Question

Calculate Eº and K at 25°C for ,02-(aq) +Cr2+ (aq) → Cr3+(aq) + 5,032-(aq) half reactions are 8,062-(aq) + 2e → 28,0,2-(aq),
0 0
Add a comment Improve this question Transcribed image text
Answer #1

Juo Cag) oC) (9) The feschien eha Can be becing C2) fm () fatoeal 2- AG Ai-A Dhsibbshe eaei pbki EUalhe Ga rat be Auahackel dplz thumbs up if found useful

Add a comment
Know the answer?
Add Answer to:
Calculate Eº and K at 25°C for ,02-(aq) +Cr2+ (aq) → Cr3+(aq) + 5,032-(aq) half reactions...
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
  • 10. You want to know the standard reduction potential for the Cr* (aq)|Cr(s) half- reaction Cr3...

    10. You want to know the standard reduction potential for the Cr* (aq)|Cr(s) half- reaction Cr3 + (aq) + 3e-→ Cr(s) y, your textbook only lists the following data for chromium: Cr3+ (aq) + e-→ Cr2+ (aq) Cr2+ (aq) + 2e-→ Cr(s) E' =-0.424 V E' =-0.900 V What is the standard reduction potential for the Cr (a)Cr(s) half-reaction at 25 °C? A)-0.741 V B)-1.324 V C) 0.382V D)-0.141 V E) 0.476 V

  • Calculate the cell potential for this voltaic cell at 25 °C: Cr(s) Cr3 (aq, 0.43 M)...

    Calculate the cell potential for this voltaic cell at 25 °C: Cr(s) Cr3 (aq, 0.43 M) || Cl2 (g, 0.771 atm) CI(aq, 0.133 M), Pt(s) Cas Ze (bp) Cl2(g)+2e 2CI (aq) Cos (aq)+e Co (aq) Co (aq)+2e Co(s) Cr3 (aq)+3e2 Cr(s) Cr3 (aq)+e Cr2 (aq) Cr2 (aq)+ 2e 2Cr(s) +1.358 3+ +1.83 -0.28 -0.744 -0.407 -0.913 AH 0 3e Cr(0H)-(s) C.o 2-(aa)

  • 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) 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...

  • Consider the following galvanic cell at 25°C. Pt | Cr2+ (0.33 M), Cr3+ (2.0 M) ||...

    Consider the following galvanic cell at 25°C. Pt | Cr2+ (0.33 M), Cr3+ (2.0 M) || Co2+ (0.18 M) I Co The overall reaction and equilibrium constant value are given below. 2 Cr2+(aq) + Co2+(aq) → 2 Cr3+ (aq) + Co(s) K = 2.79x107 Calculate the cell potential E for this galvanic cell and AG for the cell reaction at these conditions. E -2.5 ху AG Need Help? Read It

  • From a voltage calculate △ G° AND K for the reaction of Cr2+ (aq) to make...

    From a voltage calculate △ G° AND K for the reaction of Cr2+ (aq) to make Cr3+ (aq) (0.41 volts) with Cr2(aq) to make Cr3+ (aq) (1.33 volts) in acid solution. Group of answer choices G = 1010 KJ and K = e^406 G = -1010 KJ and K = e^406 G = -1010000 KJ and K = e^406 G = -1010 KJ and K = e^-406

  • Calculate AG and K at 25°C for the reactions. a. 7H,0 + 2 Cr3+ + 3Cl2...

    Calculate AG and K at 25°C for the reactions. a. 7H,0 + 2 Cr3+ + 3Cl2 Cr2O72- +6C1° +14H+ Eden 0.03 V AG" KJ K= b. Cu2+ (aq) + Ca(s) = Ca2+ (aq) + Cu(s) €° = 3.10 V AG" kJ K= c. Cl2(9) +2 Br(aq) + Br2(g) + 2 C1-(aq) ' = 0.27 V AG" kJ K = d. 3H2O(1) + 5104- (aq) + 2 Mn²+ (aq) = 5103 - (aq) + 2 MnO4 (aq) + 6H+ (aq) =...

  • Consider a galvanic electrochemical cell constructed using Cr/Cr3* and Zn/Zn2+ at 25 °C. The following half-reactions...

    Consider a galvanic electrochemical cell constructed using Cr/Cr3* and Zn/Zn2+ at 25 °C. The following half-reactions are provided for each metal: Cr3+ (aq) + 3 e → Cr(s) Eºred = -0.744 V Zna*(aq) + 2 e Zn(s) Eºred = -0.763 V What is the standard cell potential for this cell?

  • Calculate the standard Gibbs free energy changes at 25 °C for each of the reactions shown...

    Calculate the standard Gibbs free energy changes at 25 °C for each of the reactions shown below using the Eº values given. Select whether each of these reactions is nonspontaneous, at equilibrium, or spontaneous under standard conditions. (a) 2 Lit(aq) + 2 I'(aq) = 2 Li(s) + 1 12(5) ° = -3.586 V AGC= X kJ/mol nonspontaneous at equilibrium spontaneous (b) 2 CO3+ (aq) + 1 Hg(0) = 2 Co2+ (aq) + 1 Hg2+(aq) ° = 0.988 V AGº =...

  • 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)...

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