Question

Please show work Calculate the cell potential for the following reaction that takes place in an...

Please show work

Calculate the cell potential for the following reaction that takes place in an electrochemical cell at 25°C. Mn(s) ∣ Mn2+(aq, 1.28 M) ∥ Ag+(aq, 0.000837 M) ∣ Ag(s) Calculate the cell potential for the following reaction that takes place in an electrochemical cell at 25°C. Mn(s) Mn2+(aq, 1.28 M) Ag+(aq, 0.000837 M) Ag(s)

1.98 V

0.00 V

1.79 V

-0.84 V

-1.28 V

Homework Answers

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
Calculate the cell potential for the following reaction that takes place in an electrochemical cell at...
Calculate the cell potential for the following reaction that takes place in an electrochemical cell at 25
1. What is the calculated value of the cell potential at 298K for an electrochemical cell...
1. What is the calculated value of the cell potential at 298K for an electrochemical cell with the following reaction, when the Pb2+ concentration is 4.09×10-4 M and the Mg2+ concentration is 1.03 M ? Pb2+(aq) + Mg(s) ---> Pb(s) + Mg2+(aq) Answer: ___V The cell reaction as written above is spontaneous for the concentrations given____. (true or false) (From the table of standard reduction potentials: ) Pb2+(aq) + 2 e- --> Pb(s) -0.126 Mg2+(aq) + 2 e- --> Mg(s)...
What is the calculated value of the cell potential at 298K for an electrochemical cell with...
What is the calculated value of the cell potential at 298K for an electrochemical cell with the following reaction, when the Pb2+ concentration is 1.20 M and the Mn2+ concentration is 5.43×10-4 M ? Pb2+(aq) + Mn(s) Pb(s) + Mn2+(aq) Answer: in V The cell reaction as written above is spontaneous for the concentrations given: true or false?
A) When the Cu2+ concentration is 1.07 M, the observed cell potential at 298K for an...
A) When the Cu2+ concentration is 1.07 M, the observed cell potential at 298K for an electrochemical cell with the following reaction is 1.607V. What is the Mn2+ concentration? Cu2+(aq) + Mn(s)--> Cu(s) + Mn2+(aq) Answer: _____ M B) When the Cu2+ concentration is 5.71×10-4 M, the observed cell potential at 298K for an electrochemical cell with the following reaction is 2.609V. What is the Mg2+ concentration? Cu2+(aq) + Mg(s)---> Cu(s) + Mg2+(aq) Answer: _____ M C) When the Cu2+...
9 part 2 A voltaic cell employs the following redox reaction: Sn2+(aq)+Mn(s)→Sn(s)+Mn2+(aq) Calculate the cell potential...
9 part 2 A voltaic cell employs the following redox reaction: Sn2+(aq)+Mn(s)→Sn(s)+Mn2+(aq) Calculate the cell potential at 25 ∘C under each of the following conditions Part A standard conditions Ecell Part B [Sn2+]= 1.45×10−2 M ; [Mn2+]= 1.41 M . Express your answer using two significant figures. Ecell Part C [Sn2+]= 1.41 M ; [Mn2+]= 1.45×10−2 M . Ecell
Please show all work and derivations!! A voltaic cell is made by combining the following two...
Please show all work and derivations!! A voltaic cell is made by combining the following two half-reactions: Sn^(2+) (aq) + 2e- --> Sn(s) Eo red= +0.15 V AgCl(s) + e- --> Ag(s) + Cl^- (aq) Eo red= +0.22 V Calculate the cell potential E at 298K if [Sn^(2+)] = 0.050 M and [Cl-] = 4.0 M.
For the following electrochemical cells, calculate the potential and determine if the cell reaction is spontaneous...
For the following electrochemical cells, calculate the potential and determine if the cell reaction is spontaneous as written at 25°C. Pt(s) ΙSn2 (0.0055 M), Sn4 (0.12 M) ΙΙFe2 (0.0020 M), Fe3 (0.15 M) ΙPt(s) Ecell=? V
Given a potential of 0.456 V, for the following electrochemical cell, what is the concentration of...
Given a potential of 0.456 V, for the following electrochemical cell, what is the concentration of Ag+? Cu2+(aq) (1.0 M)|Cu(s)||Ag+(aq)|Ag(s)
So, I'm confused about finding E cell value Please explain me ! For example, Mn(s) ∣...
So, I'm confused about finding E cell value Please explain me ! For example, Mn(s) ∣ Mn2+(aq) ∥ Ag+(aq) ∣ Ag(s) overall reaction is Mn(s)+Ag+(aq) -> Mn2+(aq)+Ag(s) Oxidation: Mn(s)->  Mn2+(aq)+2e- E: -1.18 I think here it should be 1.18 to calcuate because on the table, Reduction Half reaction is Mn2+(aq)+2e- -> Mn(s) and the order has been changed. WHY IT IS -1.18??? Reduction: Ag+(aq)+e- -> Ag(s) E: 0.80 This makes sense following by the table. For another example, Sn(s) | Sn2+(aq)...
Calculate the cell potential for a cell operating with the following reaction at 25 degrees Celsius,...
Calculate the cell potential for a cell operating with the following reaction at 25 degrees Celsius, in which [MnO4^1-] = .01M, [Br^1-] = .01M, [Mn^2+] = .15M, and [H^1+] = 1M. The reaction is 2 MnO4^1-(aq) + 10 Br^1-(aq) + 16 H^1+(aq) --> 2 Mn^2+(aq) + 5 Br2(l) + 8 H2O(l)