Question

Using standard reduction potentials from the ALEKS Data tab, calculate the standard reaction free energy AG' for the following redox reaction. Round your answer to 3 significant digits. 2Mn2+ (aq) +8H20 (1) + 5Br2 (1) ► 2 MnO2 (aq) +16H* (aq) +10Br" (aq) kJ DxO Х 5 ?

Answer 1

In the overall reaction Mn+2 is getting oxidised to MnO4- and Br2 is reduced to Br-.

Half reaction:

At anode (oxidation) :

Mn+2 + 4 H2O $\rightarrow$ MnO4- + 8 H​​​​​​+ + 5 e- E° = - 1.51 V

At cathode ( reduction) :

Br​​​​​​2 + 2 e- $\rightarrow$ 2 Br-    E° = 1.06 V

To make overall reaction, multiply reaction at anode by 2 and reaction at cathode by 5

2 Mn+2 + 8 H2O $\rightarrow$ 2 MnO4- + 16 H+ + 10 e- E° = - 1.51 V

5 Br2 + 10 e- $\rightarrow$ 10 Br- E° = 1.06 V

So E°cell = -1. 51 + 1.06 = - 0.45 V

n ( number of moles) = 10 ( as 10 e- are involved)

∆G° = - nF E°cell

1 F = 96500 C

∆G° = - (10 mol) x 96500 C (-0.45 V)

= 434.25 KJ/mol

As ∆G° is needed in KJ only

So, ∆G° = - F E°cell

= 43.4 KJ