# How to balance this redox equation?

Can someone explain me how to balance this equation using the half-reaction method?

K2Cr2O7 + H2S + H2SO4 --> S + Cr2(SO4)3 + K2SO4 + H2O

• Cr2O7-2 + S-2 --------------------> SO4-2 + CR+3

oxidation half reaction -------------------

S-2 ----------------------> SO4-2

S-2 + 4H2O ------------> SO4-2 [add water to balance O]

S-2 +4H2O -------------> SO4-2 + 8H+ [ add H+ to balance H ]

S-2 + 4H2O ------> SO4-2 + 8H+ + 8e- --(I) [add e- to balance charge ]

reduction half reaction -------- Cr2O7-2 ----------------------> Cr+3

Cr2O7-2 ----------------------->2 Cr+3 [ balance Cr ]

Cr2O7-2 ------------------> 2Cr+3 + 7H2O [to balance O add water ]

Cr2O7-2 + 14 H+ --------> 2Cr+3 + 7 H2O [ to balance H add H+ ]

Cr2O7-2 + 14H+ + 6 e- ------> 2Cr+3 + 7H2O-----(II) [add e-to balance charge

now multiply (I) by 6 & (II) by 8 for equilising e- add----------

6 S-2 + 24 H2O+8Cr2O7-2+112 H+ -------->6SO4-2+48H+ +16Cr+3 +56H2O

or 6S-2 +8Cr2O7-2 +64H+-------> 6 SO4-2 + 16 Cr+3 + 32H2O

or 3 S-2+ 4 Cr2O7-2 + 32H+ -----> 3SO4-2 + 8 Cr+3 + 16 H2O -----------------( III)

(III) is the balanced equation

• K2cr2o7 H2s H2so4

• Cr2O7(2-)+6e+14H(+)->2Cr(3+)+7H2O

H2S-2e->S+2H(+)||x3

===============

Cr2O7(2-)+14H(+)+3H2S->

2Cr(3+)+3S+7H2O+6H(+)||-6H(+)

--------------

Cr2O7(2-)+8H(+)+3H2S->

2Cr(3+)+3S+7H2O

----------------------

K2Cr2O7+3H2S+4H2SO4->

->Cr2(SO4)3+3S+K2SO4+7H2O

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Best regards from RUSSIA!!!

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