How is pyrophosphate anion, #P_2O_7^(4-)# reduced to elemental phosphorus, with oxidation of hydrogen sulfide to sulphur?

The overal redox equation is:

#2P_2O_7^(4-)+8H^(+)+10H_2SrarrP_4 +10S+14H_2O#

Whew!! Arithmetic.

Explanation:

I take it you mean #"pyrophosphate anion"#, #P_2O_7^(4-)#. This is #P(V+)#, which is reduced to elemental phosphorus, #P^(0)#.

#P_2O_7^(4-)+14H^(+)+10e^(-) rarr1/2P_4+7H_2O# #(i)#

Charge and mass are balanced here. Why is that important?

Sulfide anion is oxidized to elemental sulfur:

#8H_2S rarrS_8 +16H^+ + 16e^-# #(ii)#

Charge and mass are balanced again!

So we cross multiply, and #16xx(i) + 10xx(ii)=#

#16P_2O_7^(4-)+64H^(+)+80H_2Srarr8P_4 +10S_8+112H_2O#

We could make this a little simpler by dividing thru by #8#, i.e. we treat #S_8# simply as #S#:

#2P_2O_7^(4-)+8H^(+)+10H_2SrarrP_4 +10S+14H_2O#

Here again, charge and mass are balanced, as required.