1-ele2Calculate the cell potential for this reaction, in the direction shown, at298 K.Hg*+ (1.17 M) → Hg²+ (9.80 × 10¬5 ME =VIn this mercury concentration cell, the reaction would proceed spontaneously in th v9.direction shownreverse direction

The given reaction is :The cell potential for the reaction can be calculated using Nernst equation. The expression for the same is represented as follows:Thus, putting known values, we get : Ecell=0-2.303×8.314×2982×96500log9.80×10-51.17=-0.02955×log8.38×10-5=-0.02955×-4.08⇒0.121 V Since Ecell is positive, thus the reaction would proceed spontaneously on the direction shown.

Science

Chemistry

Calculate the cell potential for this reaction, in the direction shown, at 298 K. M) 2+ Hgº+ (1.17 M) → Hg²+ (9.80 × 10–5 E = V In this mercury concentration cell, the reaction would proceed spontaneously in th v direction shown

Calculate the cell potential for this reaction, in the direction shown, at 298 K. M) 2+ Hgº+ (1.17 M) → Hg²+ (9.80 × 10–5 E = V In this mercury concentration cell, the reaction would proceed spontaneously in th v direction shown

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...

See similar textbooks

Similar questions

For the following reaction, predict the measured cell potential (Ecel) if the pressure of O2 is 1.59 atm, the concentration of H* is 9.67 x 103 M and the bromide ion concentration is 1.31 M. 0, (g) + 4H* (ag) + 4Br (aq) → 2H,O() + 2Br, () Half reactions: 2Br, () + 4e→ 4Br (aq) E° = +1.077 V 0,(g) + 4H* (aq) + 4e→ 2H,0(1) E° = +1.229 V
A galvanic (voltaic) cell consists of an electrode composed of aluminum in a 1.0 M aluminum ion solution and another electrode composed of gold in a 1.0 M gold(III) ion solution, connected by a salt bridge. Calculate the standard potential for this cell at 25 °C. Refer to the list of standard reduction potentials. Excell V
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)
Calculate the cell voltage under these conditions. Round your answer to 3 significant digits.
Consider the cell below at standard conditions: Zn(s) | Zn 2+ (aq) || Fe 2+ (aq) | Fe(s)Calculate the value of ΔG° for the reaction that occurs if the value of E° cell = 0.32 V for this cell.
If the concentration of Ag+ is 0.0115 M, the concentration of H+ is 0.355 M, and the pressure of H2 is 1.00 atm, calculate the cell potential at 25.0°C.
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?
You make a voltaic cell with copper and aluminum. If the cell potential is measured to be +2.50 V at room temperature, then what is the reaction quotient?
6) Calculate the potential of a cell, Eºcell, made with a standard bromine electrode as the anode and a standard chlorine electrode as the cathode. Write the shorthand cell notation.
Calculate the cell potential for this reaction, in the direction shown, at 298 K. 2+ Hg²+ (0.0000490 M) → Hg²+ 1.37 × 10⁰. E In this mercury concentration cell, the reaction would proceed spontaneously in the [ V
Calculate the standard cell potential (E°cell) of a galvanic cell made of lead and cadmium. Then determine the non-standard cell potential (Ecell) if the molarity of the cadmium solution was 0.20 M, and the molarity of the lead solution was 0.10 M. The temperature of the cell is 300 K. Cd2+(aq) + 2e– → Cd(s) E° = –0.403 V Pb2+(aq) + 2e– → Pb(s) E° = –0.126 V E°cell = ______________ Ecell = ______________
Balance the following reactions. Calculate the cell potential and the Gibbs free energy of these reactions at standard conditions. Identify the reactions with K > 1 among the following redox reactions. Write the line representations for the Galvanic cells, indicating the anode and 1. cathode. a. Cl2 + Br → Ct + Br2 b. Mnoa + H* + Ce → Cet + Mn2 + H20