Light green (Compound 'A') White Residue (B) C + D + E (i) 'D' and 'E' are two acidic gas. (ii) 'D' is passed through HgCl2 solution to give yellow ppt. (iii) 'E' is passed through water first and the

Question

Light green (Compound 'A') $\xrightarrow{\Delta}$ White Residue (B) 
$\xrightarrow{	ext{High Temp.}}$ C + D + E
(i) 'D' and 'E' are two acidic gas.
(ii) 'D' is passed through HgCl2 solution to give yellow ppt.
(iii) 'E' is passed through water first and then H2S is passed, white turbidity 
is obtained.
(iv) A is water soluble and addition of HgCl2 in it, yellow ppt is obtained but 
white ppt does not turn into grey on addition of excess solution of 'A'. [1]

Prashant Kotian · Accepted Answer

Correct answer: C. Step 1: Balance the electron transfer between the reduction of persulfate (gains 2e^- per mole) and the oxidation of manganese (loses 5e^- per mole). Step 2: To equate the electrons lost and gained, multiply the reduction half-reaction by 5 and the oxidation half-reaction by 2. This gives: 5S2O8^2- + 10e^- ---> 10SO4^2- and 2Mn^2+ + 8H2O ---> 2MnO4^- + 16H^+ + 10e^-. Step 3: This shows that 5 moles of S2O8^2- are required to oxidize 2 moles of Mn^2+. Therefore, 2.5 moles of S2O8^2- are needed per 1 mole of Mn^2+, which corresponds to option (a).

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