Question

Sn =0.14 (E0) Sn+2 = -0.15 (E0) Which form of tin is the stronger reducing agent?

Answer 1

Answer: The stronger reducing agent will be $Sn^{2+}$

Explanation:

Reducing agent is defined as the agent which helps the other substance to reduce and itself gets oxidized. Thus, it will undergo oxidation reaction.

We are given:

$E_o(Sn)=0.14V\\E_o(Sn^{2+})=-0.15V$

The electrode potential values of given substances determine the tendency to gain electrons. Higher value of positive electrode potential means that it will gain more and hence, will be a stronger oxidizing agent.

If the specie has negative electrode potential, it means that it will easily loose electrons and hence, will be considered as a stronger reducing agent.

Thus, the stronger reducing agent among the given species will be $Sn^{2+}$

Answer 2

Sn is the only form of tin that can act as a reducing agent.

Reduction: Sn²⁺ + 2e⁻ ⟶ Sn; E° = -0.15 V

The Sn²⁺ is being reduced, so it is acting as an oxidizing agent.

Oxidation: Sn ⟶ Sn²⁺ + 2e⁻; E° = +0.15 V

The Sn is being oxidized, so it is acting as a reducing agent.