1.

For a cell involving two electron changes, E_("cell")^(@) =0.3V at 25^(@)C.The cell equilibrium constant of the reaction is

Answer»

`10^(-10)`
`3xx10^(-2)`
10
`10^(10)`

Solution :`log_(10)K=(-DELTAG^@)/(2.303RT)=(nFE_(cell)^@)/(2.303RT)` (at `25^@C`)
`log_10K=(nE_(cell)^@)/(0.0591)=(2xx0.3)/(0.0591)=10impliesK=10^10`


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