, Faraday's constant: 96485 / mol e-

∆G° = -nFE°cell = -(2)(96485)(0.96)

∆G° = -185251.2 J = -185 kJ

Answer 2

Reaction at anode: Cd → Cd2+ + 2e-, Eo = -0.403 v
Reaction at cathode: 2MnO4- + 2e- → MnO4^2-, Eo = 1.5 v

Net reaction: Cd + 2MnO4- ⇆ Cd2+ + MnO4^2-

Net cell representation: Cd/Cd2+// MnO4-/MnO4^2-

Now, standard EMF of cell = E(o)cell = Er - El
= 1.5 - (-0.403)
= 1.903 v

Now, ΔGo = -nFE(o)cell
where n = number of electrons = 2
F = faraday's constant = 96500

ΔGo = - 2 X 96500 X 1.903
= 367.27 kJ