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Consider the following reaction at 298 K: 3Cu2+(aq) + 2Al(s) + 3Cu(s) + 2 A13+ (aq)...
The reducing agent in the reaction below is 3CuCl2(aq)+2Al(s) →3Cu(s)+2AlCl3(aq) a. Al3+ b. Cu2+ c. Al d. Cu
For the reaction below: (5 points) 3Cu(s) + 2NO3– (aq) + 8 H+(aq) 3Cu2+(aq) + 2NO(g) + 4H2O(퓁)What is Gº at 25 ºC?(a) −361 kJ(b) 180. kJ(c) 361 kJ(d) −120 kJ(e) none of these
For the following electron-transfer reaction: 2Cr(s) + 3Cu2+(aq) 2Cr3+(aq) + 3Cu(s) The oxidation half-reaction is: The reduction half-reaction is:
What is the calculated value of the cell potential at 298K for an electrochemical cell with the following reaction, when the Cu2+ concentration is 4.95×10-4 M and the Al3+ concentration is 1.21 M ? 3Cu2+(aq) + 2Al(s)3Cu(s) + 2Al3+(aq) Answer: V
Calculate the Ecell value at 298 K for the cell based on the reaction: Cu(s) + 2Ag+ (aq) → Cu2+ (aq) + 2Ag(s) where [Ag+] = 0.00350 M and [Cu2+] = 7.00x10-4 M. The standard reduction potentials are shown below: Ag+(aq) +e → Ag(s) E° = 0.7996 V Cu2+ (aq) + 2e -→ Cu(s) E° = 0.3419 V 2nd attempt Ecell = V
Use the tabulated electrode potentials to calculate K for the oxidation of nickel by H+: Ni(s)+2H+(aq)→Ni2+(aq)+H2(g) Express your answer using two significant figures. 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 Cu2+(aq)+2e−→Cu(s) 0.34 Mn2+(aq)+2e−→Mn(s) −1.18 2H+(aq)+2e−→H2(g) 0.00 Al3+(aq)+3e−→Al(s) −1.66 Fe3+(aq)+3e−→Fe(s) −0.036 Mg2+(aq)+2e−→Mg(s) −2.37 Pb2+(aq)+2e−→Pb(s) −0.13 Na+(aq)+e−→Na(s) −2.71 Sn2+(aq)+2e−→Sn(s) −0.14 Ca2+(aq)+2e−→Ca(s) −2.76 Ni2+(aq)+2e−→Ni(s) −0.23 Ba2+(aq)+2e−→Ba(s) −2.90 Co2+(aq)+2e−→Co(s) −0.28 K+(aq)+e−→K(s) −2.92 Cd2+(aq)+2e−→Cd(s)...
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...
1)Consider the following half-reactions: Half-reaction E° (V) Cl2(g) + 2e- > 2Cl-(aq) 1.360V Cd2+(aq) + 2e- > Cd(s) -0.403V Al3+(aq) + 3e- > Al(s) -1.660 The strongest oxidizing agent is: _______ enter formula The weakest oxidizing agent is: _______ The weakest reducing agent is: _______ The strongest reducing agent is: _______ Will Al3+(aq) reduce Cl2(g) to Cl-(aq)? _____yes or no Which species can be reduced by Cd(s)? If none, leave box blank. 2) Use the table 'Standard Reduction Potentials' located...
Candidate l: Zn(s) | Zn2+(aq,0.500 M) I Cu2+(aq, 1.00 M) Cu(s) Candidate 2: Pb(s) | Pb2+(aq, 0.500 M) || Cu2+(aq, 1.00 M) Cu(s) Candidate 3: Mg(s) | Mg2+(aq, 0.500 M) | Pb2+(aq, 1.00 M)| Pb(s) (a) 6 pts) Choose one of the candidate voltaic cells #1, #2, or #3. Draw a schematic cell diagram for the candidate voltaic cell of choice. Clearly label anode, cathode, electrodes, ions and their concentrations, salt bridge, and the flow of electrons. (b) (5 pts)...
Consider the following electrochemical cell: Al (s) I Al3+ (aq) (1.00 M) II Cu2+ (aq) (0.0020 M) I Cu (s) where Cu2+ aq + 2e- -> Cu (s) +0.34 V and Al3+ aq + 3e- -> Al (s) -1.66 V Calculate the standard cell potential for the given cell, calculate the cell potential for the given cell, and sketch the electrochemical cell using two beakers and labeling the electrodes, the cathode, the anode, the direction of electron flow in the...