Chlorine dioxide clo2 which is produced by the following...

Chlorine dioxide (ClO2), which is produced by the following reaction: 2NaClO2(aq) + Cl2(g) <-> 2ClO2(g) + 2NaCl(g) has been tested as a disinfectant for municipal water treatment. Using the following data, calculate delta G (in kJ) at 25 degrees C for the production of ClO2. ClO2(g) + 1e- <-> ClO2- (E = 0.954V) Cl2(g) + 2e- <-> 2Cl-(aq) (E = 1.36V)

Transcript

00:01 Hi, in this question, we are given with this reaction in which 2 moles of naclo2 reacts with cl2 gas to produce 2 moles of clo2 and 2 moles of nacl.

00:11 We are given with these two half reactions along with their reduction potential values.

00:16 We are asked to calculate the delta g0 value at 25 degrees celsius.

00:21 Here, in the half reactions, we can say that higher the positive value, more will be the oxidizing nature, it means the cl2 will undergo reduction and clo2 minus will undergo oxidation.

00:37 Therefore we need to reverse the first half reaction which will be called as oxidation half reaction and the second will be the reduction half reaction.

00:46 We need to balance the number of electrons for which we need to multiply the first half reaction by 2.

00:52 On reversing and multiplying we can rewrite the equations in this manner.

00:58 The first oxidation occurs at anode and the second reduction occurs at cathode.

01:07 We can say that the delta g -not is equal to minus n -f e -nodd cell.

01:16 N is the number of electrons that are exchanged in this reaction.

01:20 F is the faraday's constant and e -0 cell we need to calculate.

01:25 We can say e -0 cell is equal to e -0 oxidation plus e -0 reduction, that is, reduction potential value and oxidation potential value.

01:40 This is the e -0 oxidation and this is the e -0 reduction.

01:45 We can substitute the values here.

01:47 We can write minus 0 .954 volt plus 1 .36 volt...