InterviewSolution
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InterviewSolution
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Calculate the potential of the following half`-`cells`|` cells`:` `a. Cr|Cr^(3+)(0.1 M)||Fe^(2+)(0.01M)|Fe` Given `: E^(c-)._(Cr^(3+)|Cr)=-0.74V E^(c-)._(Fe^(2+)|Fe)=-0.44V` `b. 6e^(c-)+BrO_(3)^(c-)(aq)+3H_(2)OrarrBr^(c-)(aq)+6OH(aq)` Given `:E^(c-)._((BrO_(3)^(c-)|Br^(c-)))=0.61V,` `[BrO_(3)^(c-)]=2.5xx10^(-3)M,[Br^(c-)]=5.0xx10^(-3)M,pH=9.0` `c. Ag|Ag^(o+)(0.1M)||Cl^(c-)(0.02M)|Cl_(2)(g)(0.5atm)|Pt` Given `E_((Ag^(o+)|Ag))=0.80V,E^(c-)((Cl_(2)|2Cl^(c-))=1.36V` `d. NO_(3) ^(c-)(aq)+2H^(o+)(aq)+e^(-)rarrNO_(2)+H_(2)O` Given `: E^(c-)._(NO_(3)^(c-)|NO_(2)=0.78V` What will be the reductino potential of the half cell in neutral solution ? Assuming all the other species to be at unit concentration. |
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Answer» `a.` Anode reaction `:` `[Cr(s) rarr Cr^(3+)(0.1M)+cancel (3e^(-))]xx2` Cathode reaction `:` `[2e^(-)+Fe^(2+)(0.01M) rarr Fe(s)] x3` Cell reaction `:` `ulbar(2Cr(s)+3Fe^(2+) rarr 2Cr^(3+)+3Fe(s))` `E^(c-)._(cell)=(E^(c-)._(reduction))_(c)-(E^(c-)._(reduction))_(a)` `=-0.44-(-0.74)=0.3V` Using Nernest equation `:n_(cell)=6)` `E_(cell)=E^(c-)._(cell)-(0.059)/(6)log.([Cr^(3+)]^(2))/([Fe^(2+)]^(3))` `=0.3V-(0.06)/(6)log.((0.2)^(2))/((0.01)^(3))` `=0.3V-0.01log.(0.1xx0.1)/(0.01xx0.01xx0.01)` `=0.3V-0.01[log 10^(4)]` `=0.3V-0.01xx4=0.26 V` `b.` Using Nernst equation` : (n_(cell)=6,` take`0.059~~0.06)` `E_(cell)=E^(c-)._(cell)-(0.06)/(6)log .([Br^(c-)][overset(c-)(O)H]^(6))/([BrO_(3)^(c-)])` `(pH=9.0,pOH=14-9=5,[overset(c-)(O)H]=10^(-5)M.` Acitivity of `H_(2)O=1)` `E_((BrO_(3).^(c-)|Br.^(c-)))=E^(c-)._((BrO_(3).^(c-)|Br.^(c-)))` `-(0.06)/(6) log .(5.0xx10^(-3)xx(10^(-5)M)^(6))/(2.5xx10^(-3)M)` `=0.61V-0.01(log2xx10^(-30))` `0.61V-0.01(0.3-30),,,,(log 2 ~~0.3)` `=0.61V-0.01xx(-29.7)` `=0.61V+0.297 V~~0.907V` `c.` Anode reaction `: 2Ag(s) rarr 2 Ag ^(o+)(0.1M)+cancel(2e^(-))` Cathode reaction `:` `Cl_(2)(g)(0.5atm)+cancel(2e^(-))rarr 2Cl^(c-)(0.02M)` Cell reaction `:` `ulbar(2Ag(s)+Cl_(2)(g)(0.5atm)rarr2Ag^(o+)(0.1M)+2Cl^(c-)(0.02M))` `E_(cell)=(E^(c)._(reduction))_(c)-(E^(c-)._(reduction ))_(a)` `=1.36V-0.80V=0.56V` Using Nernst equation `: (n_(cell=2` activity of `Ag=1,` take `0.059~~0.06)` `E_(cell)=E^(c-)._(cell)-(0.06)/(2)log.([Ag^(o+)]^(2)[Cl^(c-)]^(2))/((p_(Cl_(2)(g))))` `E_(cell)=0.56V-0.03 (log.((0.1)^(2)(0.2)^(2))/(0.5 atm))` `=0.56V-0.03(log 8xx10^(-4))` `=0.56V-0.03(log 2^(3)+log 10^(-4))` `=0.56V-0.03(3 log 2 - 4) " "(log 2 ~~ 0.3)` `=0.56V-0.03(3xx0.3-4)` `=0.56 V+0.03xx3.1V` `=0.653V` `d. pH` of natural solution `=7:. [H^(o+)]=10^(-7)M` `n_(cell)=1,` concentration of `[NO_(2)]=[H_(2)O]=[NO_(3).^(c-)]=1` `E_((NO_(3).^(c-)|NO_(2)))=E^(c-)._((NO_(3).^(c-)|NO_(2)))-(0.059)/(1)log.([NO_(2)][H_(2)O])/([NO_(3).^(c-)][H^(o+)]^(2))` `=0.78V-0.059log.(1)/((10^(-7))^(2))` `=0.78V-0.059[log10^(14)]` `=0.78V-0.059xx14` `=0.78 V-0.059xx14` ltbr. `=-0.046V` |
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