Question

+ + + + + + + + + Metal Oxidation Reaction Lithium Li(s) - Li+ (aq) Potassium K(s) K+ (aq) Barium Ba(s) Ba2+ (aq) Calcium Ca(
0 0
Add a comment Improve this question Transcribed image text
Answer #1

0 I a Mg at the Mg Engati mg = -2.37. Fet toé + fers) Efedt! Fecs) - -0.441v. From top to bottom oxidising tendency is decrea

Add a comment
Know the answer?
Add Answer to:
+ + + + + + + + + Metal Oxidation Reaction Lithium Li(s) - Li+...
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for? Ask your own homework help question. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions
  • Use the tabulated electrode potentials to calculate K for the oxidation of nickel by H+: Ni(s)+2H+(aq)→Ni2+(aq)+H2(g)...

    Use the tabulated electrode potentials to calculate K for the oxidation of nickel by H+: Ni(s)+2H+(aq)→Ni2+(aq)+H2(g) Express your answer using two significant figures. Standard reduction half-cell potentials at 25∘C Half-reaction E∘ (V) Half-reaction E∘ (V) Au3+(aq)+3e−→Au(s) 1.50 Fe2+(aq)+2e−→Fe(s) −0.45 Ag+(aq)+e−→Ag(s) 0.80 Cr3+(aq)+e−→Cr2+(aq) −0.50 Fe3+(aq)+3e−→Fe2+(aq) 0.77 Cr3+(aq)+3e−→Cr(s) −0.73 Cu+(aq)+e−→Cu(s) 0.52 Zn2+(aq)+2e−→Zn(s) −0.76 Cu2+(aq)+2e−→Cu(s) 0.34 Mn2+(aq)+2e−→Mn(s) −1.18 2H+(aq)+2e−→H2(g) 0.00 Al3+(aq)+3e−→Al(s) −1.66 Fe3+(aq)+3e−→Fe(s) −0.036 Mg2+(aq)+2e−→Mg(s) −2.37 Pb2+(aq)+2e−→Pb(s) −0.13 Na+(aq)+e−→Na(s) −2.71 Sn2+(aq)+2e−→Sn(s) −0.14 Ca2+(aq)+2e−→Ca(s) −2.76 Ni2+(aq)+2e−→Ni(s) −0.23 Ba2+(aq)+2e−→Ba(s) −2.90 Co2+(aq)+2e−→Co(s) −0.28 K+(aq)+e−→K(s) −2.92 Cd2+(aq)+2e−→Cd(s)...

  • A solution containing a mixture of metal cations was treated as outlined. Dilute HClHCl was added...

    A solution containing a mixture of metal cations was treated as outlined. Dilute HClHCl was added and no precipitate formed. H2SH2S was bubbled through the acidic solution. A precipitate formed and was filtered off. The pH was raised to about 99 and H2SH2S was again bubbled through the solution. A precipitate formed and was filtered off. Finally, sodium carbonate was added to the filtered solution and no precipitate formed. What can be said about the presence of each of these...

  • A solution containing a mixture of metal cations was treated as outlined. 1. Dilute HCl was...

    A solution containing a mixture of metal cations was treated as outlined. 1. Dilute HCl was added and a precipitate formed. The precipitate was filtered off. 2. H,S was bubbled through the acidic solution. Again, a precipitate formed and was filtered off. 3. The pH was raised to about 9 and H,S was again bubbled through the solution. No precipitate formed. 4. Finally, sodium carbonate was added. A precipitate formed and was filtered off. What can be said about the...

  • A) Use tabulated electrode potentials to calculate ΔG∘ for the reaction. 2K(s)+2H2O(l)→H2(g)+2OH−(aq)+2K+(aq) B) (Refer to the...

    A) Use tabulated electrode potentials to calculate ΔG∘ for the reaction. 2K(s)+2H2O(l)→H2(g)+2OH−(aq)+2K+(aq) B) (Refer to the following standard reduction half-cell potentials at 25∘C: VO2+(aq)+Ni2+(aq)2H+(aq)++2e−e−→ →Ni(s)VO2+(aq) +H2O(l)E∘=−0.23V E∘=0.99V) An electrochemical cell is based on these two half-reactions: Oxidation:Reduction:Ni(s)VO2+(aq,0.024M)+2H+(aq,1.4M)+e−→→Ni2+(aq,1.8M)+2e−VO2+(aq,1.8M)+H2O(l) Calculate the cell potential under these nonstandard concentrations. C) Standard reduction half-cell potentials at 25∘C Half-reaction E∘ (V ) Half-reaction E∘ (V ) Au3+(aq)+3e−→Au(s) 1.50 Fe2+(aq)+2e−→Fe(s) − 0.45 Ag+(aq)+e−→Ag(s) 0.80 Cr3+(aq)+e−→Cr2+(aq) − 0.50 Fe3+(aq)+3e−→Fe2+(aq) 0.77 Cr3+(aq)+3e−→Cr(s) − 0.73 Cu+(aq)+e−→Cu(s) 0.52 Zn2+(aq)+2e−→Zn(s) − 0.76...

  • Which of the following chemical species will undergo an oxidation-reduction reaction with iron metal, Fe(s)? (Please...

    Which of the following chemical species will undergo an oxidation-reduction reaction with iron metal, Fe(s)? (Please show all works and explanations) 12. Which of the following chemical species will undergo an oxidation-reduction reaction with iron metal, Fe (s)? A) platinum, Pt (s) B) zinc bromide, ZnBr2 (a C) auric chloride, AuCl3 (aq) D) silver, Ag (s)

  • Magnesium metal is oxidized and silver ions are reduced in a voltaic cell using: Mg(s) |...

    Magnesium metal is oxidized and silver ions are reduced in a voltaic cell using: Mg(s) | Mg2+ (aq, 1M) || Ag+ (aq, 1M) | Ag(s) a. Draw the voltaic cell. Label each part of the cell. b. Write the equations for the half-reactions occurring at the anode and cathode. c. Write the overall reaction for the cell. d. Draw arrows to show the movement of electrons in the external circuit. Assuming the salt bridge contains NaNO3, indicate the movement of...

  • please help the standrad reduction provided down What is the calculated value of the cell potential...

    please help the standrad reduction provided down What is the calculated value of the cell potential at 298K for an electrochemical cell with the following reaction, when the Cu2+ concentration is 9.24x10 + M and the Cr3+ concentration is 1.47 M? 3Cu2+ (aq) + 2Cr(s) Answer: 3Cu(s) + 2Cr +(aq) v The cell reaction as written above is spontaneous for the concentrations given: What is the calculated value of the cell potential at 298K for an electrochemical cell with the...

  • A voltaic cell is based on the reduction of Ag^+(aq) to Ag(s) and the oxidation of...

    A voltaic cell is based on the reduction of Ag^+(aq) to Ag(s) and the oxidation of Sn(s) to Sn^2+(aq). (a) Write half-reactions for the cell's anode and cathode. Include the phases of all species in the chemical equation. Anode Cathode (b) Write a balanced cell reaction. Include the phases of all species in the chemical equation.

  • Need help with questions 1-5 D&#17 Determine whether each redox reaction occurs spontane- ously in the...

    Need help with questions 1-5 D&#17 Determine whether each redox reaction occurs spontane- ously in the forward direction. (a) Ca2+(aq) + Zn(s)-Ca(s) + Zr"(al) (b) 2 Ag+(aq) + Ni(s)--2 Ag(s) + N产(aq) (c) Fe(s) +Mn2 (aą)- Fe (aą)Mn(s) (d) 2 Al(s) + 3 Pb2+(aq) → 2 AP"(aq) + 3 Pb(s) Suppose you wanted to cause Pb ions to come out of solu- tion as solid Pb. What metal could you use to accomplish this? Make a sketch of an electrochemical...

  • Post Lab Questions: To receive full credit, you must SHOW ALL YOUR WORK!! Use the table...

    Post Lab Questions: To receive full credit, you must SHOW ALL YOUR WORK!! Use the table of reduction half reactions to answer the following Post-Lab questions: Table 3. Example reduction Reduction Potential Chart half reactions. The easiest to Ce+(aq) + 3e - Ce3+ (aq) reduce is at the top. The more Au3+ (aq) + 3e Au(s) difficult to reduce is at the bottom. Cl2(g) + 2e 2CH(ag) Ag+ (aq) + e- Ag(s) Fe3+ (aq) + e- Fe2+ (aq) AgCl(s) +...

ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT