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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 |
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Answer» Solution :`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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