The number of ions from the following that have the ability to liberate hydrogen from a dilute acid is . Ti2+, Cr2+ and V2+
- 1
- 2
- 3
- 0
Answer
(3) 3
Solution
E ( M3+/M2+): Ti -0.37 V, V -0.26 V, Cr -0.41 V
∵ all are below E (H+/H2) = 0.00 V ⇒ each M2+ reduces H+ to H2 .g. 2Cr2+(aq) + 2H+(aq) 2Cr3+(aq) + H2(g)
∴ all 3 ions liberate H2
Answer: (3)