Given the `E_(Fe^(3+)//Fe)^(@)` and `E_(Fe^(2+)//Fe)^(@)` are `-0.36 V` and `0.-439 V`, respectively. The value of `E_(Fe^(3+)//Fe^(2+)) ^(@)` would be :A. `(-0.36 - 0.439)V`B. `[3(-0.36) + 2(-0.439)]V`C. `(0.36 + 0.439)V`D. `[3(-0.36) - 2(-0.439)]V`

Given the `E_(Fe^(3+)//Fe)^(@)` and `E_(Fe^(2+)//Fe)^(@)` are `-0.36 V

1.	Given the `E_(Fe^(3+)//Fe)^(@)` and `E_(Fe^(2+)//Fe)^(@)` are `-0.36 V` and `0.-439 V`, respectively. The value of `E_(Fe^(3+)//Fe^(2+)) ^(@)` would be :A. `(-0.36 - 0.439)V`B. `[3(-0.36) + 2(-0.439)]V`C. `(0.36 + 0.439)V`D. `[3(-0.36) - 2(-0.439)]V`
Answer» Correct Answer - D `{:(Fe^(3+)+3e rarr Fe l-Delta G_(1)^(@)=-0.36 xxF xx3),(underset(-)(Fe^(3+))+ underset(-)(2e) rarr underset(-)(Fe),-DeltaG_(2)^(@)=underset(+)(-)0.439 xxFxx2),(bar(Fe^(3)+e rarr Fe^(2+),-Delta G^(@)=[3 xx(0.36 F)+2xx(0.439 F)])),( :. +1xxF xxE^(2)=[3xx(-0.36F)+2xx (0.439F)]),(or E^(@)=[3xx(-0.36)+2xx0.439]V):}` `:. +1 xx F xx E^(@) = [3 xx (-0.36F) + 2 xx (0.439F)]` ltbRgt or ` E^(@) = [3 xx (-0.36) + 2 xx 0.439]V`

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Given the `E_(Fe^(3+)//Fe)^(@)` and `E_(Fe^(2+)//Fe)^(@)` are `-0.36 V` and `0.-439 V`, respectively. The value of `E_(Fe^(3+)//Fe^(2+)) ^(@)` would be :A. `(-0.36 - 0.439)V`B. `[3(-0.36) + 2(-0.439)]V`C. `(0.36 + 0.439)V`D. `[3(-0.36) - 2(-0.439)]V`

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