In the reaction MnO4−+SO32−+H+→Mn2++SO42− the number of H+ ions involved is

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answer is B.

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Detailed Solution

Assign oxidation numbers for each atom in the equation. Write down the transfer of electrons. Be carefully, insert coefficients, if necessary, to make the numbers of oxidized and reduced atoms equal on the two sides of each redox couples +7 +4 2+ +6MnO4−+SO32−+H+→Mn2++SO42− MnO4−→Mn2+ ;; MnO4−+5e−→Mn2+ MnO4−+5e−+8H+→Mn2++4H2O SO32−→SO42− ; SO32−+H2O→SO42− ;;SO32−+H2O→SO42−+2H+ (SO32−+H2O→SO42−+2H++2e−)×5(MnO4−+5e−+8H+→Mn2++4H2O)×2overall reaction2MnO4−+5SO32−+5H2O+6H+→2Mn2++8H2O2MnO4−+5SO32−+6H+→2Mn2++3H2O

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