ZuoTi.Pro
Question

Another lab, similar to yours, had the following results, with the reference +2 Ni (1M)+2e Ni. half-cell being: The results o
science   chemistry   
0 0
Add a comment Improve this question Transcribed image text
Next > < Previous
Sort answers by oldest

Homework Answers

Answer #1

1) Most positive to most negative: Au3+/Au, Fe3+/Fe2+, Fe3+/Fe, Ni2+/Ni, Cr3+/Cr, Al3+/Al, Ba2+/Ba

Explanation: The order of the electrode potentials in the same order: 1.498, 0.771, -0.037, -0.257, -0.744, -1.662, -2.912 V

2) Combination (i).

Oxidation 1/2 reaction: Fe2+(aq) ----> Fe3+(aq) + e-

Reduction 1/2 reaction: Au3+(aq) + 3e- ----> Au(s)

Total net ionic equation: 3Fe2+(aq) + Au3+(aq) ----> 3Fe3+(aq) + Au(s)

Eocell = EoAu3+/Au - EoFe3+/Fe2+= 1.498 - 0.771 = 0.727 V

Combination (ii).

Oxidation 1/2 reaction: Fe(s) ---> Fe3+(aq) + 3e-

Reduction 1/2 reaction: Au3+(aq) + 3e- ----> Au(s)

Total net ionic equation: Fe(s) + Au3+(aq) ----> Fe3+(aq) + Au(s)

Eocell = EoAu3+/Au - EoFe3+/Fe= 1.498 - (-0.037) = 1.535 V

Combination (iii).

Oxidation 1/2 reaction: Ni(s) ---> Ni2+(aq) + 2e-

Reduction 1/2 reaction: Au3+(aq) + 3e- ----> Au(s)

Total net ionic equation: 3Ni(s) + 2Au3+(aq) ----> 3Ni2+(aq) + 2Au(s)

Eocell = EoAu3+/Au - EoNi2+/Ni= 1.498 - (-0.257) = 1.755 V

Combination (iv).

Oxidation 1/2 reaction: Cr(s) ---> Cr3+(aq) + 3e-

Reduction 1/2 reaction: Au3+(aq) + 3e- ----> Au(s)

Total net ionic equation: Cr(s) + Au3+(aq) ----> Cr3+(aq) + Au(s)

Eocell = EoAu3+/Au - EoCr3+/Cr= 1.498 - (-0.744) = 2.242 V

Combination (v).

Oxidation 1/2 reaction: Al(s) ---> Al3+(aq) + 3e-

Reduction 1/2 reaction: Au3+(aq) + 3e- ----> Au(s)

Total net ionic equation: Al(s) + Au3+(aq) ----> Al3+(aq) + Au(s)

Eocell = EoAu3+/Au - EoAl3+/Al= 1.498 - (-1.662) = 3.16 V

Combination (vi).

Oxidation 1/2 reaction: Ba(s) ---> Ba2+(aq) + 2e-

Reduction 1/2 reaction: Au3+(aq) + 3e- ----> Au(s)

Total net ionic equation: 3Ba(s) + 2Au3+(aq) ----> 3Ba2+(aq) + 2Au(s)

Eocell = EoAu3+/Au - EoBa2+/Ba= 1.498 - (-2.912) = 4.41 V

0 0
Add a comment
Know the answer?
Add Answer to:
Another lab, similar to yours, had the following results, with the reference +2 Ni (1M)+2e Ni....
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Log In

Sign Up

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

  • Another lab, similar to yours, had the following results, with the reference half-cell being: Nit2(1M) +...

    Another lab, similar to yours, had the following results, with the reference half-cell being: Nit2(1M) + 2e → Ni. The results obtained, with each half-cell combined, at 20.00 °C, were: + 1.203 v + - 1.206 v Alt + 36 → Al Aut? + 3€ → Au Bat2 + 2e → Ba or +3 + 36 → or $ + +2.017 v $ + +0.456 v $ + - 0.924 v Fe +3 + 36 Fe -0.198 v Ni+21 x...

  • use tabulated standard electrode potential to calculate the standard cell potential for the reaction occurring in an el...

    use tabulated standard electrode potential to calculate the standard cell potential for the reaction occurring in an electrochemical cell at 25 C. (The equation is balanced.) 3Ni^2+(aq)+2Cr(s)--->3Ni(s)2Cr^3+(aq) Express your answer to two significant figures and include the appropriate units. em 26 E (V) -0.45 -0.50 -0.73 -0.76 -1.18 Standard reduction half-cell potentials at 25°C Half-reaction E° (V) Half-reaction Aul+ (aq) + 3e +Au(s) 1.50 Fe2+ (aq) + 2eFe(s) Ag+ (aq) +e-Ag(s) 0.80 Cr3+ (aq) + Cr²+ (aq) Fe+(aq) + 3e...

  • 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)...

  • please answer all wueston Question 33 (3 points) (33) For a galvanic cell notation: (-) Ni...

    please answer all wueston Question 33 (3 points) (33) For a galvanic cell notation: (-) Ni /Ni2+ (aq) // Au3+ (aq) / Au (+), The electrode potentials: Eº (Ni2+/ Ni) = -0.257 V, E ° (AU3+ / Au ) = 1.498 V. The cell potential Eºcell (a) 1.241 V (b) - 1.755 V O (0) 1.755 V (d) 1.498 Why Question 35 (3 points) (35) For galvanic cell: (-) Cr/ Cr3+ (aq) / Cu2+ (aq) / Cu(+), the correct half...

  • 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...

  • 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...

  • Standard reduction half-cell potentials at 25°C E (V) E (V) 1.50 -0.45 0.80 -0.50 0.77 -0.73...

    Standard reduction half-cell potentials at 25°C E (V) E (V) 1.50 -0.45 0.80 -0.50 0.77 -0.73 0.52 -0.76 0.34 -1.18 Half-reaction Aut (aq) + 3e +Au(s) Ag+ (aq) + +Ag(s) Fe3+ (aq) +34 Fo+ (aq) Cut(aq) + Cu(s) Cu²+ (aq) + 2e +Cu(s) 2H+ (aq) - 2e +H2 (6) Fe3+ (aq) + 3e Fe(s) Pb2+ (aq) + 2e →Pb(s) Sn-(aq) + 2e +Sn(s) Ni2+ (aq) + 2e →Ni(s) Co2(aq) +2e + Co(s) ca? (aq) + 2e +Cd(s) 0.00 Half-reaction Fe(aq)...

  • Using standard reduction potentials from the ALEKS Data tab, calculate the standard reaction free energy AG...

    Using standard reduction potentials from the ALEKS Data tab, calculate the standard reaction free energy AG for the following redox reaction Round your answer to 3 significant digits. 2CH(OH), (-) +100H(aq) + 3Br (1) ► 2010 (92) +8H,O()+6Be (aq) cil Data E (V) 0.7996 x 5 ? -1.675 1.692 1.498 -2.912 1.066 1.35827 Half-Reaction Agt (aq) + e -- Ag (s) AP+ (aq) + 3e" -Al(s) Aut (aq) + e - Au (s) Au3+ (aq) + 3e" - Au (5)...

  • Half-reaction Cr3+ (aq) + 3e--Cr(s) Fe2+ (aq) + 2e - Fe(s) Fe3+ (aq) - Fe2+ (5)...

    Half-reaction Cr3+ (aq) + 3e--Cr(s) Fe2+ (aq) + 2e - Fe(s) Fe3+ (aq) - Fe2+ (5) Sn+ (aq) + 2e - Sn2(aq) E (V) -0.74 -0.440 +0.771 +0.154 1. Calculate the standard cell potential for the voltaic cell based on the reaction below, given the table above: 35nt(s) + 2Cr (s) - 2 C (s) + 3 Sn () ANSWER: 2. Calculate the standard cell potential for the voltaic cell based on the reaction below, Riven the table above 3Feb...

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