On the basis of the folwing `E^@` values, the strongest oxidizing agent is `[Fe(CN)_6 ]_4^(-) rarr [Fe(CN)_5]^(3-) +3^- , E^@ =- 0.35 V` `Fe^(2+) rarr Fe^(3+) + e^(- : E^@ =- 0. 77 V`.A. ` Fe^(2+)`B. ` Fe^(3+)`C. ` [Fe(CN)_6]^(3-)`D. ` [Fe(CN)_6]^(4-)`

On the basis of the folwing `E^@` values, the strongest oxidizing agen

1.	On the basis of the folwing `E^@` values, the strongest oxidizing agent is `[Fe(CN)_6 ]_4^(-) rarr [Fe(CN)_5]^(3-) +3^- , E^@ =- 0.35 V` `Fe^(2+) rarr Fe^(3+) + e^(- : E^@ =- 0. 77 V`.A. ` Fe^(2+)`B. ` Fe^(3+)`C. ` [Fe(CN)_6]^(3-)`D. ` [Fe(CN)_6]^(4-)`
Answer» Correct Answer - B `Fe^(3+)` and ` Fe (CN)_(6)^(3)` are oxidants. Higher si `E_(RP)^(@)` stronger is oxidant . ` fe^(3+) = e rarr Fe^(2+) , E_(RP)^2 = 0.77 V` `[ Fe (CN)_6 ]^(3-) + e rarr [Fe(CN)_6]^(4-) , E_(RP)^@ = 0.35 V`.

Discussion

No Comment Found

Related InterviewSolutions

Define molar conductivity of a solution and explain how molar conductivity changes with change in concentration of a solution for a weak and a strong electrolyte.
Limiting molar conductivity of `NH_(4)OH` [i.e., `Lambda_(m)^(@)(NH_(4)OH)`] is equal to:A. `Lambda_(m (NH_(4)Cl ))^(@) + Lambda_(m (NaCl))^(@) - Lambda_(m(NaOH))^(@)`B. `Lambda_(m (NaOH))^(@) + Lambda_(m(NaCl))^(@) - Lambda_(m(NH_(4)Cl))^(@)`C. `Lambda_(m(NH_(4)OH))^(@) + Lambda_(m (NH_(4)Cl))^(@) - Lambda_(m(HCl))^(@)`D. `Lambda_(m (NH_(4)Cl))^(@) + Lambda_(m (NaOH))^(@) - Lambda_(m(NaCl))^(@)`
Limiting molar conductivity of `NH_(4)OH` [i.e., `Lambda_(m)^(@)(NH_(4)OH)`] is equal to:A. `Lambda_(m)^(@)(NH_(4)Cl)+Lambda_(m)^(@)(Na_(4)Cl)-Lambda_(m)^(@)(NaOH)`B. `Lambda_(m)^(@)(NaOH)+Lambda_(m)^(@)(NaCl)-Lambda_(m)^(@)(NH_(4)Cl)`C. `Lambda_(m)^(@)(NH_(4)OH)+Lambda_(m)^(@)(NH_(4)Cl)-Lambda_(m)^(@)(HCl)`D. `Lambda_(m)^(@)(NH_(4)Cl)+Lambda_(m)^(@)(NaOH)-Lambda_(m)^(@)(NaCl)`
For the cell reaction `Cu^(2+)(C_(1),aq.)+Zn(s)hArrZn^(2+)(C_(2),aq)+Cu(s)` of an electrochemical cell, the change in free energy `(DeltaG)` of a given temperature is a function ofA. `lnC_(1)`B. `lnC_(2)//C_(1)`C. `ln (C_(1) + C_(2))`D. `ln C_(2)`
Use of lithium metal as an electrode in high energy density batteries is due to:A. Lithium is highest elementB. Lithium has highest oxidation potentialC. Lithium is quite reactiveD. Lithium does not corrode readily
The Gibbs energy for the decomposition of `Al_(2)O_(3)` at `500^(@)C` is as follow : `(2)/(3)Al_(2)O_(3) rarr (4)/(3)Al+O_(2), Delta_(r)G= +960 kJ mol^(-1)` The potential difference needed for the electrolytic reduction of aluminium oxide `(Al_(2)O_(3))` at `500^(@)C` isA. `5.0 V`B. `4.5 V`C. `3.0 V`D. `2.5 V`
Based on the data given below, the correct order of reducing power is: `Fe_((aq.))^(3+) + e rarr Fe_((aq.))^(2+), E^(@) = +0.77 V` `Al_((aq.))^(3+) + 3e rarr Al_((s)), E^(@) = -1.66 V` `Br_(2(aq.)) + 2e rarr 2Br_((aq.))^(-), E^(@) = +1.08 V`A. `Br^(-) lt Fe^(2+) lt A1`B. `Fe^(2+) lt Al lt Br^(-)`C. `Al^(-) lt Br^(-) lt Fe^(2+)`D. `Al^(-) lt Fe^(2+) lt Br^(-)`
Based on the data given below, the correcy order of reducing power is: `Fe_((aq.))^(3+) + e rarr Fe_((aq.))^(2+), E^(@) = +0.77 V` `Al_((aq.))^(3+) + 3e rarr Al_((s)), E^(@) = -1.66 V` `Br_(2(aq.)) + 2e rarr 2Br_((aq.))^(-), E^(@) = +1.08 V`A. `Br^(-) lt Fe^(2+) lt Al`B. `Fe^(2+) lt Al lt Br^(-)`C. `Al lt Br^(-) lt Fe^(2+)`D. `Al lt Fe^(2+) lt Br^(-)`
Based on the data given below, the correcy order of reducing power is: `Fe_((aq.))^(3+) + e rarr Fe_((aq.))^(2+), E^(@) = +0.77 V` `Al_((aq.))^(3+) + 3e rarr Al_((s)), E^(@) = -1.66 V` `Br_(2(aq.)) + 2e rarr 2Br_((aq.))^(-), E^(@) = +1.08 V`A. `Br^(-) lt Fe^(2+) lt Al`B. `Fe^(2+) lt Al lt Br^-`C. `Al lt Br^(-) lt Fe^(2+)`D. `Al lt Fe^(2+) lt Br^(-)`
What are elementary reactions ?

On the basis of the folwing `E^@` values, the strongest oxidizing agent is `[Fe(CN)_6 ]_4^(-) rarr [Fe(CN)_5]^(3-) +3^- , E^@ =- 0.35 V` `Fe^(2+) rarr Fe^(3+) + e^(- : E^@ =- 0. 77 V`.A. ` Fe^(2+)`B. ` Fe^(3+)`C. ` [Fe(CN)_6]^(3-)`D. ` [Fe(CN)_6]^(4-)`

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment