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Determine the equilibrium constant for t...

Determine the equilibrium constant for the reaction :
`Fe^(2+)(aq)+Ce^(4+)(aq)rarr Fe^(3+)(aq)+Ce^(3+)(aq)`
Givne : `E_(Ce^(4+)|Ce^(3+))^(@)=1.44V & E_(Fe^(3+)|Fe^(2+))^(@)=0.77V`.

Text Solution

Verified by Experts

`{:("Anode reaction : "Fe^(2+)(aq)" "rarr" "Fe^(3+)(Aq)+e),("Cathode reaction : "Ce^(4+)(aq)+e" "rarr" "Ce^(3+)(aq)+e),(bar(" Cell reaction : "Fe^(2+)(aq)+Ce^(4+)(aq)rarrFe^(3+)(aq)+Ce^(3+)(aq))):}`
`E_("cell")^(@)=E_("cathode")^(@)-E_("anode")^(@)=E_(Ce^(4+)|Ce^(3+))^(@)-E_(Fe^(3+)|Fe^(2+))^(@)=1.44-0.77=0.67V`
As 1 mol of electrons is involved in the cell reaction, n = 1
`therefore" "logK=(nFE^(@))/(2.303RT)=(1xx96500xx0.67)/(2.303xx8.314xx298)=11.33`
`therefore" "K=2.137xx10^(11)`
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