Use cell potential to calculate an equilibrium constant.Calculate the cell potential and the equilibrium constant for the following reaction at 298 K:Cu2+(aq) + Fe2+(aq) Cu+(aq) + Fe3+(aq)Hint: Carry at least 5 significant figures during intermediate calculations to avoid round off error when taking theantilogarithm.Equilibrium constant: Use measured cell potential to calculate concentration.When [Ag] 1.18 M, the observed cell potential at 298K for an electrochemical cell with the reaction shown below is 3.283V. What is the Mg2+ concentration in this cell?2Ag (aq) + Mg(s) 2Ag(s) + Mg2+(aq)[Mg2+]=M

E0 Cell = E0 Cathode - E0 Anode = E0 reduction - E0 oxidation = 0.16 - 0.77…

Answered: Use cell potential to calculate an…

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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When [Hg2+] = 1.16 M, the observed cell potential at 298 K for an electrochemical cell with the reaction shown below is 2.583 V. What is the Al3+ concentration in this cell? 3H92*(aq) + 2Al(s) – 3Hg(8) + 2AI3*(aq) [A|3*] = mol/L 6.724x10^-4 Show Approach Hide Tutor Steps Resubmit TUTOR STEP The first step is to separate the overall reaction into oxidation and reduction half-reactions. These are shown here. Hg2*(aq) + 2e" – Hg(e) reduction Al(s) → A3*(aq) + 3e oxidation Next (1 of 5)
Calculate the cell potential for the following reaction that takes place in an electrochemical cell at 25°C. Mg(s) ∣ Mg2+(aq, 1.74 M) || Cu2+(aq, 0.0053 M) ∣ Cu(s)
When the Ag concentration is 1.00 M, the observed cell potential at 298 K for an electrochemical cell with the following reaction is 2.530 V. What is the A1³+ concentration? 3+ 3Ag+ (aq) + Al(s) → 3Ag(s) + Al³+ (aq) red Ag+ (aq) + e → Ag(s) Ed Al³+ (aq) + 3e → Al(s) Ee red [A1³+] = M 0.799 V = -1.660 V
Suppose the left half cell of a galvanic cell apparatus is filled with a 5.00 M solution of A certain metal M forms a soluble sulfate salt M₂ M₂(SO4)3 and the right half cell with a 2.50 mM solution of the same substance. Electrodes made of M are dipped into both solutions and a voltmeter is connected between them. The temperature of the apparatus is held constant at 30.0 °C. Which electrode will be positive? What voltage will the voltmeter show? Assume its positive lead is connected to the positive electrode. Be sure your answer has a unit symbol, if necessary, and round it to 2 significant digits. 0 left right × S
For the cell: Zn(s) + 2H† (aq) →→ Zn2+(aq) + H2 (g) If [Zn2+] = 1 M, PH2 = 1 atm, and E = 0.123 V what is pH?
Consider a galvanic electrochemical cell constructed using Cr/Cr³⁺ and Zn/Zn²⁺ at 25 °C. The following half-reactions are provided for each metal: Cr³⁺(aq) + 3 e⁻ → Cr(s) E°red = -0.744 V Zn²⁺(aq) + 2 e⁻ → Zn(s) E°red = -0.763 V Given that the standard cell potential is 0.019 V and the overall equation is 2 Cr³⁺ (aq) + 3 Zn (s) → 2 Cr (s) + 3 Zn²⁺ (aq), what is the cell potential for this cell at 25 °C when [Zn²⁺] = 0.0323 M and [Cr³⁺] = 0.1736 M?
When [Pb2+] = 1.49 M, the observed cell potential at 298 K for an electrochemical cell with the reaction shown below is 0.749 V. What is the Zn2+ concentration in this cell? Pb2+(aq) + Zn(s) → Pb(s) + Zn²+(aq) [Zn2+] = mol/L
A galvanic cell consists of a Ni/Ni²+ half-cell and a Co/Co²+ half-cell with initial concentrations: [Ni²+] = 0.80M and [Co²+] = 0.20 M at 298.15 K. half-reaction Eº (V) Ni²+ (aq) +2e→Ni(s) -0.25 Co²+ (aq) +2e→Co(s) -0.28
Consider the following voltaic cellat 25°C: Al(s) | Al3*(0.20 M, aq) || Al3+(0.75 M, aq) | Al(s) If the standard reduction potential for the Al3+(aq) /Al(s) redox couple is -1.66 V, what is the overall cell reaction and the cell potential of the voltaic cell shown above? (Reference the cell as written above.) The overall cell reaction is Al3+(0.75 M, aq) AIS*(0.20 M, ag) and E°cell = +0.0113 V. There is no overall cell reaction and E°cell = 0.00 V. The overall cell reaction is Al3+(0.20 M, aq) Als*(0.75 M, aq) and E°cell = +0.0113 V. The overall cell reaction is Al3+(0.20 M, aq) →A13+(0.75 M, ag) and E°cell| = -0.0113 V. The overall cell reaction is Al3*(0.75 M, aq) →AI3*(0.20 M, aq) and E°cell = -0.0113 V. None of these
The free energy change for the following reaction at 25 °C, when [Hg²+] = 1.20 M and [Pb2+] = 7.55x10-3 M, is -202 kJ: Hg2+(1.20 M) + Pb(s) Hg(l) + Pb2+(7.55×10-3 M) AG = -202 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? forward or reverse
You have a cell that uses the following overall reaction. Calculate the cell potential at 298 K if [Au3+] = 0.010 M and [Mg2+] = 0.150 M 2 Au3+(aq) + 3 Mg(s) ---> 2 Au(s) + 3 Mg2+(aq)
When [Hg2+] = 1.37 M, the observed cell potential at 298 K for an electrochemical cell with the reaction shown below is 3.321 V. What is the Mg2+ concentration in this cell? Hg2+(aq) + Mg(s) Hg(t) + Mg2+(aq) [Mg2+] = mol/L Submit

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