In thermodynamics, we determine the spontaneity of a reaction by the sign of Del
ID: 726143 • Letter: I
Question
In thermodynamics, we determine the spontaneity of a reaction by the sign of Delta G. In electrochemistry, spontaneity is determined by the sign of Ecell 0. The values of Delta G and E cell 0 are related by the following formula: Deta G 0= -nFE cell 0 where n is the number of moles of electrons and F = 96,500J/V mol e- is the Faraday constant. The standard reduction potentials of lithium metal and chlorine gas are as follows: Reaction Reduction potential (V) Li+(aq) + e- rightarrow Li(s) -3. 04 Cl2(g) + 2e- rightarrow 2Cl-(aq) + 1. 36 In a galvanic cell, the two half-reactions combine to 2Li(s) + Cl2(g) rightarrow 2LiCl(aq) Calculate the cell potential of this reaction under standard reaction conditions. Express your answer with the appropriate units.Explanation / Answer
Combining these half-reactions, we can identify four possible reactions 2 Li+ (aq) + 2 Cl- (aq) = Li (s) + Cl2 (g) E o = - 3.04 - 1.36 = - 4.40 Cell Potential= 4.40 V n is the number of moles of electrons transferred in the reaction, AS WRITTEN. In this case, n = 2 because you multiplied the Li half-reaction by 2 (before switching it around and adding it to the Cl reaction). Then 2 Li lose 2 moles of electrons while the the two moles of Cl in one mole of Cl2 gain 2 moles of electrons in being converted to 2 Cl?. So, ?G° = –(2 mol)(96.485 kJ/V•mol)(4.4 V) = – 849.1 kJ FreeEnergy = – 849.1 kJ