Question

Calculate Ecell (the reduction potential for the balanced chemical equation under nonstandard conditions) for the electrochemical...

Calculate Ecell (the reduction potential for the balanced chemical equation under nonstandard conditions) for the electrochemical cell shown below: Pt | Fe2+ (4.25x10-3 M), Fe3+ (1.50x10-3 M) || MnO4 - (6.50x10-3 M), Mn2+ (2.00x10-2 M), H+ (0.100 M) | Pt. Use the following standard reduction potentials for each half reaction. The Eo for Fe3+/Fe2+ = 0.77 V and Eo for MnO4 - /Mn2+ = 1.51 V.

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

Pt /pt50), FeC50OH (eso Co-lcchy Anodic al Oeaclse calladic Lal dearby Fe +le Cag Fe2t CaL Fansele 0.77V Fablede 51V Ecell al4. CFett150NO Cana(650AoM CHO 00KID CHJ Co. 100 H) 5 e: C1-50x10)2.00 xio C6. 50x0)425 o (o100 .15 = 7.59375 XIO x2.00 X10 1

Add a comment
Know the answer?
Add Answer to:
Calculate Ecell (the reduction potential for the balanced chemical equation under nonstandard conditions) for the electrochemical...
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
  • A voltaic cell consists of an Al/Al3+ half-cell and an Fe2+/Fe3+ half-cell. Calculate Ecell when conc....

    A voltaic cell consists of an Al/Al3+ half-cell and an Fe2+/Fe3+ half-cell. Calculate Ecell when conc. of Fe2+= 0.402 M, Fe3+= 0.073 M and Al3+= 0.239 M. Use the reduction potentials for Al3+ is -1.66 V and for Fe3+ is 0.77 V. How do you decide which is oxi, red.

  • 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...

  • Calculate the potential of the electrochemical cell and determine if it is spontaneous as written at...

    Calculate the potential of the electrochemical cell and determine if it is spontaneous as written at 25 °C Cu(s) Cu2+ (0.15 M) Fe2+ (0.0039 M) Fe(s) E =-0.440 V E+Cu = 0.339 V Fe2+/Fe Is the electrochemical cell spontaneous or not spontaneous -0.779 Ecell = as written at 25 C? not spontaneous spontaneous о Calculate the potential of the electrochemical cell and determine if it is spontaneous as written at 25 'C Pt(s) Sn2 (0.0024 M), Sn4+ (0.12 M) |...

  • Calculate the potential of the electrochemical cell and determine if it is spontaneous as written at...

    Calculate the potential of the electrochemical cell and determine if it is spontaneous as written at 25 °C. Cu(s) Cu2+(0.14 M) | Fe2+(0.0044 M) Fe(s) Ecu?+Icu = 0.339 V Efez lfe = -0.440 V Ecell = v Is the electrochemical cell spontaneous or not spontaneous as written at 25 °C? O not spontaneous O spontaneous Calculate the potential of the electrochemical cell and determine if it is spontaneous as written at 25 °C. Pt(s) Sn2+(0.0048 M), Snº+(0.11 M) || Fe3+(0.12...

  • 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...

  • Calculate the potential of the electrochemical cell and determine if it is spontaneous as written at 25 °C Cu(s) Cu2 (0...

    Calculate the potential of the electrochemical cell and determine if it is spontaneous as written at 25 °C Cu(s) Cu2 (0.12 M |Fe2 (0.0012 M) Fe(s) E2 =-0.440 V Efe/Fe = 0.339 V Cu2t/Cu Is the electrochemical cell spontaneous or not spontaneous Ecell V as written at 25 °C? not spontaneous spontaneous Calculate the potential of the electrochemical cell and determine if it is spontaneous as written at 25 °C. Pt(s) Sn2(0.0060 M), Sn4+(0.14 M) Fe3+(0.13 M), Fe2+(0.0056 M) Pt(s)...

  • Consider the following electrochemical cell: Pt | Cu2+ (aq) | Cu+ (aq) || Fe2+ (aq) |...

    Consider the following electrochemical cell: Pt | Cu2+ (aq) | Cu+ (aq) || Fe2+ (aq) | Fe3+ (aq)| Pt The cell is constructed by preparing one half-cell with a solution containing 8.33 * 10-3 M FeCl3 and 1.67*10-2 M FeCl2, and preparing the other half-cell with a solution containing 8.33*10-3 M CuCl2 and 0.025 M CuCl. Calculate the cell potential once the half-cells are connected to each other. E°(Cu2+/Cu+) = 0.16 V; E°(Fe2+/Fe3+) = 0.77 V

  • Calculate the potential of the electrochemical cell and determine if it is spontaneous as written at...

    Calculate the potential of the electrochemical cell and determine if it is spontaneous as written at 25 ∘C . Pt(s) ∣∣ Sn2+(0.0022 M),Sn4+(0.13 M) ‖‖ Fe3+(0.11 M),Fe2+(0.0014 M) ∣∣ Pt(s) ?∘Sn4+/Sn2+=0.154 V?∘Fe3+/Fe2+=0.771 V ?cell=?

  • 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)...

  • Electrochemistry - Equilibrium 1. The electrochemical cell described by the balanced chemical equation has a standard...

    Electrochemistry - Equilibrium 1. The electrochemical cell described by the balanced chemical equation has a standard emf (electromotive force) of -0.08 V. Calculate the value (J) for the Wmax that the cell can do under standard conditions. Round your answer to 3 significant figures. St. Red. Pot. (V) Faraday's Constant Hg2+/Hg +0.85 F = 96485 C Fe3+/Fe2+ +0.77 2Fe3+(aq) + Hg(l) → 2Fe2+(aq) + Hg2+(aq) 2. The voltaic cell described by the balanced chemical equation has a standard emf of...

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