Are the questions asked in the annonyomous valid ? 2 NaCIO2 (aq)+C12 (g) >2 CIO2

Question

Are the questions asked in the annonyomous valid ? 2 NaCIO2 (aq)+C12 (g) >2 CIO2 (g)+2 NaCI (aq) ClO2(aq) + e--> ClO2^-E-0.954 V (anode) Cl2(g) + 2 e > 2 Cl- E0 = +1.36 v (cathode) EOcell = Ec-Ea = 1.36-0.954 = 0.406 v -2*96500*0.406 =-78358 joule DG0 =-RTInKeq -78358 =-8.314"298Ink k 5.44*10^13 Hide comments (1) Comments Anonymous posted 1 hour ago Edit Delete in order to get the same amount of e in the equations wouldnt you mutiply the 1st equation by 2 to g = amount of electrons. ? You commented ual?

Explanation / Answer

Ans: Yes, It is correct u must multiply 1st equation by factor 2 to balance the net number of electron in the redox reaction i.e Zero and value of E0cell doesn't change it is fixed value (E0cell = 0.954 V)

half cell reaction:

At anode : Multiply !st reaction by 2

2ClO2^- ---> 2ClO2(aq) + 2e ( oxidation reaction)

At cathode : Cl2(g) + 2e ---> 2Cl- ( Reduction reaction)

Net reaction = 2ClO2^- + Cl2 ----> 2ClO2 + 2Cl- ( net electron is zero )

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