Given the following redox reaction conducted in an acidic electrochemical cell, calculate ΔGo, ΔG, Eocell, and Ecell using the conditions provided.
Al (s) + VO2+ (aq) → Al3+ (aq) + VO2+ (aq)
VO2+(aq) + 2H+(aq) + e- → VO2+(aq) + H2O(l) Eo = 1.00V
Al3+(aq) + 3e- → Al (s) Eo = -1.66V
[VO2+] = 1.2M
[Al3+] = 0.025M
[VO2+] = 0.05M
[H+] = 2.1M
Given the following redox reaction conducted in an acidic electrochemical cell, calculate ΔGo, ΔG, Eocell, and...
22) Given the following redox reaction conducted in an acidic electrochemical cell, calculate ΔG®, ΔG, Eocell, and Ecell using the conditions provided. Al (s)VO2 (aq)Al (aqVo2 (aq) VO2(a)2H (aq)e VO2(a)H20() E 1.00V Al3+(aq) + 3e - Al (s) Eo 1.66V [VO2]- 1.2M [Al+0.025M vo21 0.05M [H ] 2.1M
22) Given the following redox reaction conducted in an acidic electrochemical cell, calculate ΔG®, ΔG, Eocell, and Ecell using the conditions provided. Al (s)VO2 (aq)Al (aqVo2 (aq) VO2(a)2H (aq)e VO2(a)H20() E 1.00V...
A) Use tabulated electrode potentials to calculate ΔG∘ for the reaction. 2K(s)+2H2O(l)→H2(g)+2OH−(aq)+2K+(aq) B) (Refer to the following standard reduction half-cell potentials at 25∘C: VO2+(aq)+Ni2+(aq)2H+(aq)++2e−e−→ →Ni(s)VO2+(aq) +H2O(l)E∘=−0.23V E∘=0.99V) An electrochemical cell is based on these two half-reactions: Oxidation:Reduction:Ni(s)VO2+(aq,0.024M)+2H+(aq,1.4M)+e−→→Ni2+(aq,1.8M)+2e−VO2+(aq,1.8M)+H2O(l) Calculate the cell potential under these nonstandard concentrations. C) Standard reduction half-cell potentials at 25∘C Half-reaction E∘ (V ) Half-reaction E∘ (V ) Au3+(aq)+3e−→Au(s) 1.50 Fe2+(aq)+2e−→Fe(s) − 0.45 Ag+(aq)+e−→Ag(s) 0.80 Cr3+(aq)+e−→Cr2+(aq) − 0.50 Fe3+(aq)+3e−→Fe2+(aq) 0.77 Cr3+(aq)+3e−→Cr(s) − 0.73 Cu+(aq)+e−→Cu(s) 0.52 Zn2+(aq)+2e−→Zn(s) − 0.76...
Answer the following questions related to the given
electrochemical cell.
Answer the following questions related to the given electrochemical cell 2NO(g) H20(l) 2e N20(g) 20H (aq) 0.760 V NO aq) H2O(l) 2e NO2 (aq) 20H (aq) Eo 0.010 V 1. Answer the following questions under standard conditions (a) The half cell containing N20/NO is the (b) The half cell containing NO3 /NO2 s the (c) What is Eocell (in V)? Report your answer to three decimal places in standard notation...
8) Calculate the cell potential for the following reaction that takes place in an electrochemical cell at 25°C. 8) 10 Al(s) Al3+(aq, 0.1 15 M) I Al3 (aq, 3.89 M) I Al(s) A13+(aq)+3 e Al(s) E =-1.66 V A) 0,030 V B) 0.090 V 9,1.66 V D) 0.00 V E) 0.060 V
calculate ecell for the following electrochemical cell at 25 C. Pt(s)| H2 (g,1.00 atm) | H+ (aq, 1.00 M) || (Pb2+ (aq, 0.150 M) | Pb (s) Pb^2+(aq) + 2e- ---> Pb (s) Eo=-0.126 V 2H^+ (aq) + 2e- ---> H2 (g) Eo=0.00 V Thank you!
1. A voltaic cell is constructed based on the following redox reaction: Sn2+ (aq) + Mn (s)àSn (s) + Mn2+ (aq) Calculate Ecell at 25 oC under the following conditions: (a) Standard conditions (b) [Sn2+] = 0.0200 M; [Mn2+] = 4.00 M (c) [Sn2+] = 0.500 M; [Mn2+] = 0.00250 M 2. Consider the following redox reaction at 25oC: MnO2 (s)àMn2+ (aq) + MnO4- (aq) (a) Balance the equation in acid (b) Calculate Eocell (c) CalculateDGorxn (d) Calculate K Hint!!!...
1)Calculate the cell potential for the following reaction that
takes place in an electrochemical cell at 25°C.
Al(s) Al3+(aq, 0.115 M) Al3+(aq, 3.89 M) Al(s)
Question options: a) +0.090 V b)+1.66 V c)+0.030 V d)+0.060 V
e)0.00 V
2) Determine the identity of the daughter nuclide from the
positron emission of F.
Question options:
a)
N
b)
Na
c)
Ne
d)
O
e)
F
D Question 14 3 pts The following redox reaction is conducted with [A13+] = 0.80 M and [Mn2+] = 0.30 M. 2 Al(s) + 3 Mn2+(aq) + 2 A13+(aq) + 3 Mn(s) Ecell = 0.48 V Determine the moles of electrons transferred for the reaction as written (n), Q, and the cell potential (cell) at 298 K. n= (Select] Q = (Select] Ecell = (Select)
Conceptual: Consider a Sn(s)|Sn2+(aq) || Cu2+(aq)|Cu(s) cell. If the Sn2+ concentration is increased, what will happen to the measured Ecell value? • Calculation, full-reaction Nernst equation: Use the full Nernst equation to calculate Ecell for the conditions described… • Easier: Ni | Ni2+(0.300 M) || Cu2+(0.002 M) | Cu • Harder: Al | Al3+(0.002 M) || Cu2+(4.00 M) | Cu • Calculation, half-reaction Nernst equation: Use the Nernst equation to calculate E at pH 3.00 and [Cl- ] = 0.0035...
19. Calculate EceLL (in V to two decimal places) for an electrochemical cell based on the following half-reactions at equilibrium. In addition, determine ΔG° (in kJ mol-1 to two decimal places) for the reaction under standard conditions (i.e. all concentrations are 1.0 M) and predict the magnitude of K (e.g. very small, very large, etc.) E (o)0.34 V; E red) 0 9 = 1.68 V oxidation : Cu (s) → Cu2+ (aq, .010 M) + 2e- Reductin: MnO4 (aq, 2.0...