Nernst equation for pair NO3 |NO is:RT a(NO, Ja*(H*)E = E° +- InnFa(NO)RT a(NO, )a*(H*)E = E° –InnFa(NO)RT a(NO, )a*(H*)E = E° +nFa(NO)RT a(NO, )a (H')- InnFE = E° +a(NΟ)

Answered: Image /qna-images/answer/41fa2081-2031-4193-a4a9-b6a46ce1a0db.jpg

Nernst equation for pair NO3|NO is: RT a(NO, )a*(H*) E = E° + In nF a(NO) RT a(NO, )a*(H*) E = E° - In nF a(NO) RT a(NO, )a*(H*) E = E° + -lg- nF a(NO) RT n a(NO, )a (H') E = E° + nF a(NO)

Nernst equation for pair NO3|NO is: RT a(NO, )a(H) E = E° + In nF a(NO) RT a(NO, )a(H) E = E° - In nF a(NO) RT a(NO, )a(H) E = E° + -lg- nF a(NO) RT n a(NO, )a (H') E = E° + nF a(NO)

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter18: Electrochemistry

Section: Chapter Questions

Problem 9RQ: What characterizes an electrolytic cell? What is an ampere? When the current applied to an...

See similar textbooks

Similar questions

R = 8.314 mol-K F = 96,485 mol AG = AG° + RT · In(Q) ΔΕ - (E) · In(Q) ΔΕ. Half Reaction (Note: All given as reduction) E° (V) 02 (g) + 4 H*(aq) + 4 e 2 H20 (1) 1.229 Z2 (s) + 2 e 3+ (aq) + 3 e 2z (aq) 0.426 A (s) 0.292 2 H20 (1) + 2 e G2+ (aq) + 2 e M²+ (aq) + 2 e¯ H2 (g) + 2 OH¯ (aq) - 0.828 G (s) - 1.245 M (s) - 1.893 A student constructs a voltaic electrochemical cell with two metal electrodes [metal G and metal A] in their respective aqueous nitrate solutions [G(NO3)2 and A(NO3)3]. Use this information, as well as the reduction potentials in the table above to complete each statement below. Consider the same cell from the above prompt. Calculate A E°. cell for this galvanic cell given the reference information given at the top of the quiz. Report your answer with 3 decimal places. You do not need to report units with your answer.
R = 8.314 mol·K F = 96, 485 moi AG = AG° + RT · In(Q) ΔΕΔΕ-().In(Q) RT Half Reaction (Note: All given as reduction) E° (V) 02 (g) + 4 H*(aq) + 4 e 2 H20 (1) 1.229 - Z2 (s) + 2 e 3+ (aq) + 3 e 2Z (aq) 0.426 → A (s) 0.292 2 H20 (1) + 2 e G2+ (aq) + 2 e H2 (g) + 2 OH (aq) - 0.828 G (s) - 1.245 > M2+ (aq) + 2 e M (s) - 1.893 A student constructs a voltaic electrochemical cell with two metal electrodes [metal G and metal A] in their respective aqueous nitrate solutions [G(NO3)2 and A(NO3)3]. Use this information, as well as the reduction potentials in the table above to complete each statement below. Consider the same cell from the above prompt. If the A 3+ (aq) solution is diluted to 0.100 M and the G<* (aq) solution is diluted to 2+ 0.500 M, what is the initial value of Q under these new conditions? Report your answer with the correct significant figures.
A galvanic cell is powered by the following redox reaction: 3 Cu (aq) + Cr(OH);(s) + 5 OH (aq) → 3 Cu" (aq) + Cro (aq) + 4H,0(1) Answer the following questions about this cell. If you need any electrochemical data, be sure you get it from the ALEKS Data tab. Write a balanced equation for the half-reaction that takes place at the cathode. Write a balanced equation for the half-reaction that takes place at the anode. Calculate the cell voltage under standard conditions. E = Ov Round your answer to 2 decimal places.
3c₂2- 1 Tap here or pull up for additional resources + 2 (s) OH- Complete and balance the following redox reaction in acidic solution H₂O₂(aq) + Cr₂O72- (aq) → O₂(g) + Cr³+ (aq) 3 O Reset e (1) H* Question 14 of 14 e 5 LO Cr 6 MacBook Air H 7 个 • x H₂O 8 H₂O* 9 (aq) H₂O O Submit
R = 8.314 mol·K F = 96,485 mot AG = AG° + RT · In(Q) ΔΕΔΕ-).In(O) Half Reaction (Note: All given as reduction) E° (V) 02 (g) + 4 H*(aq) + 4 e¯ → 2 H20 (1) 1.229 2 z" (aq) Z2 (s) + 2 e 3+ (aq) + 3 е 0.426 A (s) 0.292 2 H20 (1) + 2 e G2+ (aq) + 2 e H2 (g) + 2 OH (aq) - 0.828 G (s) - 1.245 > M2+ (aq) + 2 e - 1.893 → M (s) A student constructs a voltaic electrochemical cell with two metal electrodes [metal G and metal A] in their respective aqueous nitrate solutions [G(NO3)2 and A(NO3)3]. Use this information, as well as the reduction potentials in the table above to complete each statement below. Consider the same cell from the above prompt. Calculate AG° in kJ/mol for this galvanic cell. Report your answer with 4 significant figures. You do not need to report units with your answer. If your value is negative, make sure to include a "-" symbol.
1:41 PM Mon Apr 3 Complete and balance the following redox reaction in acidic solution. Be sure to include the proper phases for all species within the reaction. 2+ Mn²+ (aq) + BiO3(aq) →MnO4 (aq) + Bi³+ (aq) 3C₂ 1)[ 2- + Question 3 of 24 (s) H₂O 3 4 2 ( Reset (1) 8 507 90 (g) Submit ㅈ H H3O+ O X = (aq) H* Mn
Find the AHn for the following reaction: 20F,(g) + 2 S(s) → SO2(9) + SF4(9) given the following reactions and subsequent AH° values $$ OF2 (9) + H2O(1) → O2 (g) + 2 H F(g) SF.(9) + 2 H2O(I) → 4 H F (g) + S02(g) S(s) + 0,(g) → SO2(g) AH° -272.8 kJ ΔΗ' = -827 kJ AH° = -292.5 kJ $$
1. What is the conjugate base of the formic acid HCO,H? 2. You have a 5.43 x 10 M solution of HNO, at 25 °C. Calculate the [OH] of this solution? 3. A solution that has a pH of 4.5 is (acidic, basic, neutral)? 4. A solution has a hydronium ion concentration of 3.2 x 10 M. Calculate the pOH.
Complete and balance the following redox reaction in acidic solution NO2(g) - NO, (aq) + NO(g) 04-03-D]2-| D- D2+ 03+ D4+ 1 2 3 4 5 6 7 8 9 O, 0 0, 0, 0. 12 (s) (g) (aq) + H2O OH- e- N H* H Reset • x H,O Delete
Pt(s) |X ELECTROCHEMISTRY: Pt(s) |X"(aq). X2+ (aq)I/ Y3+ Given the line notation: (aq) I Y(s) where: x2+ + e x+ E° = 0.85 V -> y3+ + 3 e → Y E° = -2.44 V P4+ + 4 e - Pt E° = 1.20 V Calculate the EMF. Input values only with 2 decimal places. Do not include the unit. If positive value: DO NOT include the positive (+) sign If negative value: Please include the negative (-) sign
er Reactions STATION 3 - NERNST -0.14 -1.66 in a EQUATION 1. Fe(s) + Cu²+(aq) → Fe2+(aq) + Cu(s) E°= +0.78 V a) What is n? moles b) If [Cu2+] = 0.10 M and [Fe2+] = 1.5 M, Q= [ ] [ ] c) Calculate the Ecell. x+1.00 NO3)2 and Al in 0.25 M Al(NO3)3 at 25°C.
Calculate the f.e.m. obtained considering the 2 following reduction reactions 2H+(aq) + 2e- → H2(g) Eo = 0 Cu2+(aq) + 1e- → Cu+(aq) Eo = 0.15 Eocell ← please insert your value