Question

In: Chemistry

Ascorbate and copper ions have the following electrochemical potentials: Dehydroascorbate+2H++2e−→Ascorbate ε∘′=+0.08V Cu2++e−→Cu+ ε∘′=+0.159V 10 mL of...

Ascorbate and copper ions have the following electrochemical potentials:

Dehydroascorbate+2H++2e→Ascorbate ε′=+0.08V

Cu2++e→Cu+ ε′=+0.159V

10 mL of a 0.02 M solution of ascorbate in a buffered solution at pH7 at 25∘C is mixed with 10 mL of 0.02 M of Cu2+. Figure out the equilibrium concentration of dehydroascorbate, ascorbate, Cu2+, and Cu+.

Solutions

Expert Solution

Usin EO values of half cell reactions Eo value for the complete cell reaction is calculated and with the help of this value

Equilibrium constant for the reaction is calculated . Now with the help of this equilibrium constant concentration of various

species is calculated at pH 7  as shown in images

  


Related Solutions

Ascorbate and Cu ions have the following electrochemical potentials; Dehydroascorbate + 2H+ + 2e- --> Ascorbate...
Ascorbate and Cu ions have the following electrochemical potentials; Dehydroascorbate + 2H+ + 2e- --> Ascorbate Eo' = +0.08V Cu2+ + e- --> Cu+ Eo' = 0.159 V 10 mL of a 0.02M solution of ascorbate in a buffered solution at pH 7.0; 25 degrees Celsius is mixed with 10mL of 0.02 M of Cu2+. Write a balanced chemical equation for the reaction, and figure out the equilibrium concentration of dehydroascorbate, ascorbate, Cu2+, and Cu+.
Copper can be electroplated at the cathode of an electrolysis cell by the half-reaction. Cu2+(aq)+2e−→Cu(s)Cu2+(aq)+2e−→Cu(s) Part...
Copper can be electroplated at the cathode of an electrolysis cell by the half-reaction. Cu2+(aq)+2e−→Cu(s)Cu2+(aq)+2e−→Cu(s) Part A How much time would it take for 338 mgmg of copper to be plated at a current of 7.1 AA ?
NO3-(aq)+4H+(aq)+3e->NO(g)+2H2O(l) E=0.96V ClO2(g)+e->ClO2-(aq) E=0.95V Cu2+(aq)+2e->Cu(s) E=0.34V 2H+(aq)+2e->H2(g) E=0.00V Pb2+(aq)+2e->Pb(s) E=-0.13V Fe2+(aq)+2e->Fe(s) E=-0.45V Use appropriate data to...
NO3-(aq)+4H+(aq)+3e->NO(g)+2H2O(l) E=0.96V ClO2(g)+e->ClO2-(aq) E=0.95V Cu2+(aq)+2e->Cu(s) E=0.34V 2H+(aq)+2e->H2(g) E=0.00V Pb2+(aq)+2e->Pb(s) E=-0.13V Fe2+(aq)+2e->Fe(s) E=-0.45V Use appropriate data to calculate E?cell for the reaction. 3Cu(s)+2NO3-(aq)+8H+(aq)->3Cu2+(aq)+2NO(g)+4H2O(l) Express your answer using two decimal places.
Reduction half-reaction E∘ (V) Ag+(aq)+e−→Ag(s) 0.80 Cu2+(aq)+2e−→Cu(s) 0.34 Sn4+(aq)+4e−→Sn(s) 0.15 2H+(aq)+2e−→H2(g) 0 Ni2+(aq)+2e−→Ni(s) −0.26 Fe2+(aq)+2e−→Fe(s) −0.45...
Reduction half-reaction E∘ (V) Ag+(aq)+e−→Ag(s) 0.80 Cu2+(aq)+2e−→Cu(s) 0.34 Sn4+(aq)+4e−→Sn(s) 0.15 2H+(aq)+2e−→H2(g) 0 Ni2+(aq)+2e−→Ni(s) −0.26 Fe2+(aq)+2e−→Fe(s) −0.45 Zn2+(aq)+2e−→Zn(s) −0.76 Al3+(aq)+3e−→Al(s) −1.66 Mg2+(aq)+2e−→Mg(s) −2.37 1) Use the table of standard reduction potentials given above to calculate the equilibrium constant at standard temperature (25 ∘C) for the following reaction: Fe(s)+Ni2+(aq)→Fe2+(aq)+Ni(s) 2) Calculate the standard cell potential (E∘) for the reaction X(s)+Y+(aq)→X+(aq)+Y(s) if K = 3.80×10−4. Express your answer to three significant figures and include the appropriate units.
Calculate the cell potential for the following electrochemical cell: Zn | [Zn2+]=4.2M || [Cu2+]=0.004M | Cu
Calculate the cell potential for the following electrochemical cell: Zn | [Zn2+]=4.2M || [Cu2+]=0.004M | Cu  
Consider a galvanic cell based upon the following half reactions: Cu2+ + 2e- → Cu 0.34...
Consider a galvanic cell based upon the following half reactions: Cu2+ + 2e- → Cu 0.34 V Fe3+ + 3e- → Fe -0.0036 V How many of the following responses are true? 1. Decreasing the concentration of Cu2+ (assuming no volume change) will decrease the potential of the cell 2. Increasing the concentration of Fe3+ (assuming no volume change) will increase the potential of the cell 3. Adding equal amounts of water to both half reaction vessels will increase the...
Find the standard reduction potentials (voltages) expected for the following half reactions Cu+2 + 2e- ------------->...
Find the standard reduction potentials (voltages) expected for the following half reactions Cu+2 + 2e- -------------> Cu Pb+2 + 2 e- ------------> Pb Zn+2 + 2 e- -------------> Zn Fe+3 + e- ---------------> Fe+2 Br2 + 2e- --------------> 2Br- Calculate expected voltages for the electrochemical cells made by connecting the copper(II) half reaction to each of the others in the list above. Note that one of the listed reductions must be reversed to become an oxidation for anything tooccur and...
Consider the following reduction potentials: Mg2+ + 2e- → Mg E° = -2.37 V V2+ +...
Consider the following reduction potentials: Mg2+ + 2e- → Mg E° = -2.37 V V2+ + 2e- → V E° = -1.18 V Cu2+ + e- → Cu+ E° = +0.15 V Which one of the following reactions will proceed spontaneously? • Mg2+ + V → V2+ + Mg • Mg2+ + 2Cu+ → 2Cu2+ + Mg • V2+ + 2Cu+ → V + 2Cu2+ • V + 2Cu2+ → V2+ + 2Cu+
Given: Pb2+(aq)+2e–⇌Pb(s);E°=–0.13 Mg2+(aq)+2e–⇌Mg(s);E°=–2.38V Ag+(aq)+e–⇌Ag(s);E°=0.80V 2H+(aq)+2e–⇌ H2(g);E°=0.00V Under standard-state conditions, which of the following species is the...
Given: Pb2+(aq)+2e–⇌Pb(s);E°=–0.13 Mg2+(aq)+2e–⇌Mg(s);E°=–2.38V Ag+(aq)+e–⇌Ag(s);E°=0.80V 2H+(aq)+2e–⇌ H2(g);E°=0.00V Under standard-state conditions, which of the following species is the best oxidizing agent? a. H b.Mg2+ c. Ag+ d. Pb e. Ag
Given the following standard reduction potentials: Pb2+ (aq) +2e- ---> Pb (s) E= -.126V PbSO4(s) +...
Given the following standard reduction potentials: Pb2+ (aq) +2e- ---> Pb (s) E= -.126V PbSO4(s) + 2e- ---> Pb(s) + SO42- (aq) E= -.356V Determine the Ksp for PbSO4(s) at 25 degrees C
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT