HomeworkLib
Question

4. Calculate the cell potential for the half cells Fe3+/Fe2+ and MnO4- /Mn2+, where the Mn...

4. Calculate the cell potential for the half cells Fe3+/Fe2+ and MnO4- /Mn2+, where the Mn process occurs at the cathode, under the following conditions and predict whether the reaction is spontaneous: [Fe3+] = 1.0 M, [Fe2+] = 0.1 M, [MnO4 - ] = 0.01 M [Mn2+] = 1 x 10-4 M [H+ ] = 1 x 10-3 M Fe3+ + e− → Fe2+ +0.700 V MnO4 − (aq) + 8 H+ (aq)+ 5e− → Mn2+ (aq) + 4 H2O(l) +1.49V

science   chemistry   
0 0
Add a comment Improve this question Transcribed image text
Next > < Previous
Sort answers by oldest

Homework Answers

Answer #1

0 0
Add a comment
Know the answer?
Add Answer to:
4. Calculate the cell potential for the half cells Fe3+/Fe2+ and MnO4- /Mn2+, where the Mn...
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

  • Show all steps please :) A galvanic cell is prepared using the following two half-cells: (i)...

    Show all steps please :) A galvanic cell is prepared using the following two half-cells: (i) MnO4 (0.273 M), Mn2+ (0.167 M), and H+ (1.0 M), and (ii) Fe2+ (0.247 M) and Fe3+ (0.389 M). The standard reduction potentials for the two half cells are: Mnoa + 8H+ + 5e – Mn2+ + 4H20 E° = 1.507 V Fe3+ + e- Fe2+ E° = 0.770 V a) Calculate the voltage for this galvanic cell. b) Write the balanced equation for...

  • A galvanic cell is prepared using the following half-cells: (i) MnO4-(0.283M), Mn2+(0.081M), and H+(1.0 M), and...

    A galvanic cell is prepared using the following half-cells: (i) MnO4-(0.283M), Mn2+(0.081M), and H+(1.0 M), and (ii) Sn2+(0.236M) and Sn4+(0.145M). MnO4-+ 8H++ 5e-↔Mn2++ 4H2O Eo= 1.507 V Sn4++ 2e-↔Sn2+Eo= 0.151 V a)Calculate the galvanic cell voltage. b)Write the complete cell description. c)Write the cell reaction indicating clearly the direction of spontaneous reaction. d)Name the electrode (anode or cathode), give the polarity (positive or negative) of the electrode, and state the reaction (oxidation or reduction) occurring in the tin half-cell. Please...

  • An electrochemical cell involves the Au3+/Au and the acidic MnO4-/Mn2+ half-cells. For standard conditions, the anode...

    An electrochemical cell involves the Au3+/Au and the acidic MnO4-/Mn2+ half-cells. For standard conditions, the anode is the gold cell. Calculate the pH (to two decimal places) required to make the gold cell the cathode. Assume an Ecell of 0.01 V. E (Au +) = 1.42V E (8H+ +MnO4-) = 1.55V

  • Some 1L electrolysis cells have a Fe3+ solution with a concentration of 0.0250M Mn2+. It is made up of Mn and Pt electr...

    Some 1L electrolysis cells have a Fe3+ solution with a concentration of 0.0250M Mn2+. It is made up of Mn and Pt electrodes and the next electrochemical reaction occurs in the cell. Mn (s) -> Mn2+ + 2e- Fe3+ + 3e- -> Fe (s) (1) Is Mn electrode anode or cathode (2) If a current of 2.60A is applied to this cell for 18 minutes and 0.504g of Fe metal is released on the surface of the Pt electrode, calculate...

  • Consider the following UNBALANCED reaction IN ACIDIC SOLUTION: Fe2+(aq) + MnO4-(aq) ⟶ Fe3+(aq) + Mn2+(aq) a....

    Consider the following UNBALANCED reaction IN ACIDIC SOLUTION: Fe2+(aq) + MnO4-(aq) ⟶ Fe3+(aq) + Mn2+(aq) a. (5) Species that is oxidized (be specific – i.e. identify which atom and if there are multiple atoms with different charges, identify the correct one) b. (5) Species that is reduced (same instructions as above) c. (10) Full Balanced oxidation ½ reaction (‘full’ means with regard to mass (atoms) and charges.) d. (10) Full Balanced reduction ½ reaction e. (9) Complete Balanced Reaction Using...

  • A voltaic cell is constructed from a standard Cr3Cr half cell (Ered=-0.740V) and a standard Fe3+|Fe2+...

    A voltaic cell is constructed from a standard Cr3Cr half cell (Ered=-0.740V) and a standard Fe3+|Fe2+ half cell (Ered=0.771V). (Use the lowest possible coefficients. Use the pull-down boxes to specify states such as (aq) or (s). If a box is not needed, leave it blank.) The anode reaction is: The cathode reaction is: + The spontaneous cell reaction is: The cell voltage is V.

  • For the following cell:                      Zn | Zn2+ (1.0 M) || Fe3+ (1.0 M), Fe2+ (1.0...

    For the following cell:                      Zn | Zn2+ (1.0 M) || Fe3+ (1.0 M), Fe2+ (1.0 M) | Pt Write the half-reactions and the balanced net reaction for the cell. Identify which species is oxidized. Which is reduced? Which is the reducing reagent? Oxidizing reagent? Draw the electrochemical cell. Label the anode and cathode. Indicate the direction of electron and ion flow. Calculate the standard cell potential. Is the reaction spontaneous?

  • Question 31 4 pts 31. Determine which is reduced agent? MnO4 (aq) +5Fe2(aq) + 8 H(aq)...

    Question 31 4 pts 31. Determine which is reduced agent? MnO4 (aq) +5Fe2(aq) + 8 H(aq) 5 Fe(aq) + Mn2 (aq) + 4H2O(1) Mn in MnO4- H: Fe3+ Fe2+ Question 30 30. Calculate pH of 0.100 M HNO2(aq). K-2.8 x 10-5 9.01 6.21 2.78 5.21

  • Problem Set #34: Galvanic Cells & Standard Reduction Potential Practice Sheet Directions: (a) Sketch the galvanic...

    Problem Set #34: Galvanic Cells & Standard Reduction Potential Practice Sheet Directions: (a) Sketch the galvanic cell based on the following overall reaction OR half- reactions. o Show the direction of electron flow o Identify the cathode and anode (b) Give the overall balanced reaction (c) Determine the Evalues for each of the galvanic cells. (d) Give the standard line notation for each cell. (Assume that all concentrations are 1.0M and that all partial pressures are 1.0 atm) 1. 103(aq)...

  • help me solve this correctly Calculate the value of E for the galvanic cell made from...

    help me solve this correctly Calculate the value of E for the galvanic cell made from these two half reactions when MnO4) = 0.010 M. [H] =0.20 M, [Mn?] = 0.020 M, [Fe3+] = 0.75 M, [Fe2+] = 0.30 M. Ece - OTO nF 75 Q: 1.02m] 6.30m 5 (-010[0.1073747 (0.05 .000 4 MnO4 +8H+ +5e → Mn2+ + 4H20 E° = 1.51 V Fe3+ + + Fe2+ Ecelli Ecell - (8. (8./4.J/mol) (E° = 0.77 V Product S496485) Treactants...

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