You are titrating 25.025.0 mL of 0.0100 M Sn2 Sn2 in 1 M HCl HCl with 0.0500 M Tl3 Tl3 resulting in the formation of Sn4 Sn4 and Tl Tl . A PtPt indicator electrode and a saturated Ag|AgCl Ag|AgCl reference electrode are used to monitor the titration. What is the balanced titration reaction

Sn²⁺ + Tl³⁺ → Sn⁴⁺ + Tl⁺

Explanation:

The Sn²⁺ is oxidized to Sn⁴⁺. Whereas Tl³⁺ is reduced to Tl⁺. The half-reactions are:

Sn²⁺ → 2e⁻ + Sn⁴⁺ (Oxidation, loosing electrons)

Tl³⁺ + 2e⁻ →Tl⁺ (Reduction, gaining electrons)

The sum of the reactions gives:

Sn²⁺ + Tl³⁺ + 2e⁻ → 2e⁻ + Sn⁴⁺ + Tl⁺

Subtracting the electrons in both sides of the reaction:

Sn²⁺ + Tl³⁺ → Sn⁴⁺ + Tl⁺

Answerthe half-reaction having the greatest tendency to occur at the anode:     i₂(aq) + 2e⁻  ⇄ 2i⁻  ;   e = +0.53 v======oxidation occurs at the anode. we must therefore find the half-reaction with the greatest tendency to oxidize.the given values of e are reduction potentials (the form of the half-reactions are of reduction).the lower the reduction potential, the higher the potential to oxidize. the half-cell with the lowest reduction potential will have the greatest tendency to oxidize at the anode.the half-reaction having the greatest tendency must be    i₂(aq) + 2e⁻  ⇄ 2i⁻    e = +0.53 vbecause its reduction potential is the lowest out of the three.