Question

In: Chemistry

An electrochemical cell is based on the following two half-reactions: Ox: Sn(s)→Sn2+(aq, 1.65 M )+2e− Red:...

An electrochemical cell is based on the following two half-reactions:

Ox: Sn(s)→Sn2+(aq, 1.65 M )+2e−

Red: ClO2(g, 0.165 atm )+e−→ClO−2(aq, 1.90 M )

Compute the cell potential at 25 ∘C.

Solutions

Expert Solution

Ans:- Compute the cell potential at 25 ∘C.

         Ox:   Sn(s)→Sn2+(aq, 1.65 M )+ 2e− : E = -0.14v

          Red: ClO2(g, 0.165 atm )+ e−→ClO−2(aq, 1.90 M): E = 0.954v

E0 = EC - EA

= 0.954-(-0.14)

= 1.094v

              NERNST EQUTION:   E = E0 -0.0591/2*log (cathode)/(anode)

                                 E = 1.094-0.029*log(1.90)2/(1.65)

                             E = 1.094-0.029*log 2.1878

                             E = 1.094-0.029*0.34

                             E = 1.0842 Volts


Related Solutions

An electrochemical cell is based on the following two half-reactions: Ox: Pb(s)→Pb2+(aq,Pb(s)→Pb2+(aq, 0.15 MM )+2e−)+2e− Red:...
An electrochemical cell is based on the following two half-reactions: Ox: Pb(s)→Pb2+(aq,Pb(s)→Pb2+(aq, 0.15 MM )+2e−)+2e− Red: MnO−4(aq,MnO4−(aq, 1.80 MM )+4H+(aq,)+4H+(aq, 1.9 MM )+3e−→)+3e−→ MnO2(s)+2H2O(l) Compute the cell potential at 25 ∘C∘C.
An electrochemical cell is based on the following two half-reactions: Ox: Pb(s)?Pb2+(aq, 0.21M )+2e? Red: MnO?4(aq,...
An electrochemical cell is based on the following two half-reactions: Ox: Pb(s)?Pb2+(aq, 0.21M )+2e? Red: MnO?4(aq, 1.25M )+4H+(aq, 2.5M )+3e?? MnO2(s)+2H2O(l). Compute the cell potential at 25 ?C.
1. An electrochemical cell is based on the following two half-reactions: Ox: Pb(s)→Pb2+(aq, 0.18 M )+2e−...
1. An electrochemical cell is based on the following two half-reactions: Ox: Pb(s)→Pb2+(aq, 0.18 M )+2e− Red: MnO−4(aq, 1.50 M )+4H+(aq, 1.8 M )+3e−→ MnO2(s)+2H2O(l) Compute the cell potential at 25 ∘C. 2.A concentration cell consists of two Sn/Sn2+ half-cells. The cell has a potential of 0.16 V at 25 ∘C. What is the ratio of the Sn2+ concentrations in the two half-cells? Express your answer using two significant figures. 3.Metal plating is done by passing current through a metal...
An electrochemical cell is based on the following two half-reactions: Ox: Pb(s)→Pb2+(aq, 0.15 M )+2e− Red:MnO−4(aq,...
An electrochemical cell is based on the following two half-reactions: Ox: Pb(s)→Pb2+(aq, 0.15 M )+2e− Red:MnO−4(aq, 1.70 M )+4H+(aq, 1.5 M )+3e−→ MnO2(s)+2H2O(l) Compute the cell potential at 25 ∘C.
QUESTION 5 An electrochemical cell is based on the following two half-reactions: Ox: Pb(s)→Pb2+(aq, 0.21 M...
QUESTION 5 An electrochemical cell is based on the following two half-reactions: Ox: Pb(s)→Pb2+(aq, 0.21 M )+2e− Red: MnO−4(aq, 1.80 M )+4H+(aq, 1.5 M )+3e−→ MnO2(s)+2H2O(l) Compute the cell potential at 25 ∘C.?
Explain why the sum of the potentials for the half-reactions Sn2+(aq) + 2e− → Sn(s) and...
Explain why the sum of the potentials for the half-reactions Sn2+(aq) + 2e− → Sn(s) and Sn4+(aq) + 2e− → Sn2+(aq) does not equal the potential for the reaction Sn4+(aq) + 4e−→ Sn(s). What is the net cell potential? Compare the values of ΔG° for the sum of the potentials and the actual net cell potential.
Comprehensive thermochemical problem: Consider an electrochemical cell with the following half- cells: Pb2+(aq)(0.01 M)/Pb(s) and Sn2+(aq)(2M)/Sn(s)...
Comprehensive thermochemical problem: Consider an electrochemical cell with the following half- cells: Pb2+(aq)(0.01 M)/Pb(s) and Sn2+(aq)(2M)/Sn(s) at 25 °C. a. Find the potential of the cell. b.Determine the oxidizing agent and reducing agent. c. Calculate the standard free energy change for the reaction. d. Find the free energy change for the cell under the current conditions. e. Determine the equilibrium concentrations of the solutions. f. Use ΔH°f data to determine ΔH°rxn. ΔH°f (Sn2+ (aq)) = -8.8 kJ/mol ΔH°f (Pb2+ (aq))...
Half-reaction E° (V) Br2(l) + 2e- 2Br-(aq) 1.080V Sn2+(aq) + 2e- Sn(s) -0.140V Al3+(aq) + 3e-...
Half-reaction E° (V) Br2(l) + 2e- 2Br-(aq) 1.080V Sn2+(aq) + 2e- Sn(s) -0.140V Al3+(aq) + 3e- Al(s) -1.660V (1) The weakest oxidizing agent is: enter formula (2) The strongest reducing agent is: (3) The strongest oxidizing agent is: (4) The weakest reducing agent is: (5) Will Al(s) reduce Br2(l) to Br-(aq)? (6) Which species can be oxidized by Sn2+(aq)? If none, leave box blank.
1. A voltaic cell is constructed that is based on the following reaction: Sn2+(aq)+Pb(s)→Sn(s)+Pb2+(aq). a. If...
1. A voltaic cell is constructed that is based on the following reaction: Sn2+(aq)+Pb(s)→Sn(s)+Pb2+(aq). a. If the concentration of Sn2+ in the cathode compartment is 1.50 M and the cell generates an emf of 0.22 V , what is the concentration of Pb2+ in the anode compartment? b. If the anode compartment contains [SO2−4]= 1.50 M in equilibrium with PbSO4(s), what is the Kspof PbSO4? 2. A voltaic cell is constructed with two silver-silver chloride electrodes, each of which is...
A voltaic cell is constructed that is based on the following reaction: Sn2+(aq)+Pb(s)→Sn(s)+Pb2+(aq). 1. If the...
A voltaic cell is constructed that is based on the following reaction: Sn2+(aq)+Pb(s)→Sn(s)+Pb2+(aq). 1. If the concentration of Sn2+ in the cathode compartment is 1.50 M and the cell generates an emf of 0.25 V , what is the concentration of Pb2+ in the anode compartment? 2.If the anode compartment contains [SO2−4]= 1.00 M in equilibrium with PbSO4(s), what is the Ksp of PbSO4?
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT