What will separate first at the cathode and what at the anode during molten electrolysis a mixture of NaCl and CaBr2.

Answer 1

1. Possible catholic reactions: Na_(l)⁺ +. e^- ====>. Na_(l)^. E^o. =. -2.71 V
2. Ca_(l)²⁺. +. 2e^- ====> Ca_(l) E^o. =. -2.76 V

Na⁺ has slightly higher standard reduction potential, therefore, reduction of Na⁺ is favored.

1. Possible anodic reactions : 2Cl_(l)^- ===>. Cl_2(g). +. 2e^- E⁰_ox = - E^o. = -1.36 V
2. 2Br_(l)^- ===> Br_2(l). +. 2e^- E^o_ox. = - E^o. = - 1.07 V

Br^- has higher standard oxidation potential, therefore, oxidation of Br^- is favored.

P.S. Standard reduction potentials are for aqueous solutions at standard state conditions (conc. = 1.0 M, Temp = 25^oC), and we are dealing with molten salts in this question.

1. 
   

Answer 2

Check chlorine and bromine: the one with the least positive reduction potential will come out as non-metal at anode

Check Ca²⁺ and Na⁺ : the one with the least negative reduction potential (least positive oxidizing potential) will come out as metal at the cathode.

Br unsurprising. Na⁺ is less negative than Ca²⁺!