The free energy change for an oxidation-reduction reaction occurring at constant

Question

The free energy change for an oxidation-reduction reaction occurring at constant temperature and pressure is given by Delta G = -nFE where n is the moles of electrons transferred in the reaction, E is the cell potential of the reaction, and F is Faraday's constant. Use this relationship to answer the problem below. In one half-cell of a voltaic cell, the following reaction occurs: Be(s) rightarrow Be^2+(aq) + 2e^-. Is the other half-cell of voltaic cell, the following reaction occurs: 2 H^+(aq) + 2e^- rightarrow H_2(g). When [Be^2+] = 0.038 M, [H^+] = 4.9 M, and the H_2 pressure = 0.34 atm, E = 1.83 V. What is the free energy change for electron flow in this voltaic cell? (Include the sign of the value in your answer.)

Explanation / Answer

Addition of given reactions , Be +2H+ ---àBe+2+ H2(g), Eo= 1.83V

E= Eo- (0.0591/n)* logQ, n= no of electrosn exchanged= 2

Q= [Be+2] [PH2]/ [H+]2 = 0.038* 0.34/ (4.9)2= 0.000538

E= 1.83- (0.0591/2)* log (0.000538)= 1.93V

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