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⁺.

Expert Solution

Usin E^O values of half cell reactions E^o 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

queen_honey_blossom answered 2 years ago

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...

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Consider a galvanic cell based upon the following half reactions: Cu2+ + 2e- → Cu 0.34...

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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...

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