R = 8.314mol·KF = 96,485 motAG = 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)1.2292 z" (aq)Z2 (s) + 2 e3+(aq) + 3 е0.426A (s)0.2922 H20 (1) + 2 eG2+ (aq) + 2 eH2 (g) + 2 OH (aq)- 0.828G (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 tocomplete 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 youranswer. If your value is negative, make sure to include a "-" symbol.

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Calculate AG° in kJ/mol for this galvanic cell. Report vOur answer with 4 significant figures You do

Chemistry: Principles and Practice

3rd Edition

ISBN:9780534420123

Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer

Publisher:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer

Chapter18: Electrochemistry

Section: Chapter Questions

Problem 18.103QE

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Actually, the carbon in CO2(g) is thermodynamically unstable with respect to the carbon in calcium carbonate(limestone). Verify this by determining the standardGibbs free energy change for the reaction of lime,CaO(s), with CO2(g) to make CaCO3(s).
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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.
b) Determine the standard enthalpy change and std. Gibbs free energy change of D reaction at 400 k for the reaction CO(g) +2H2(g) → CH;OH (g) At 298.15 K, AH CO (9)= -26.41 kcal/mol, AH.CH,0H(9)= -48.08 kcal/mol, (9)= -32.8079 kcal/mol, AG CH30H(g)= -38.69 kcal/mol, The standard heat capacity of various components is given by, C = a + bT + cT2 + dT³, where C is in cal/mol-K and T is in K b x10² c x105 d x10º Сomponent CH3OH a 4.55 2.186 -0.291 -1.92 CO 6.726 0.04 0.1283 -0.5307 H2 6.952 -0.0457 0.09563 -0.2079
R = 8.314 mol·K F = 96,485 mol AG° + RT · In(Q) AE° () · In(Q) AG ΔΕ %3D Half Reaction (Note: All given as reduction) E° (V) 02 (g) + 4 H*(aq) + 4 e → 2 H20 (I0) 1.229 Z2 (s) + 2 e - → 2Z (aq) 0.426 (aq) + 3 e A (s) 0.292 2 H20 (1) + 2 e G2+ (aq) + 2 e M2+ (aq) + 2 e Н2 (в) + 2 ОН" (aq) - 0.828 G (s) - 1.245 M (s) - 1.893 Using a U-tube, a student sets up a non-spontaneous electrochemical cell with a battery connected to two carbon electrodes that are submerged in 1 M MZ2 (aq) solution (M is a metal and Z is an anion composed of the newly discovered element Z). Use the reference information given in the table above to answer the following three questions. Question 10 Oxidation will occur at the while reduction will occur at the Possible answers are "anode" and "cathode".
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How long would it take (in minutes) to reduce 1 mole of each of the following ions using the current indicated? (Ampere (A) = Coulomb/second (C/s), I = C/s, F= 96500 C/mol e-) (C = Charge unit Coulomb, n = electron mole number) Q=I×t=n×F(a) Fe3+ , 2.344 A (b) Cu2+ , 25.260 A
(a) In the electrolysis of aqueous NaClO3, how many liters of O₂(g) (at STP) are generated by a current of 53.9 A for a period of 73.9 min? The unbalanced chemical reaction representing this electrolysis is shown below. NaClO3(aq) + H₂O(1) <→ Cl₂(g) + O₂(g) + NaOH(aq) 7.93 X liters of O₂(9) is generated by this electrolysis. (b) How many moles of NaOH(aq) are formed in the solution in this process? 5.614 x moles of NaOH(aq) are formed.
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