Chemistry

posted by .

For there reaction system involving the species in the table below, what conclusion is
justified for standard state conditions?
Standard Reduction Potentials E0
2Fe3+ + Sn2+ ¡ú 2Fe2+ + Sn4+ +0.62 V
Sn2+ + I2 ¡ú Sn4+ + 2I¨C +0.38 V
(A) I¨C would react with Fe3+
(B) Fe2+ would react with I2
(C) Fe3+ is a weaker oxidizing agent than I2
(D) Fe2+ is a stronger reducing agent than I¨C

  • Chemistry -

    My screen is gibberish again on both .

  • Chemistry -

    I'm wondering if you are using a Mac.

  • Chemistry -

    2Fe3+ + Sn2+ ¡ú 2Fe2+ + Sn4+ +0.62 V
    Sn2+ + I2 ¡ú Sn4+ + 2I¨C +0.38 V

  • Chemistry -

    man this sucks well hold on
    2Fe3+ + Sn2+ --> 2Fe2+ + Sn4+ +0.62 V
    Sn2+ + I2 --> Sn4+ + 2I– +0.38 V
    i hope this is better

  • Chemistry -

    Yes. That's better but I don't have time to look at it now. I'll get back later.

  • Chemistry -

    But you also need to redo the answers because they contain some of the same strange symbols. especially the A answer.

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Chemistry

    in the reaction:Ni+SN4+-->Ni2+ (+)Sn2+ which species has been reduced?
  2. chemistry

    A balanced equation that represents an overall cell reaction is shown. Choose the cell notation which corresponds to the electrochemical cell that is described by this equation. a) Sn2+(aq) + 2Cu+(aq) → Sn4+(aq) + 2Cu(s) b) Pt(s) …
  3. CHEMISTRY

    I haven't learned half reactions for oxidation and reduction so please teach me by leading me from step to step.... I don't get this.. Sn2+ +2Fe3+---> Sn4+ +2Fe2+ The answer key says Oxidation: Sn2+--->Sn4+ +2e- Reduction: 2Fe3+ …
  4. CHEMISTRY

    I haven't learned half reactions for oxidation and reduction so please teach me by leading me from step to step.... I don't get this.. Sn2+ +2Fe3+---> Sn4+ +2Fe2+ The answer key says Oxidation: Sn2+--->Sn4+ +2e- Reduction: 2Fe3+ …
  5. Chem Check Please!!!

    1) Assuming the following reaction proceeds in the forward direction, 3 Sn4+(aq) + 2 Cr(s) ---> 3 Sn2+(aq) + 2 Cr3+(aq) a.Sn4+(aq) is the reducing agent and Cr(s) is the oxidizing agent. b.Cr(s) is the reducing agent and Sn2+(aq) …
  6. CHEMISTRY HELP !!!!

    Assuming the following reaction proceeds in the forward direction, 3 Sn4+(aq) + 2 Cr(s)  3 Sn2+(aq) + 2 Cr3+(aq) a. Sn4+(aq) is the reducing agent and Cr(s) is the oxidizing agent. b. Cr(s) is the reducing agent and Sn2+(aq) …
  7. Chemistry

    Consider the titrtation of 25.0mL of 0.0100M Sn2+ by 0.0500M Ti3+ in 1 M HCL, using Pt and Saturated Calomel Electrodes to find the endpoint. F=96486.7 Cmol E0 Sn4+/Sn2+ =0.15V E0 TI3+/TI+ = 1.28 Esce = 0.241V R=8.3141 J K Mol (a) …
  8. Chemistry (Electrochemistry)

    Consider the titrtation of 25.0mL of 0.0100M Sn2+ by 0.0500M Ti3+ in 1 M HCL, using Pt and Saturated Calomel Electrodes to find the endpoint. F=96486.7 Cmol E0 Sn4+/Sn2+ =0.15V E0 TI3+/TI+ = 1.28 Esce = 0.241V R=8.3141 J K Mol (a) …
  9. Chemistry re-post

    Consider the titrtation of 25.0mL of 0.0100M Sn2+ by 0.0500M TI3+ in 1 M HCL, using Pt and Saturated Calomel Electrodes to find the endpoint. F=96486.7 Cmol E0 Sn4+/Sn2+ =0.15V E0 TI3+/TI+ = 1.28 Esce = 0.241V R=8.3141 J K Mol (a) …
  10. Chemistry (Electrochemistry)

    Consider the titrtation of 25.0mL of 0.0100M Sn2+ by 0.0500M TI3+ in 1 M HCL, using Pt and Saturated Calomel Electrodes to find the endpoint. F=96486.7 Cmol E0 Sn4+/Sn2+ =0.15V E0 TI3+/TI+ = 1.28 Esce = 0.241V R=8.3141 J K Mol (a) …

More Similar Questions