ChemistryANSWER= S= -70.7 J/K, H= -112. Question 8For the reaction Ag (a + I)→ Agl (s).(a) Calculate AG° for the above reaction at 298 K.AG= nFE ceyQG₁² = -(imol X(96₁485) (0.951)AG= 917572 Jonoga suneva sbancE 298 +0.951 V and K128= 1.88 x 10¹14=COMGM 19022!ins8 Isvonoisemitno(b) Calculate AH° and AS° for this reaction at 298 K. Assume that AH° and AS are independent of temperature.91.7kJTOLOMS tids G: 19dmun nolissimorituA28ae00uoy InsrT-bsvonqqA2 1 1 yoo zint nisteR-306310mlapso amor invM of mutq96.TYAM :01 nes

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For the reaction Ag'(a) + I)→ Agl (), (a) Calculate AGº for the above reaction at 298 K. AG FE сеч vonopA suneva E 298 +0.951 V and K328 = 1.88 × 10¹4 como 1922 ins8 Isvon Das G₁ = -(1malX96,485)(0.951) AGF 917572 J = 91.7KJ (b) Calculate AH° and AS° for this reaction at 298 K. Assume that AH° and AS° are independent of temperature. TOLOMs sidst :19dmun nolissimoritu Paaed :noissmitno

For the reaction Ag'(a) + I)→ Agl (), (a) Calculate AGº for the above reaction at 298 K. AG FE сеч vonopA suneva E 298 +0.951 V and K328 = 1.88 × 10¹4 como 1922 ins8 Isvon Das G₁ = -(1malX96,485)(0.951) AGF 917572 J = 91.7KJ (b) Calculate AH° and AS° for this reaction at 298 K. Assume that AH° and AS° are independent of temperature. TOLOMs sidst :19dmun nolissimoritu Paaed :noissmitno

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 3RQ: Table 17-1 lists common half-reactions along with the standard reduction potential associated with...

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Table 17-1 lists common half-reactions along with the standard reduction potential associated with each half-reaction. These standard reduction potentials are all relative to some standard. What is the standard (zero point)? lf is positive for a half-reaction, what does it mean? If is negative for a half-reaction, what does it mean? Which species in Table 17-1 is most easily reduced? Least easily reduced? The reverse of the half-reactions in Table 17-1 are the oxidation half-reactions. How are standard oxidation potentials determined? In Table 17-1, which species is the best reducing agent? The worst reducing agent? To determine the standard cell potential for a redox reaction, the standard reduction potential is added to the standard oxidation potential. What must be true about this sum if the cell is to be spontaneous (produce a galvanic cell)? Standard reduction and oxidation potentials are intensive. What does this mean? Summarize how line notation is used to describe galvanic cells.
Calculate AG° f( in kJ) or the reaction of iron(II) ions with permanganate ions. (F =96500 C) MnO4 (aq) + 8H*(aq) + 5e Fe3+(aq) + e Mn2+(ag) + 4H,O() E° = 1.51 V E°= 0.77 V → Fe2+(aq) -357.05
J R = 8.314 mol·K F = 96,485 mol AG = AG° + RT · In(Q) ΔΕ ΔΕ-).In(Q) 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 2z° (aq) 0.426 (aq) + 3 е A (s) 0.292 2 H20 (1) + 2 e G2* (aq) + 2 e → H2 (g) + 2 OH¯ (aq) - 0.828 2+ 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. The number of electrons transferred per mole of reaction in this cell is.. Your answer should be a number with no units. Question 6 Diluting the half-cell solution will decrease the measured cell potential. o cathode O anode
J R = 8.314 mol·K F = 96,485 mol AG = AG° + RT · In(Q) ΔΕ ΔΕ-). In(Q) 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 → 2Z° (aq) 0.426 3+ (aq) + 3 е A' → A (s) 0.292 2 H20 (I) + 2 e H2 (g) + 2 OH¯ (aq) - 0.828 G2+ (aq) + 2 e G (s) - 1.245 (aq) + 2 e¯ → 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 MMZ2 (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 11 The electrode connected to the red wire (negative terminal on the battery) will... o produce bubbles and the pH will increase. o produce M (s) on the electrode and the pH will not change. o produce bubbles and the pH will decrease. o produce M (s) on the electrode and the pH will decrease. Question 12 The electrode…
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.
At 298 K, AG° = +20.9 kJ/mol for the reaction CH;CO,H (aq) = H"(aq) + CH;CO, Calculate AG for this reaction if the [H"]= 3.19 x 10³, [CH;CO,]= 7.67 x 10, and [ CH;CO,H] = 7.47 x 10-5.
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 reaction at 400 k for the reaction Co(g) +2H2(g) → CH;OH (g) At 298.15 K, AH?.co )= -26.41 kcal/mol, AH?.CH,OH()= -48.08 kcal/mol, AG.co )= -32.8079 kcal/mol, AG.CH,OH(@)= -38.69 kcal/mol, The standard heat capacity of various components is given by. CS = a + bT + cT2 + dT³, where C is in cal/mol-K and T is in K |Component b x10 e x105 d x10° a -0.291 CH3OH CO 4.55 2.186 -1.92 6.726 0.04 0.1283 -0.5307 H2 6.952 -0.0457 0.09563 -0.2079
ASAP
For the reaction A + B* - A* + B = -2,4 V Which of these conclusions is TRUE at standard conditions? (A) Equilibrium lies far on the reactant side (B) K is a large number (C) The reactant A is reduced (D) AG° is negative (E) The reactant B* is the reducing agent Submit
Use data to compute AG° at 89.0°C for the following reaction. 3N2H4() + 403(g) → 6NO(g) + 6H2O(1) AH°¢ (N2H4(1)) = 50.6 kJ mol 1 AH°t (O3(g)) = 142.7 kJ mol 1 AH°; (NO(g)) = 91.3 kJ mol-1 AH°f (H2O(1)) = -285.83 kJ mol1 AS° (N2H4()) = 121.2 J mol-1 K-1 AS° (O3(g)) = 238.9 J mol-1 K1 AS° (NO(g)) = 210.8 J mol-1 K-1 AS° (H20(1)) = 69.95 J mol-1 K-1 AG°f (N2H4(1)) = 149.3 kJ mol 1 AG°t (O3(g)) = 163.2 kJ mol1 AG°f (NO(g)) = 87.6 kJ mol-1 AG°t (H2O(1)) = -237.1 kJ mol 1 %3D %3D %3D AG° = i kJ
If the value of E°cell is 0.59 V for the reaction C12(g) + 2FE2+(aq) → 2FE3+(aq) + 2Cl-(aq), what is the value of E°cell for 2Cl-(aq) + Fe3+(aq) → Fe2+(aq) + ½C12(g)? -0.30 ?