Homework Answers
Add Answer to:
16. What is the value of E for the half-cell: Mn04 (0.010 M) +8H+ (0.20 M)...
help me solve this correctly Calculate the value of E for the galvanic cell made from...
help me solve this correctly
Calculate the value of E for the galvanic cell made from these two half reactions when MnO4) = 0.010 M. [H] =0.20 M, [Mn?] = 0.020 M, [Fe3+] = 0.75 M, [Fe2+] = 0.30 M. Ece - OTO nF 75 Q: 1.02m] 6.30m 5 (-010[0.1073747 (0.05 .000 4 MnO4 +8H+ +5e → Mn2+ + 4H20 E° = 1.51 V Fe3+ + + Fe2+ Ecelli Ecell - (8. (8./4.J/mol) (E° = 0.77 V Product S496485) Treactants...
A galvanic cell is prepared using the following half-cells: (i) MnO4-(0.283M), Mn2+(0.081M), and H+(1.0 M), and...
A galvanic cell is prepared using the following half-cells: (i) MnO4-(0.283M), Mn2+(0.081M), and H+(1.0 M), and (ii) Sn2+(0.236M) and Sn4+(0.145M). MnO4-+ 8H++ 5e-↔Mn2++ 4H2O Eo= 1.507 V Sn4++ 2e-↔Sn2+Eo= 0.151 V a)Calculate the galvanic cell voltage. b)Write the complete cell description. c)Write the cell reaction indicating clearly the direction of spontaneous reaction. d)Name the electrode (anode or cathode), give the polarity (positive or negative) of the electrode, and state the reaction (oxidation or reduction) occurring in the tin half-cell. Please...
A solution containing 0.190.19 M Pb2+Pb2+ , 1.5×10−61.5×10−6 M Pb4+Pb4+ , 1.5×10−61.5×10−6 M Mn2+Mn2+ , 0.190.19...
A solution containing 0.190.19 M Pb2+Pb2+ , 1.5×10−61.5×10−6 M Pb4+Pb4+ , 1.5×10−61.5×10−6 M Mn2+Mn2+ , 0.190.19 M MnO−4MnO4− , and 0.900.90 M HNO3HNO3 was prepared. For this solution, the balanced reduction half‑reactions and overall net reaction shown can occur. 5[Pb4++2e−↽−−⇀Pb2+]5[Pb4++2e−↽−−⇀Pb2+] ?∘+=1.690 VE+°=1.690 V 2[MnO−4+8H++5e−↽−−⇀Mn2++4H2O]2[MnO4−+8H++5e−↽−−⇀Mn2++4H2O] ?∘−=1.507 VE−°=1.507 V 5Pb4++2Mn2++8H2O↽−−⇀5Pb2++2MnO−4+16H+5Pb4++2Mn2++8H2O↽−−⇀5Pb2++2MnO4−+16H+ A. Determine ?∘cellEcell∘, Δ?∘ΔG∘, and ?K for this reaction. ?∘cell=Ecell∘= V Δ?∘=ΔG∘= J ?=K= B. Calculate the value for the cell potential, ?cellEcell, and the free energy, Δ?ΔG, for the given conditions....
Please state the answer clearly for a thumbs up! D Question 10 Refer to the galvanic...
Please state the answer clearly for a thumbs up!
D Question 10 Refer to the galvanic cell below (the contents of each half-cell are written beneath each compartment): 0.10 M MnO4 0.40 M Cr3+ 0.20 M Mn2+ 0.30 M Cr20,2- 0.010 MH 0.010 MH+ The standard reduction potentials are as follows: MnO4 +8H+ +5e Mn2+ + 4H20, <= 1.51 V Cr2O72- + 14H* +6e2Cr3+ +7H20, € = 1.33 V When current is allowed to flow, which species is oxidized? o...
Show all steps please :) A galvanic cell is prepared using the following two half-cells: (i)...
Show all steps please :)
A galvanic cell is prepared using the following two half-cells: (i) MnO4 (0.273 M), Mn2+ (0.167 M), and H+ (1.0 M), and (ii) Fe2+ (0.247 M) and Fe3+ (0.389 M). The standard reduction potentials for the two half cells are: Mnoa + 8H+ + 5e – Mn2+ + 4H20 E° = 1.507 V Fe3+ + e- Fe2+ E° = 0.770 V a) Calculate the voltage for this galvanic cell. b) Write the balanced equation for...
3. Calculate the redox potential (in volts) for a LiMnO2/graphite battery. MnO4- + 8H + 5e-...
3. Calculate the redox potential (in volts) for a
LiMnO2/graphite battery.
MnO4- + 8H + 5e- →
Mn2+ (aq) + 4H2O (+1.51 V)
MnO2 + 4H +2e-
→ Mn2+ (aq) + 2H2O (+1.22 V)
xLi+ + C6 +
e- → LixC5 (-3.00 V)
1. Write balanced equations for the following processes: a. The reaction of potassium with water. C. Thermal decomposition of sodium azide. d. The reaction of potassium peroxide with water. e. Calcium hydride reacting with water. f. The reaction...
3. Calculate the redox potential (in volts) for a LiMnO2/graphite battery. (Redox reaction) MnO4- + 8H...
3. Calculate the redox potential (in volts) for a
LiMnO2/graphite battery.
(Redox reaction)
MnO4- + 8H + 5e- →
Mn2+ (aq) + 4H2O (+1.51 V)
MnO2 + 4H +2e-
→ Mn2+ (aq) + 2H2O (+1.22 V)
xLi+ + C6 + e- → LixC5 (-3.00
V)
1. Write balanced equations for the following processes: a. The reaction of potassium with water. C. Thermal decomposition of sodium azide. d. The reaction of potassium peroxide with water. e. Calcium hydride reacting with water. f....
3. A student connects a Cd?"(0.20 M)|Cd(s) half-cell to a Cu?"(1 M) Cu(s) electrode. When the...
3. A student connects a Cd?"(0.20 M)|Cd(s) half-cell to a Cu?"(1 M) Cu(s) electrode. When the red lead is attached to the Cu electrode, the cell potential read by the voltmeter, E is +0.77 V. a. What is the reduction half-reaction at the cathode (red lead)? b. What is the oxidation half-reaction at the anode (black lead)? c. What is the overall cell reaction? d. Write the expression for the thermodynamic reaction quotient, Q, and calculate its value for this...
What is E for a concentration cell where [Ag+]=0.00012 M in the diluted half-cell and 1.50...
What is E for a concentration cell where [Ag+]=0.00012 M in the diluted half-cell and 1.50 M in the more concentrated one at SATP? -0.56 V 0.56 V 0.24 V -0.24 V
An electrochemical cell is expressed as Cu(s) | Cu^2+ (0.20 M) || I^- (0.10 M) |...
An electrochemical cell is expressed as Cu(s) | Cu^2+ (0.20 M) || I^- (0.10 M) | I_3^- (0.20 M) | Pt. I_3^- + 2e^- rightarrow 3I^-, E^0 = 0.535 V Cu^2+ + 2e^- rightarrow Cu(s) E^0 = 0.339 V Please answer the following questions: a. Write half-cell reactions at anode and at cathode. b. Write whole-cell reaction. c. Calculate the potential/voltage of the cell. d. Calculate equilibrium constant for this whole-cell reaction.
help me solve this correctly
Calculate the value of E for the galvanic cell made from these two half reactions when MnO4) = 0.010 M. [H] =0.20 M, [Mn?] = 0.020 M, [Fe3+] = 0.75 M, [Fe2+] = 0.30 M. Ece - OTO nF 75 Q: 1.02m] 6.30m 5 (-010[0.1073747 (0.05 .000 4 MnO4 +8H+ +5e → Mn2+ + 4H20 E° = 1.51 V Fe3+ + + Fe2+ Ecelli Ecell - (8. (8./4.J/mol) (E° = 0.77 V Product S496485) Treactants...
Please state the answer clearly for a thumbs up!
D Question 10 Refer to the galvanic cell below (the contents of each half-cell are written beneath each compartment): 0.10 M MnO4 0.40 M Cr3+ 0.20 M Mn2+ 0.30 M Cr20,2- 0.010 MH 0.010 MH+ The standard reduction potentials are as follows: MnO4 +8H+ +5e Mn2+ + 4H20, <= 1.51 V Cr2O72- + 14H* +6e2Cr3+ +7H20, € = 1.33 V When current is allowed to flow, which species is oxidized? o...
Show all steps please :)
A galvanic cell is prepared using the following two half-cells: (i) MnO4 (0.273 M), Mn2+ (0.167 M), and H+ (1.0 M), and (ii) Fe2+ (0.247 M) and Fe3+ (0.389 M). The standard reduction potentials for the two half cells are: Mnoa + 8H+ + 5e – Mn2+ + 4H20 E° = 1.507 V Fe3+ + e- Fe2+ E° = 0.770 V a) Calculate the voltage for this galvanic cell. b) Write the balanced equation for...
3. Calculate the redox potential (in volts) for a
LiMnO2/graphite battery.
MnO4- + 8H + 5e- →
Mn2+ (aq) + 4H2O (+1.51 V)
MnO2 + 4H +2e-
→ Mn2+ (aq) + 2H2O (+1.22 V)
xLi+ + C6 +
e- → LixC5 (-3.00 V)
1. Write balanced equations for the following processes: a. The reaction of potassium with water. C. Thermal decomposition of sodium azide. d. The reaction of potassium peroxide with water. e. Calcium hydride reacting with water. f. The reaction...
3. Calculate the redox potential (in volts) for a
LiMnO2/graphite battery.
(Redox reaction)
MnO4- + 8H + 5e- →
Mn2+ (aq) + 4H2O (+1.51 V)
MnO2 + 4H +2e-
→ Mn2+ (aq) + 2H2O (+1.22 V)
xLi+ + C6 + e- → LixC5 (-3.00
V)
1. Write balanced equations for the following processes: a. The reaction of potassium with water. C. Thermal decomposition of sodium azide. d. The reaction of potassium peroxide with water. e. Calcium hydride reacting with water. f....
3. A student connects a Cd?"(0.20 M)|Cd(s) half-cell to a Cu?"(1 M) Cu(s) electrode. When the red lead is attached to the Cu electrode, the cell potential read by the voltmeter, E is +0.77 V. a. What is the reduction half-reaction at the cathode (red lead)? b. What is the oxidation half-reaction at the anode (black lead)? c. What is the overall cell reaction? d. Write the expression for the thermodynamic reaction quotient, Q, and calculate its value for this...
An electrochemical cell is expressed as Cu(s) | Cu^2+ (0.20 M) || I^- (0.10 M) | I_3^- (0.20 M) | Pt. I_3^- + 2e^- rightarrow 3I^-, E^0 = 0.535 V Cu^2+ + 2e^- rightarrow Cu(s) E^0 = 0.339 V Please answer the following questions: a. Write half-cell reactions at anode and at cathode. b. Write whole-cell reaction. c. Calculate the potential/voltage of the cell. d. Calculate equilibrium constant for this whole-cell reaction.
Most questions answered within 3 hours.
-
In your opinion, did anyone "win" the Cold War? If no, why? If
yes, who won...
asked 1 minute ago -
Please complete the implementation of the four functions IN
C.
//************************************************************************************/
//
// countNumberofOnes
//
//...
asked 1 minute ago -
Shown below are the budgeted sales for ABC Company for
the next six months:
Sales...
asked 6 minutes ago -
A balance sheet shows exactly where a business stands at any
given point in time. It...
asked 7 minutes ago -
For the reaction between ethylene and chlorine, the equilibrium
constant is 2.00x10^6. If the initial concentrations...
asked 14 minutes ago -
What is the energy released in this β+ nuclear reaction
23/12Mg→23/11Na+0/1e (The atomic mass of 23Mg...
asked 29 minutes ago -
Choose one:
Periodic Inventory Accounting is less expensive to maintain
than Perpetual Inventory Accounting
Periodic Inventory...
asked 26 minutes ago -
4. Describe the role played by glutamine in the transport of
nitrogen.
asked 25 minutes ago -
1. In IntelliJ create a new project called F1_2
2. In the Project window create a...
asked 30 minutes ago -
30 mL of 1 M HCl were added to 200 mL of 0.1 M lysine solution...
asked 38 minutes ago -
The market portfolio has an expected return of 9.2% with a
volatility of 16.7% while the...
asked 42 minutes ago -
Personal selling is of special importance in B2B marketing. What
has been the impact of the...
asked 49 minutes ago