1.

Given E_(Cr^(3+)//Cr)^(@) = -0.72 V, E_(Fe^(2+)//Fe)^(@) = -0.439 V. The value of standard electrode potential for the change, Fe_(aq.)^(3+) + e^(-) rarr Fe_(aq.)^(2+) will be:

Answer»

`-0.072 V`
`0.385 V`
`0.770 V`
`-0.270 V`

SOLUTION :As `E_(Cr^(3+)//Cr)^(@) = -0.72 V` and
`E_(Fe^(2+)//Fe)^(@) = -0.42 V`
`2Cr + 3FE^(2+) rarr 3Fe + 2Cr^(3+)`
Six ELECTRONS `(n = 6)` are used in redox CHANGE.
`E_(cell) = E_(cell)^(@) - (0.0591)/(n)log'([Cr^(3+)]^(2))/([Fe^(2+)]^(3))`
`= (-0.42 + 0.72) - (0.0591)/(6)log'((0.01)^(2))/((0.01)^(3))`
`= 0.30 - (0.0591)/(6)log.((0.1)^(2))/((0.01)^(3))`
`= 0.30 - (0.0591)/(6)log10^(4)`
`E_(cell) = 0.2606 V`


Discussion

No Comment Found

Related InterviewSolutions