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.

Explanation

`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`