Given, standard electrode potentials `Fe^(2+) +2e^(-) rarr Fe, E^(@)=-0.440 V` `Fe^(3+)+3e^(-) rarr Fe, E^(@)=- 0.036 V` The standarde potential `(E^(@))` for `Fe^(2+)+e^(-) rarr Fe^(2+)`, isA. `+0.772 V`B. `-0.772 V`C. `+0.417 V`D. `-0.417 V`

Given, standard electrode potentials `Fe^(2+) +2e^(-) rarr Fe, E^(@)=-

1.	Given, standard electrode potentials `Fe^(2+) +2e^(-) rarr Fe, E^(@)=-0.440 V` `Fe^(3+)+3e^(-) rarr Fe, E^(@)=- 0.036 V` The standarde potential `(E^(@))` for `Fe^(2+)+e^(-) rarr Fe^(2+)`, isA. `+0.772 V`B. `-0.772 V`C. `+0.417 V`D. `-0.417 V`
Answer» Correct Answer - A `Fe^(2+)+2e^(-) rarr Fe, DeltaG^(@)=-nFE^(@)` …(i) `DeltaG^(@)=-2xxFxx(-0.440V)=0.880 F` `Fe^(3+)+3e^(-) rarr Fe` …(ii) `DeltaG^(@)=-3xxFxx(-0.036)` `=0.108 F` On subtracting Eq. (ii) from Eq. (i) we get `Fe^(3+) +e^(-) rarr Fe^(2+)` `DeltaG^(@)=0.108 F-0.880 F=-0.772 F` `E^(@)=(-DeltaG^(@))/(nF)=((-0.772 F))/(1xxF)= +0.772 V`

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Given, standard electrode potentials `Fe^(2+) +2e^(-) rarr Fe, E^(@)=-0.440 V` `Fe^(3+)+3e^(-) rarr Fe, E^(@)=- 0.036 V` The standarde potential `(E^(@))` for `Fe^(2+)+e^(-) rarr Fe^(2+)`, isA. `+0.772 V`B. `-0.772 V`C. `+0.417 V`D. `-0.417 V`

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