VO2^+ can be reduced by tin, in which case the metal is oxidized to Sn^2+. However, this reduction stops at V^3+ (V^2+ does not form). Write a balanced net ionic equation to show this reaction in an acidic medium
So I start with: VO2^+ + SN ---> Sn^2+ + V^3+
I don't know where to go from there, and how do you identify the reduction and oxidation halves of the equation ?
Net Ionic Equation?
2017-10-25 11:04 am
回答 (1)
2017-10-25 11:42 am
The question states that VO₂⁺ is reduced to V³⁺ while Sn is oxidized to Sn²⁺.
In an acidic medium, H⁺ and H₂O can takes place in reaction.
Reduction half-equation: VO₂⁺(aq) + 4H⁺(aq) + 2e⁻ → V³⁺(aq) + 2H₂O(l)
Oxidation half-equation: Sn(s) → Sn²⁺(aq) + 2e⁻
Sum up the above two half-equations, and cancel 2e⁻ on the both sides. The overall equation is :
VO₂⁺(aq) + Sn(s) + 4H⁺(aq) → Sn²⁺(aq) + V³⁺(aq) + 2H₂O(l)
In an acidic medium, H⁺ and H₂O can takes place in reaction.
Reduction half-equation: VO₂⁺(aq) + 4H⁺(aq) + 2e⁻ → V³⁺(aq) + 2H₂O(l)
Oxidation half-equation: Sn(s) → Sn²⁺(aq) + 2e⁻
Sum up the above two half-equations, and cancel 2e⁻ on the both sides. The overall equation is :
VO₂⁺(aq) + Sn(s) + 4H⁺(aq) → Sn²⁺(aq) + V³⁺(aq) + 2H₂O(l)
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