The standard electrode potential for V, Cr, Mn & Co are V, V, V and V, respectively. The metal ions which can liberate from a dilute acid are
- A.and
- B.and
- C.and
- D.and
Show answer and solution
Correct option: A
Solution
The M²⁺ ion can liberate H₂ from dilute acid if M²⁺ is a strong enough reducing agent, i.e., it can reduce H⁺ to H₂. This requires the E°(M³⁺/M²⁺) to be negative (below 0 V), meaning M²⁺ → M³⁺ + e⁻ has a positive oxidation potential, making the overall cell reaction (M²⁺ reduces H⁺) spontaneous.
V³⁺/V²⁺ = −0.26 V and Cr³⁺/Cr²⁺ = −0.41 V are both negative, so V²⁺ and Cr²⁺ are oxidized to their 3+ states while reducing H⁺ to H₂ (ΔG < 0). In contrast, Mn³⁺/Mn²⁺ = +1.57 V and Co³⁺/Co²⁺ = +1.97 V are large and positive, meaning Mn²⁺ and Co²⁺ have no tendency to get oxidized to 3+; they cannot reduce H⁺. Therefore, V²⁺ and Cr²⁺ are the ions that liberate H₂ from dilute acid.
V³⁺/V²⁺ = −0.26 V and Cr³⁺/Cr²⁺ = −0.41 V are both negative, so V²⁺ and Cr²⁺ are oxidized to their 3+ states while reducing H⁺ to H₂ (ΔG < 0). In contrast, Mn³⁺/Mn²⁺ = +1.57 V and Co³⁺/Co²⁺ = +1.97 V are large and positive, meaning Mn²⁺ and Co²⁺ have no tendency to get oxidized to 3+; they cannot reduce H⁺. Therefore, V²⁺ and Cr²⁺ are the ions that liberate H₂ from dilute acid.
[JEE Main 2023 · 29 Jan · Shift 1]