PLEASE,COULD ANYONE HELP ME to balance this redox reaction : Cr2S3 PbO2 + H2SO4 + -------> PbSO4 H2Cr2O7+H20
I REALLY wish you would spell chemistry right. You've posted the last several days with the wrong spelling.
I've done all but the water. You can fill that in.
2Cr2S3 + 30PbO2 + 24H2SO4 --> 30PbSO4 + 2H2Cr2O7 + ?H20
Check it out. Make sure I made no typos.
please,sir,If you give me the answers briefly means with oxidation state and half reaction,I will be grateful to you.Because i am suffering so much to find out oxidation state of Cr2S3 and half reaction.
To balance the redox reaction: Cr2S3 + PbO2 + H2SO4 -> PbSO4 + H2Cr2O7 + H2O, we need to follow some steps:
Step 1: Assign oxidation numbers to each element in the equation.
Cr2S3 + PbO2 + H2SO4 -> PbSO4 + H2Cr2O7 + H2O
Cr: +3
S: -2
Pb: +4
O: -2
H: +1
S: +6
Step 2: Identify which elements are being reduced and which are being oxidized.
In this equation, Cr is being oxidized from +3 to +6, while Pb is being reduced from +4 to +2.
Step 3: Write the half-reactions for oxidation and reduction.
Reduction half-reaction:
PbO2 + 4H+ + 2e- -> Pb2+ + 2H2O
Oxidation half-reaction:
Cr2S3 -> Cr2O7^2- + S
Step 4: Balance the atoms in each half-reaction.
Reduction half-reaction:
PbO2 + 4H+ + 2e- -> Pb2+ + 2H2O
(Pb: 1 Pb -> 1 Pb; O: 2 O -> 2 H2O; H: 4 H -> 4 H2O)
Oxidation half-reaction:
Cr2S3 -> Cr2O7^2- + S
(2 Cr: 2 Cr -> 2 Cr; S: 3 S -> 1 S)
Step 5: Balance the charges in each half-reaction.
Reduction half-reaction:
PbO2 + 4H+ + 2e- -> Pb2+ + 2H2O
(2-: 2- + 4+ + 2e- -> 2+ + 0)
Oxidation half-reaction:
Cr2S3 -> Cr2O7^2- + S
(0: 0 -> 2- + 1-)
Step 6: Balance the number of electrons transferred.
To balance the electrons, we need to multiply the oxidation half-reaction by 2:
2Cr2S3 -> Cr2O7^2- + S + 12e-
(PbO2 + 4H+ + 2e- -> Pb2+ + 2H2O; 2Cr2S3 -> Cr2O7^2- + S + 12e-)
Step 7: Combine the two half-reactions.
To balance the electrons, we need to multiply the reduction half-reaction by 12:
12PbO2 + 48H+ + 24e- -> 12Pb2+ + 24H2O
(2Cr2S3 -> Cr2O7^2- + S + 12e-; 12PbO2 + 48H+ + 24e- -> 12Pb2+ + 24H2O)
Now, let's combine the two half-reactions:
12PbO2 + 48H+ + 24e- + 2Cr2S3 -> 12Pb2+ + 24H2O + Cr2O7^2- + S + 12e-
Step 8: Simplify the equation by canceling out the species that appear on both sides.
12PbO2 + 48H+ + 2Cr2S3 -> 12Pb2+ + 24H2O + Cr2O7^2- + S
Finally, the balanced redox reaction is:
12PbO2 + 48H+ + 2Cr2S3 -> 12Pb2+ + 24H2O + Cr2O7^2- + S