How would dichromate ion, #Cr_2O_7^(2-)#, oxidize sulfide ion, #S^(2-)#, to give elemental sulfur? The reduction product is #Cr^(2+)#.

Answer 1

Are you sure that the dichromate is reduced to #Cr(+II)#.........?

Normally, we would expect dichromate to give #Cr^(3+)# as its reduction product:
#Cr_2O_7^(2-) +14H^+ + 6e^(-) rarr 2Cr^(3+) +7H_2O#

Elemental sulfur is produced by oxidizing the sulfide anion.

#S^(2-) rarr S(s) + 2e^(-)#

All things considered.

#Cr_2O_7^(2-) + 14H^(+) + 3S^(2-) rarr 2Cr^(3+) + 3S + 7H_2O#

PS: I would double-check your sources and lab manual; I'm not saying you're incorrect.

We could formulate reduction to #"chromous ion"# as follows:
#Cr_2O_7^(2-) +14H^+ + 8e^(-) rarr 2Cr^(2+) +7H_2O(l)#
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Answer 2

The dichromate ion (Cr₂O₇²⁻) oxidizes sulfide ion (S²⁻) to elemental sulfur, with the reduction product being Cr²⁺. The balanced equation is:

[3 \text{S}^{2-} + 14 \text{H}^+ + \text{Cr}_2\text{O}_7^{2-} \rightarrow 3\text{S} + 2\text{Cr}^{2+} + 7\text{H}_2\text{O}]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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