Chemistry - Redox Reactions Question with Solution | TestHub
Match the equivalent weight of HCl in Column I with the chemical change(s) in Column II.
Column I | Column II |
|---|---|
A) Greater than its molecular weight | P) Neutralization reaction |
B) Equal to molecular weight | Q) MnO₂ + HCl MnCl₂ + Cl₂ + H₂O |
C) Less than molecular weight | R) HClO HCl |
| S) HCl HClO₃ |
| T) Cu + HCl H₂[CuCl₄] + H₂ |
Options:
Answer:
Solution:
E(HCl) = (moles of HCl consumed per mole of electrons) x 36.5. (Q) MnO₂ + 4HCl MnCl₂ + Cl₂ + 2H₂O: 4 mol HCl per 2 e⁻ (only 2 of the 4 HCl are oxidized), E = 73 36.5. (T) Cu + 4HCl H₂[CuCl₄] + H₂: 4 mol HCl consumed per 2 e⁻ transferred to H₂ (the extra Cl⁻ complex Cu²⁺), E = 73 M. So A maps to Q, T. (P) Neutralization: n = 1, E = 36.5 = M, so B maps to P. (R) HClO HCl: Cl(+1) -1, 2 e⁻ per HCl formed, E = 36.5/2 M. (S) HCl HClO₃: Cl(-1) +5, 6 e⁻, E = 36.5/6 M. So C maps to R, S.
