) c =-1.595+1.0652=-0.529V

) c =- . + . =- .529V ∴ The species undergoing disproportionation is HBrO (Option D) Short answer . Given C = .01M S cm mol cation o λ K= . S cm S cm mol o λ anion . Molar conductivity Λ m K K (sm C (in M) mol m = . S cm S mol m c m = m = . S m mol = . . Degree of dissociation α = Λ Λ   ∞ Λ λ λ ∞ o cation o anion o S cm mol S cm mo ( . ) l s m mol - - - - α α S m mol S m mol K = c - = ( . ) ( . ) - . a α α . Given I = .608A; t = min = V= mL C = . M = 3000S η = % Calculate the number of faradays of electricity passed through the CuSO solution ⇒ ∴ Q=It Q = . Q = 4824C number of Faradays of electicity = F 96500 = . Electrolysis of CuSO Cu (aq)+2e Cu(s). + − → The above equation shows that 2F electricity will deposit mole of Cu + to Cu. ∴ .05F electricity will deposit 1mol 2F F = . mol Initial number of molar of Cu in ml of solution = . mL mL = . mol ∴ number of moles of Cu afer electrolysis = . - . = . mol ∴ Concentration of Cu mol mL mL = . M . Required half cell reaction Br Br E V Fe ox o