4-ele The free energy change for the following reaction at 25 °C, when [Cr³*] = 4.83×10-³ M and [Hg²+] = 1.13 M, is 937 kJ: 2Cr*(4.83×103 M) + 3Hg(1)- $2Cr(s) + зHg2*(1.13 M) ДG%3 937 kJ What is the cell potential for the reaction as written under these conditions? Answer: V Would this reaction be spontaneous in the forward or the reverse direction v forward reverse

4-ele The free energy change for the following reaction at 25 °C, when [Cr³] = 4.83×10-³ M and [Hg²+] = 1.13 M, is 937 kJ: 2Cr(4.83×103 M) + 3Hg(1)- $2Cr(s) + зHg2*(1.13 M) ДG%3 937 kJ What is the cell potential for the reaction as written under these conditions? Answer: V Would this reaction be spontaneous in the forward or the reverse direction v forward reverse

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

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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[References) What are the reaction quotient (Q) and the calculated value of the cell potential (Ecell) at 298 K for an electrochemical cell with the following reaction, when the Hg2+ concentration is 7.35x104 M and the Mg²* concentration is 1.25 M? Hg"(aq) + Mg(s) Hg(l) + Mg²+(aq) E°cell = 3.225 V Q= Ecel =|
1) Consider the following electrochemical cell. S.C.E. (satd. KCI) || CuCl2 (aq) (A=0.0046 M) / Cu (s) Hg2Cl2 (s) + 2e --------> 2 Hg() +2 Cl(aq) Cu²+ (aq) + 2 e------- -> Cu (s) a) What is the Ecell (V)? b) What is the net (overall) reaction? c) What is AG for the reaction in (b)? E° (V) +0.241 V (satd. KCI) +0.339 V
9. Write a Nernst expression, at 25 °C, for the following half-reactions: (a) Zn* (aq) + 2 e- 5 Zn (s) (b) Fe* (aq) + e- s Fe2* (aq) (c) 2 H* (aq) + 2 e- H2 (g) (d) Cr2O2- (aq) + 6 e + 14 H* (aq) S 2 Cr* (aq) + 7 H2O (I)
The free energy change for the following reaction at 25 °C, when [Ag+] = 1.12 M and [Co²+] M, is -221 kJ: 2Ag+ (1.12 M) + Co(s)—2Ag(s) + Co2+ (7.66x10-3 M) AG = -221 kJ What is the cell potential for the reaction as written under these conditions? Answer: V Would this reaction be spontaneous in the forward or the reverse direction? reverse = 7.66x10-3
A concentration cell running at 25°C consists of two Al(s)/Al3+(aq) cells. If the concentration of Al3+ at the anode is 4.12 x 10-8 M and the concentration of Al3+ at the cathode is 1.27 M, what is the value of Ecell? (Two decimal places)
The overall reaction and equilibrium constant value for a hydrogen-oxygen fuel cell at 298 K is given below. 2 H₂(g) + O₂(g) → 2 H₂O(l) K = 1.34 x 1083 (a) Calculate and AG at 298 K for the fuel-cell reaction. V k] 80 AGO (b) Predict the signs of AH° and AS for the fuel-cell reaction. AH°: O positive. O negative AS°: positive O negative (c) As temperature increases, does the maximum amount of work obtained from the fuel-cell reaction increase, decrease, or remain the same? Explain. Since AS is ---Select---, as T increases, AG becomes more ---Select--- . Therefore, the maximum work obtained will ---Select--- as T increases. ASK YOUR TEACH
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7>) True or False
Calculate E for the following unbalanced reactions. Indicate if the reactions are spontaneous as written. (a) Al3+(aq) + Mg(s) → Al(s) + Mg2+(aq)(b) ClO4-(aq) +Mn2+(aq) + H2O(l) → MnO4-(aq) + H+(aq) + ClO3-(aq) Given data:Al3+ + 3e-→Al Ered =-1.66 v Mg2+ + 2e-→Mg Ered= -2.37 vMnO4- + 8H+ +5e-→Mn2+ + 4H2O Ered= 1.51 v ClO4- + 2H+ +2e- →ClO3- + H2O Ered= 1.19 v
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Calculate the cell voltage at 25°C for this cell when [H+] = [HSO4–] = 4.5 M. At 25°C, E° = 2.04 V for the lead-acid battery. For this reaction, ∆H° = –315.9 kJ and ∆S° = 263.5 J/mole K. Calculate E° at –20° C and Ecell when [H+] = [HSO4–] = 4.5 M.
The potential, AE°, for an electrochemical cell for the reaction shown below is 0.339 V: Cu2*(aq) + H2(g) = Cu(s) + 2H*(aq) (1) What is AE° for reaction 2? ½Cu2*(aq) + ½H2(g) = ½Cu(s) + H*(aq) (2) Select your response from the list below. i) 0.339 V ii) 0.678 V iii) 0.1695 V