Consider the following standard reduction potentials Fe^(3+)(aq) + e^(-) iff Fe^(3+) (aq) , E^(@) = 0.77 V , H_2O_2(aq) + 2e^(-) iff 2OH^(-)(aq) , E^(@) = 0.88 V For the voltaic cell reaction below , calculatethe Fe^(2+) concentration ( in M) that would be needed to produce a cell potential equal to 0.16 V at25^@ Cwhen [OH^(-)] = 0.1 M , [Fe^(3+)] = 0.5 M and [H_2O_2]= 0.35M 2Fe^(2+)(aq) + H_2O_2(aq) iff 2Fe^(3+) (aq) + 2OH^(-) (aq)

Consider the following standard reduction potentials Fe^(3+)(aq) + e^

1.	Consider the following standard reduction potentials Fe^(3+)(aq) + e^(-) iff Fe^(3+) (aq) , E^(@) = 0.77 V , H_2O_2(aq) + 2e^(-) iff 2OH^(-)(aq) , E^(@) = 0.88 V For the voltaic cell reaction below , calculatethe Fe^(2+) concentration ( in M) that would be needed to produce a cell potential equal to 0.16 V at25^@ Cwhen [OH^(-)] = 0.1 M , [Fe^(3+)] = 0.5 M and [H_2O_2]= 0.35M 2Fe^(2+)(aq) + H_2O_2(aq) iff 2Fe^(3+) (aq) + 2OH^(-) (aq)
Answer» `0.3 M` `0.6 M` `0.41 M` `0.354 M` Solution :`E_("cell")^(0) = 0-88-0.77 = 0.11 V ` `E=E^(0) + (0.0591)/(2) LOG ""([Fe^(+2)]^(2)[H_2O_2]^(1))/([Fe^(+3)]^(2) xx [OH^(-)]^(2)), 0.16 = 0.11 + (0.0591)/(2) log""[((Fe^(+2))^(2) xx (0.35)^(1))/((0.5)^(2) xx (10^(-1))^(2))]` `[log""([Fe^(+2)]^(2) [0.35])/(0.25 xx 0.01) ] = (0.05 xx 2)/(0.0591) = log""[([Fe^(+2)]^(2)(0.35))/(0.00875)]=1.6920` `([Fe^(+2)]^(2)(0.35))/(0.00875) = 50 , (Fe^(+2)) = SQRT((50 xx 0.000875)/(0.35)) = sqrt((0.04375)/(0.35)) = 0.35355 = 0.354M`