In the electrolysis of a molten mixture of NaCl and CaBr2, identify the product that forms at the anode and the cathode.?

In the molten mixture, NaCl gives Na⁺ and Cl⁻ ions while CaBr₂ gives Ca²⁺ and Br⁻ ions.

At the anode :
Cl⁻ and Br⁻ ions migrate to the anode. The reduction potential of Cl₂(g)|Cl⁻(l) is more positive than that of Br₂(g)|Br⁻(l), and thus Br⁻ ion undergoes oxidation more readily than Cl⁻ ion. Hence, Br⁻ ion is preferentially discharged (oxidized) to gives Br₂ vapor.
2Br⁻(l) → Br₂(g) + 2e⁻

At the cathode :
Na⁺ and Ca²⁺ ions migrate to the cathode. The reduction potential of Na⁺(l)|Na(l) is less negative than that of Ca²⁺(l)|Ca(l), and thus Na⁺ ion undergoes reduction more readily than Ca²⁺ ion (although Na is more reactive than Ca). Hence, Na⁺ ion is preferentially discharged (reduced) to give molten liquid of Na.
Na⁺(l) + e⁻ → Na(l)
(Na⁺ ion is preferentially discharged. Therefore, in electrolysis of NaCl, CaCl₂ is added to lower the boiling point of NaCl, and the presence of Ca²⁺ ion does not affect the product formed at the cathode.)

Conclusively, bromine vapor is formed at the anode while molten sodium is formed at the cathode.