Question

for a cell reaction
2Fe3+ +2I- --------> 2Fe2+ +I2
E=0.24 V at 298K. The standard Gibbs energy of the cell reaction is? (F= 96500 C/mol)
a) 23.16kJ/mol.
B) -46.32kJ/mol
c) -23.16kJ/mol.
D) 46.32kJ/mol

Answer 1

Answer:

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Answer 2

Answer:

option B) -46.32 kJ/mol

Explanation:

Given :

For a cell reaction  $\rm 2Fe^{+3}+2I^{-} \longrightarrow 2Fe^{+2}+I_2$ , E = 0.24 V at 298 K

To find :

the standard Gibbs energy of the cell reaction

Solution :

The standard Gibbs energy of a cell reaction is given by,

 $\longmapsto \boxed{\sf \triangle G^{\circ}=-nFE^{\circ} _{cell}}$

where

n is the number of electrons transferred

Given cell reaction,

$\rm 2Fe^{+3}+2I^{-} \longrightarrow 2Fe^{+2}+I_2$

Oxidation half reaction :

$\rm 2I^{-} \longrightarrow I_2+2e^{-}$

Reduction half reaction :

$\rm 2Fe^{+3} +2e^{-} \longrightarrow 2Fe^{+2}$

⇒ The number of electrons transferred = 2

So,

 $\rm \triangle G^{\circ}=-nFE^{\circ} _{cell} \\\\ \rm \triangle G^{\circ}=-2 \times 96500 \times 0.24 \\\\ \rm \triangle G^{\circ}=-46320 \ J/mol \\\\ \underline{ \rm \triangle G^{\circ}=-46.32 \ kJ/mol}$