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The cell in which the following reaction...

The cell in which the following reaction occurs, `2Fe^(3+) (aq) + 2I^(-)(aq) rarr 2Fe^(2+) (aq) + I_(2)(s)`
has `E_("cell")^(@) = 0.236V` at 298K , calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.

Text Solution

Verified by Experts

Two half reactions for the given redox reaction may be written as`:`
`2Fe^(3+) (aq) + 2e^(-) rarr 2Fe^(2+) ( aq)`
`2I^(-) rarr I_(2) + 2e^(-)`
2 moles of electrons are involved in the reaction, so n =2.
`Delta_( r ) G^(@) = -n FE_("cell")^(@)`
`= -( 2 "mol") xx( 96500 C "mol"^(-1)) xx ( -.236 V ) `
`= - 45548 CV = - 45548J`
`Delta _(r )G^(@) = - 45.55kJ`
`log K_(C ) = - ( Delta G^(@))/( 2.303 R T )`
`= - ( ( - 45.55kJ ))/( 2.303 xx ( 8.314 xx 10^(-3)kJ K^(-1)) xx ( 298 K))`
= 7.983
`K_( C ) =` antilog ( 7.983) `= 9.616 xx 10^(7)`
`K_( C ) = 9.616 xx 10^(7)`
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