InterviewSolution
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InterviewSolution
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The standard reduction potential data at `25^(@)C` is given below `E^(@) (Fe^(3+), Fe^(2+)) = +0.77V`, `E^(@) (Fe^(2+), Fe) = -0.44V`, `E^(@) (Cu^(2+),Cu) = +0.34V`, `E^(@)(Cu^(+),Cu) = +0.52 V`, `E^(@) (O_(2)(g) +4H^(+) +4e^(-) rarr 2H_(2)O] = +1.23V` `E^(@) [(O_(2)(g) +2H_(2)O +4e^(-) rarr 4OH^(-))] = +0.40V`, `E^(@) (Cr^(3+), Cr) =- 0.74V`, `E^(@) (Cr^(2+),Cr) = - 0.91V`, Match `E^(@)` of the redox pair in List-I with the values given in List-II and select the correct answer using the code given below teh lists: `{:(List-I,List-II),((P)E^(@)(Fe^(3+),Fe),(1)-0.18V),((Q)E^(@)(4H_(2)O hArr 4H^(+)+4OH^(+)),(2)-0.4V),((R)E^(@)(Cu^(2+)+Curarr2Cu^(+)),(3)-0.04V),((S)E^(@)(Cr^(3+),Cr^(2+)),(4)-0.83V):}` Codes:A. `{:(P,Q,R,S),(4,1,2,3):}`B. `{:(P,Q,R,S),(2,3,4,1):}`C. `{:(P,Q,R,S),(1,2,3,4):}`D. `{:(P,Q,R,S),(3,4,1,2):}` |
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Answer» Correct Answer - D `(P) e^(-) + Fe^(3+) rarr Fe^(+2)` `DeltaG_(1)^(ul(@)) =- 1 xx F xx 0.77` `2e^(-) +Fe^(2+) rarr Fe` `DeltaG_(2)^(ul(@))=- 2 xx F xx (-0.44)` Required equation `Fe^(3+) +3e^(-) rarr Fe` `DeltaG_(3)^(ul(@)) =- 3 xx F xx E^(@)` `DeltaG_(3)^(ul(@)) = DeltaG_(2)^(ul(@)) +DeltaG_(1)^(ul(@))` `- 3 xx F xx E^(ul(@)) =- 1 xx F xx 0.77 - 2 xx F xx (-0.44)` `E^(ul(@)) =- (0.11)/(3) =- 0.04V` `(Q) O_(@)(g) +4H^(+) +4e^(-) rarr 2H_(2)O` `E_(1)^(ul(@)) = 1.23` ...(i) `O_(2)(g) +2H_(2)O +4e^(-) rarr 40H^(-) E_(2)^(ul(@)) = 0.40V` ...(ii) Required equation `4H_(2)O hArr 4H^(+) + 4OH^(-) (E_(2)^(ul(@))-E_(1)^(ul(@))) =- 0.83V` `(R) {:(Cu^(+2)+e^(-),rarr,Cu^(+)),(E_(1)^(ul(@))=0.34V,,),(Cu,rarr,Cu^(+)+e^(-)),(E_(2)^(ul(@))=-0.52,,):}` `bar(Cu^(+2) +Cu rarr 2Cu^(+1)` `-0.18 V` `(S) Cr^(+3) +3e^(-) rarr Cr` `DeltaG_(1)^(ul(@)) =- 3 xx F xx (-0.74)` `Cr^(+2) +2e^(-) rarr Cr` `DeltaG_(2)^(ul(@)) = - 2xx F xx (-0.91)` Required equation `Cr^(3+) +e^(-) rarr Cr^(+2)` `DeltaG_(3)^(ul(@)) =- 1 xx F xx (E^(@))` `DeltaG_(1)^(ul(@)) = DeltaG_(2)^(ul(@)) +DeltaG_(3)^(ul(@))` `- 3 xx F xx (-0.74) =- 2 xx F xx (-0.91) - 1 xx F xx E^(@)` `E^(@) = 1.82 - 2.22 =- 0.4V` |
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