**Learning Goal:**To learn how to use the Nernst equation.The standard reduction potentials listed in any reference table are only valid at the common reference temperature of 25 °C and standard conditions of 1 M for solutions and 1 atm for gases. To calculate the cell potential at nonstandard conditions, one uses the Nernst equation:\[ E = E^o - \frac{2.303RT}{nF} \log_{10} Q \]Where:- \( E \) is the potential in volts.- \( E^o \) is the standard potential at 25 °C in volts.- \( R = 8.314 \, \text{J/(K·mol)} \) is the gas constant.- \( T \) is the temperature in kelvins.- \( n \) is the number of moles of electrons transferred.- \( F = 96,500 \, \text{C/mol e}^-\) is the Faraday constant.- \( Q \) is the reaction quotient.At the common reference temperature of 298 K, substituting each constant into the equation the result is:\[ E = E^o - \frac{0.0592 \, \text{V}}{n} \log_{10} Q \] **Title: Determining Electrochemical Cell Potential****Introduction:**In this task, we aim to calculate the cell potential for a given reaction under specific conditions.**Reaction:**\[ \text{Mg(s)} + \text{Fe}^{3+}(\text{aq}) \rightarrow \text{Mg}^{2+}(\text{aq}) + \text{Fe(s)} \]**Conditions:**- Temperature: \(4^\circ \text{C}\)- Concentration of \( \text{Fe}^{3+} \) = \(3.80 \, \text{M} \)- Concentration of \( \text{Mg}^{2+} \) = \(0.210 \, \text{M} \)**Objective:**Express the answer to three significant figures and include the appropriate units.**Instructions:**1. Use the standard reduction potentials for magnesium and iron to calculate the overall cell potential.2. Consider Nernst equation adjustments for the given concentrations and temperature.**Input Fields:**- Enter the calculated potential value and select the proper units in the input boxes provided.**Resources:**- View available hints for guidance.**Graph/Diagram:**There are no graphs or diagrams associated with this calculation task.**Tools:**A toolset including undo, redo, calculator, and help is available to facilitate the calculation.**Submission:**- After calculating, enter the value in the form \(E =\) [Value] [Units].**Conclusion:**To solve, apply electrochemical principles, considering concentration effects on the cell potential at the given temperature. Click "Submit" once done.

In the given question we have to calculate the cell potential of the reaction, first of all, we…

What is the cell potential for the reaction at 49 °C when [Fe²+] = 3.80 M and [Mg²+] = 0.210 M. Express your answer to three significant figures and include the appropriate units. ▸ View Available Hint(s) μA E= Value Submit y Units Mg(s) + Fe²+ (aq) →Mg²+ (aq) + Fe(s) ?

What is the cell potential for the reaction at 49 °C when [Fe²+] = 3.80 M and [Mg²+] = 0.210 M. Express your answer to three significant figures and include the appropriate units. ▸ View Available Hint(s) μA E= Value Submit y Units Mg(s) + Fe²+ (aq) →Mg²+ (aq) + Fe(s) ?

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

The free energy change for the following reaction at 25 °C, when [H*] = 3.34x103 M and [Hg*] = 1.15 M, is 180. kJ: 2H*(3.34x10-3 M) + Hg() H2(g) + Hg²*(1.15 M) AG = 180. kJ What is the cell potential for the reaction as written under these conditions? Answer: -0.932 Would this reaction be spontaneous in the forward or the reverse direction? forward v
Be sure to answer all parts. (a) Balance the skeleton reaction. Include the states of matter for each substance. (b) Calculate E (c) Determine whether the reaction is spontaneous. A table of standard electrode potentials is given below for reference. Cl₂(g) + Fe2+ (aq) → CI (aq) + Fe³+ (aq) (a) (b) (c) O Strength of oxidizing agent cell V not spontaneous spontaneous Selected Standard Electrode Potentials (298 K) Half-Reaction F2(g) + 2e7 Cl₂(g) + 2e7 MnO₂(s) + 4H+ (aq) + 2e7 NO3(aq) + 4H+ (aq) + 3e¯ Ag+ (aq) + e Fe³+ (aq) + e O₂(g) + 2H₂O() + 4e Cu²+ (aq) + 2e7 2H+ (aq) + 2e7 N₂(g) + 5H+ (aq) + 4e¯ Fe2+ (aq) + 2e7 2+ Zn²+ (aq) + 2e 2H₂O() +2e7 Na+ (aq) + e 2F (aq) 2C1 (aq) 2+ Mn²+ (aq) + 2H₂O(l) NO(g) + 2H₂O(l) Ag(s) Fe²+ (aq) 4OH(aq) Cu(s) H₂(g) N₂Hs+ (aq) Fe(s) Zn(s) H₂(g) + 2OH(aq) Na(s) Strength of reducing agent E half-cell (V) +2.87 +1.36 +1.23 +0.96 +0.80 +0.77 +0.40 +0.34 0.00 -0.23 -0.44 -0.76 -0.83 -2.71
Pls help ASAP.
Part A If a voltaic cell is created between a Am"/Am electrode and a Fe*/Fe electrode, what species is being oxidized? O Am"(ag) O Fe(s) O Fe"(ag) O Am(s)
A voltaic cell employs the following redox reaction: 2 Fe3+ Calculate the cell potential at 25 °C under each of the following conditions. (ag) + 3 Mg (s) → 2 Fe (s) + 3 Mg²+ (aq) Part C [Fe+] = 3.10 M ; [Mg²+] = 1.2x10-3 M Ecell Submit Previous Answers Request Answer X Incorrect; Try Again; 5 attempts remaining
Given the balanced electrochemical reaction below (M and N represent generic metals), what is the [M3*], in moles per liter, required to generate a cell potential (Ecell) of 0.584 V for this reaction at 58.1 °C when [N*] = 1.246 M? Enter your answer to the thousandths place. Do not include units. M(s) + 3N*(aq) –→ M3+(aq) + 3N(s) E°cell = 0.572 V
Given the balanced electrochemical reaction below (M and N represent generic metals), what is the cell potential (Ecell), in volts, for this reaction at 35.4 °C when [M+] = 0.00533 M and [N2+] = 0.725 M? Enter your answer to the thousandths place. Do not include units. 2M+(aq) + N(s) → 2M(s) + N2+(aq) E°cell = 0.128 V
+2 t- 4) Consider the following redox reactions and their standard cell potentials: (acidic conditions) (basic conditions) S,0, (aq) + Cl2(g) = HCI0(aq) + So,"(aq) (acidic conditions) cell =-1.31 V E°cell = +0.10 V E° cell = +0.37 V E°. C2(g) + Bio*(aq) = Bi(s)+ HCIO(aq) So(aq) + 03(g) = S20 (aq) + 02(g) a) Using the ½ rxn method, balance the three redox reactions under the conditions indicated. b) Write each balanced redox reaction in the spontaneous direction and indicate its corresponding standard cell potential. c) Between these 3 redox reactions there are FOUR oxidizing agents and FOUR reducing agents. (Hint: You may need to consider the balanced equations.) i) Place the FOUR oxidizing agents in order of decreasing oxidizing strength. (From the strongest OA to the weakest OA.) ii) Place the FOUR reducing agents in order of decreasing reducing strength. ( (From the strongest RA to the weakest RA.)
+ Cell Potential and Equilibrium The equilibrium constant, K, for a redox reaction is related to the standard potential, E, by the equation In K FE RT where 72 is the number of moles of electrons transferred, F (the Faraday constant) is equal to 96,500 C/(mol e-). R (the gas constant) is equal to 8.314 J/(mol K), and T is the Kelvin temperature. Standard reduction potentials ▾ Part A Use the table Express your answer numerically. ▸ View Available Hint(s) K= ΥΠΙ ΑΣΦ 9 Reduction half-reaction Ag+ (aq) +eAg(s) Cu²+ (aq) +201 >Cu(s) Sn+ (aq) + 4e →→Sn(s) 2H+ (aq) +2e →H₂(g) Ni2+ (aq) + 2e →Ni(s) Fe²+ (aq) + 2e →Fe(s) Zn²+ (aq) +2e →Zn(s) Al³+ (aq) + 3e →Al(s) Mg²+ (aq) + 2e →Mg(s) standard reduction potentials given above to calculate the equilibrium constant at standard temperature (25 °C) for the following reaction: Fe(s) + Ni²+ (aq) →>Fe²+ (aq) + Ni(s) ? E (V) 0.80 0.34 0.15 0 0.26 -0.45 0.76 -1.66 2.37 29 o Review | Constants | Period
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
Use the Nernst equation to find the cell potential ECell, of a galvanic cell with the following half cells: Ag+ (aq, 0.75M) + e- => Ag(s) Cr3+ (aq, 0.0012 M) + 3e- => Cr(s) Then calculate E°cell