Question

Find the best combination of half-cell pair from the following list, which will give the highest...

Find the best combination of half-cell pair from the following list, which will give the highest voltage. What is the voltage for that Galvanic cell?

Given that Reduction Half-reaction Standard Potential (Eredo)

  Zn2+(aq) + 2e → Zn(s) -0.763 (V)

Fe2+(aq) + 2e → Fe(s) -0.44 (V)

Cu2+(aq) + 2e → Cu(s) +0.34 (V)

Sn2+(aq) + 2e → Sn(s) -0.14 (V)

Cu2+(aq) + e → Cu+(aq) + 0.153 (V)

Ag+(aq) + e → Ag(s) + 0.80 (V)

  Cu+(aq) + e → Cu(s) + 0.52 (V)

0 0
Add a comment Improve this question Transcribed image text
Answer #1

Higher the reduction potential, mou is the tendency to reduce s lower the reduction podential, more is the tendency to oxidiz

Add a comment
Know the answer?
Add Answer to:
Find the best combination of half-cell pair from the following list, which will give the highest...
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
  • Using the information in the table: Which combination of metals, if used to create an electrochemical...

    Using the information in the table: Which combination of metals, if used to create an electrochemical cell, would produce the largest voltage? Liu lur the reaction between Zn and Cu2+ ions is 1.1030 V, we can use the known value for the half-cell potential for zinc to determine the half-cell potential for copper: Zn(s) → Zn2+(aq) + 2e + Cu2+(aq) + 2e → Cu(s) Zn(s) + Cu2+ (aq) → Zn2+ (aq) + Cu(s) E half-cell = 0.7628 V Eºhalf-cell =...

  • Half-Reaction Fe2+(aq) + 2e Fe(s) Hg2+ (aq) + 2e Hg() Ag+ (aq) + e + Ag(s)...

    Half-Reaction Fe2+(aq) + 2e Fe(s) Hg2+ (aq) + 2e Hg() Ag+ (aq) + e + Ag(s) Cu2+ (aq) + 2e + Cu(s) Zn2+ (aq) + 2e → Zn(s) E (V) -0.44 0.86 0.80 0.34 - 0.76 Using the table, calculate Eºcell for the following electrochemical cell under standard conditions voltmeter a) 1.24 V Fe. salt bridge Ag b) -1.24 V c) 2.04 V d) - 2.04 V Ag a b С

  • Separate galvanic cells are made from the following half-cells: cell 1: H+(aq)/H2(g) and Pb2+(aq)/Pb(s) cell 2:...

    Separate galvanic cells are made from the following half-cells: cell 1: H+(aq)/H2(g) and Pb2+(aq)/Pb(s) cell 2: Fe2+(aq)/Fe(s) and Zn2+(aq)/Zn(s) Which of the following is correct for the working cells? Standard reduction potentials, 298 K, Aqueous Solution (pH = 0): Cl2(g) + 2e --> 2C1-(aq); E° = +1.36 V Fe3+(aq) + e --> Fe2+(aq); E° = +0.77 V Cu2+(aq) + 2e --> Cu(s); E° = +0.34 V 2H+(aq) + 2e --> H2(g); E° = 0.00 V Pb2+(aq) + 2e --> Pb(s);...

  • Using the table below: 19. Three combinations of metals are listed below, which combination would produce...

    Using the table below: 19. Three combinations of metals are listed below, which combination would produce the largest voltage if they were used to construct an electrochemical cell? Copper (Cu) with zinc (Zn) Lead (Pb) with zinc (Zn) Lead (Pb) with cadmium (Cd) Liu lur the reaction between Zn and Cu2+ ions is 1.1030 V, we can use the known value for the half-cell potential for zinc to determine the half-cell potential for copper: Zn(s) → Zn2+(aq) + 2e +...

  • a • Which of these half-cells would combine with the SHE to give the largest possible...

    a • Which of these half-cells would combine with the SHE to give the largest possible Eºcell value? • Fe3+/Fe, Ni2+/Ni, Cu2+/Cu = = = Zn2+ (aq) + 2e Cr+ (aq) + 3e Fe2+ (aq) + 2 Cd + (aq) + 2e Ni”* (aq) + 2e Sn²+ (aq) + 2e Pb2+ (aq) + 2 Fe+ (aq) + 3 2H(aq) + 2 Sn+ (aq) + 2e Cu²+ (aq) + e Cu²(aq) + 2e Zn(s) Cr(s) Fe(s) Cd(s) Ni(s) Sn(s) Pb(s) Fe(s)...

  • I need help with questione 1-12 and discussion question 1 and 2. The previous pictures help...

    I need help with questione 1-12 and discussion question 1 and 2. The previous pictures help determine the chart. Please Show Work thank you so much An oxidation half-reaction is characterized by electrons appearing on the product side. The oxidation of aluminum for instance would be represented thusly: Al(s) → Al3+ + 3e- (1) An reduction half-reaction is characterized by electrons appearing on the reactant side. The reduction of ferrous iron for instance would be represented thusly: Fe2+ + 2e...

  • 12. Using two half reactions that have NEGATIVE standard reduction potentials results results in a battery...

    12. Using two half reactions that have NEGATIVE standard reduction potentials results results in a battery that... Reduction Half-Reaction F2(g) + 2e →2F(aq) S2082 (aq) + 2e- → 25042 (aq) O2(g) + 4H+ (aq) + 4e → 2H2O(1) Br2(1) + 2e + 2Br (aq) Agt(aq) + e → Ag(s) Fe3+ (aq) + e- → Fe2+ (aq) 126) + 2e → 21 (aq) Cu2+ (aq) + 2e → Cu(s) Sn4+ (aq) + 2e → Sn2+ (aq) S(s) + 2H+ (aq) +...

  • 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...

  • 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)...

  • 4. Write a balanced equation from each cell notation, then calculate thee of the cell, is...

    4. Write a balanced equation from each cell notation, then calculate thee of the cell, is it spontaneous or nonspontaneous? A. Cr(s) Crot || Cui Cu(s) B. Al(s) Al* | Ce Ce Pt C. Cu (s) Cu? || Al Al (s) Useful information: F= 96485 J/V mol electrons, AG ----F-Ecell AG --RTINK: Ece 0.0592/n*logK; EcellEcett -0.0592/n*logQ Ce (aq) + Ag+ (aq) + e Fet (aq) + Cu" (aq) + Cu (aq) + 2e 2H(aq) + 2 Pb (aq) + 2e...

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