Calculate Standard Gibbs Energy Change Of the Cell Reaction (using log table) -Electrochemistry

The cell in which the following reaction occurs:

2Fe3+(aq) + 2I-(aq)→ 2Fe2+(aq) + I2(s)

Has Eocell = 0.236 V at 298 K. Calculate the standard Gibbs energy of the cell reaction (Given: 1F = 96,500 C/mol) (CBSE 2017)

The standard Gibbs free energy (∆G) can be obtained from the equation,

                                                                                             ∆G = -nFEcell

Where,

n is the number of electrons taking part in the reaction

Ecell is the EMF of the cell Electrode potential

F is Faraday’s Constant (1F= 96500 C/mol)

 

On Splitting the cell reaction into two half reactions, we get,

Reduction half-reaction

Fe3+  + e-  → Fe2+         (here n=1)

So, for 2 moles of Fe3+

2Fe3+ + 2e- → 2Fe2+      (n =2)  

Similarly, for the Oxidation Half-reaction

I- → ½ I2 + e-        (n=1)

For 2 moles of I-

2I- → I2 + 2e-      (n=2)

 

Using the obtained values in the equation, we get,

-∆G = nFE = 2 x 96500 x 0.236 = -45548 kJ/mol

 

Solving the equation using the log table,

Log x= Log (2 x 96500 x 0.236) = Log 2 + Log 96500 + log (2.36 x 10-1)

= Log 2 + Log 96500 + log 2.36 + log 10-1 = 0.3010 + 4.9845 + 0.3729 -1

Separating the characteristic and the mantissa and solving further,

= 0.3010 + 4 + 0.9845 + 0.3729 -1

= 3 + 1.6584

Log (x) = 4.6584

x= Antilog (4.6584) = 45540

-∆G= 45540 kJ/mol or ∆G=  -45540 kJ/mol 

The reaction is exothermic and spontaneous.

 

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