Answer:-
for the reaction is -58 kJ/mol.
Solution:- Oxidation half reaction taking place at the anode is--
= 0.14 V
Reduction half equation taking place at the cathode is---
= -0.04V
for the cell = 
for the cell = -0.04V + 0.14V = 0.10V
Now we could easily calculate
by using the below formula--
= -nF
where n is the number of electrons transferred in the over all reaction. Looking at two half equations the value of n is 6.
F is Farady constant and it's value is 96500 C/mol.
Plugging in the values in the formula...
= -(6)(96500)(0.10)
= -57900 J
Since, 1000 J = 1 kJ
So,
= -58 kJ/mol
Answer:
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Explanation:
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Explanation: