University:
Boston University
Course:
CAS CH 214 | Organic Chemistry with Qualitative Analysis
Academic year:

2022
Views:

345

Pages:

2
Author:

Annabel Combs

Electrochemistry example problems 0.763 V. A galvanic cell voltage is positive Problem 1: At 25 , the standard reduction potential of Zn2 is E o is made by coupling a standard Zn/Zn2 electrode to a standard hydrogen electrode at 25 . The hydrogen electrode is then buffered to have pH 5.57. Calculate the resulting cell potential. The reduction potential of the standard hydrogen electrode is by definition exactly 0 V. Since this is greater than standard reduction potential of Zn2 , E o 0.763 V, we conclude the following: H is reduced to H2 at the cathode, 2 H aq, 10 pH M 2 e ⟶ H2 g, 1 atm , and Zn is oxidized to Zn2 at the anode, Zn s ⟶ Zn2 aq, 1 M 2e . From these half reactions, we see that the number of electrons per reaction unit is 2. The full cell redox reaction is 2 H3O aq, 10 pH M Zn s ⟶ H2 g, 1 atm Zn2 aq, 1 M 2 H2O l , and therefore the we can evaluate the reaction quotient to be Zn H O 1 H2 10 The standard cell potential is ° With °cathode °anode 0 0.763 0.763 V and °, we can use the Nernst equation at 25 ° 0.05912 V log 1 0.763 V 0.05912 V log 2 0.763 V 0.05912 V pH 0.763 V 0.05912 V 5.57 0.434 V 10 to evaluate the cell potential, Problem 2: This question concerns a concentration cell with positive voltage at 25 , constructed from hydrogen half cells of different hydronium ion concentration and sealed so that their hydrogen gas pressure can be controlled. In one half cell the hydronium concentration is 2.0 M and the hydrogen gas pressure is 0.50 atm, and in the other half cell the hydronium concentration is 0.020 M and the hydrogen gas pressure is 5.0 atm. Platinum electrodes are used to connect the cell to an external circuit. a. Write the line notation of the cell. Indicate for each species its concentration or pressure. For the cell to be spontaneous, the skeleton process must be from high concentration to low concentration, H3O aq, 2.0 M ⟶ H3O aq, 0.020 M The corresponding half reactions are 2 H3O aq, 2.0 M H2 g, 5.0 atm 2 e ⟶ H2 g, 0.50 atm 2 H2O l , cathode 2 H2O l ⟶ 2 H3O aq, 0.020 M 2 e , anode and so the line notation of the cell is Pt(s) | H2 g, 5.0 atm | H3O aq, 0.020 M || H3O aq, 2.0 M | H2 g, 0.50 atm | Pt s b. Calculate the cell voltage at 25 The standard potential is ° 0 V. From the half reactions, the full cell reaction is 2 H3O aq, 2.0 M H2 g, 5.0 atm ⟶ H2 g, 0.50 atm From the full cell reaction, the reaction quotient is 0.50 0.020 2.0 5 1.0 10 And from the reaction quotient, the cell voltage is ° 0 0.05912 V log 0.05912 V log 1.0 10 2 0.05912 V 2 0.15 V 5.0 2 H3O aq, 0.020 M