The standard electrode potential (M³⁺/M²⁺) for V, Cr, Mn Co are -0.26 V, -0.41 V, + 1.57 V and + 1.97 V, respectively. The metal ions which can liberate H₂ from a dilute acid are :

Solution

Chemistry ma’am · 3 steps

Only ions with a negative E° give up electrons more readily than hydrogen and free H₂.

1. What happens. To liberate H₂, the ion M²⁺ must reduce H⁺. While doing so it is oxidised: M²⁺ → M³⁺ + e⁻, and 2H⁺ + 2e⁻ → H₂ with E° = 0 V.

   E°_(cell) = E°_(H+/H₂) - E°_(M³⁺/M²⁺) = 0 - E°_(M³⁺/M²⁺)

2. Sign decides. The reaction is spontaneous only when E°_(cell) > 0. That happens only when the reduction potential of M³⁺/M²⁺ is negative.

   Ion	E°_(M³⁺/M²⁺)	E°_(cell)	Liberates H₂?
   V²⁺	-0.26 V	+0.26 V	yes
   Cr²⁺	-0.41 V	+0.41 V	yes
   Mn²⁺	+1.57 V	-1.57 V	no
   Co²⁺	+1.97 V	-1.97 V	no

3. So. Only V²⁺ and Cr²⁺ are strong enough reducing agents to liberate H₂ from dilute acid, which is option (c).

The idea to keep. A negative reduction potential means the lower oxidation state is a good reducing agent. Anything below 0 V (the hydrogen electrode) can reduce H⁺ to H₂.

Answer from the NTA final answer key. Solution by PrepWiser.