A) Yes, this is the balanced equation for the reaction between Sn2+ and Tl3+.

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Discussion Overview

The discussion centers around the balanced equation for the redox reaction between Sn2+ and Tl3+, as well as the calculation of the half-cell potential during a titration process. The scope includes theoretical and mathematical reasoning related to electrochemistry.

Discussion Character

  • Technical explanation, Mathematical reasoning

Main Points Raised

  • One participant proposes the balanced equation Sn2+ + Tl3+ → Sn4+ + Tl+ and asks for confirmation of its correctness.
  • Another participant agrees that the redox reaction is correct, noting that thallium is most stable at the +1 oxidation state and that tin(II) ions reduce it to tin(IV).
  • A participant then calculates the half-cell potential after adding 5.00 mL of Tl3+ and presents a formula for this calculation, seeking validation.
  • A later reply acknowledges the formula but questions the logarithm ratio used in the calculation, indicating some uncertainty about its correctness.

Areas of Agreement / Disagreement

Participants generally agree on the balanced equation for the reaction, but there is uncertainty regarding the logarithm ratio in the half-cell potential calculation, indicating that the discussion remains unresolved on that point.

Contextual Notes

There is a lack of clarity regarding the specific logarithm ratio used in the half-cell potential calculation, which may depend on additional context or definitions not provided in the discussion.

Mag
consider the titration of 50 mL on 0.10M Sn2+ with 0.10M Tl3+

A) Wite the equation for the reaction that occurs between Sn2+ and Tl3+

[tex]Sn^2^++Tl^3^+\longrightarrow\ Sn^4^++Tl^+[/tex]

Correct?
 
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Thallium is most stable at 1+ oxidation state, and tin(II) ions are present to reduce it. Tin(IV) ions will be produced in this redox reaction, so your redox reaction is correct.
 
Last edited:
ok good.

what is the half cell potential after 5.00mL of Tl[tex]^3^+[/tex]?

[tex]E_1_/_2=0.139-\frac{0.05196}{2}log\frac{1}{4.5}[/tex]
[tex]E_1_/_2=0.156v[/tex]

this right also?
 
The formula is something like that, but I don't remember the logarithm ratio in which you wrote 4.5. However, it seems okay.
 

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