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Find the equilibrium constant for the re...

Find the equilibrium constant for the reaction,
`In^(2+)+Cu^(2+) rarr In^(3+)+Cu^(+)`, at `298 K`
Given: `E_(Cu^(2+)//Cu^(+))^(@) = 0.15 V , E_(In^(3+)//In^(+))^(@) = -0.42 V`
and `E_(In^(2+)//In^(+))^(@) = -0.40V`

Text Solution

Verified by Experts

`Cu^(2+) + e rarr Cu^(+) E_(1)^(@) = 0.15 V`...(1)
`In^(3+) + 2e rarr In^(+) E_(2)^(@) = -0.42 V` ...(2)
`In^(2+) + 2e rarr In^(+) E_(3)^(@) = -0.40 V` ...(3)
By eqs. `(2)` and (3), a third half cell reaction can be given as
`In^(3) ++ e rarr In^(2+)`
`1 xx F xx E_(4)^(@) = E_(2)^(@) xx 2 F - e_(3)^(@) xx 1 xx F`
or `E_(4)^(@) = 2 xx (-0.42) - (-0.40) xx 1 = -0.44 V`
For two half reactions, the net redox change as asked is
`Cu^(2+) + In^(+) rarr Cu^(+) + In^(3+)`
`E_(cell)^(@) = E_(1_(RP_(Cu^(2+)//Cu^(+)))^(@)) + E_(4OP_(IN^(2+)//IN^(3+)))^(@)`
`[E_(OP_(In^(2+)//In^(3+))) = -E_(RP)^(@) = + 0.44]`
`= 0.15 + 0.44 = 0.59V`
Also, `E_(cell)^(@) = (0.059)/(1)logK_(c)`
`0.59 = (0.059)/(1)logK_(c)`
`:. K_(c) = 10^(10)`
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