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S2–(aq) + CrO42–(aq) → S(s) + Cr(OH)3(s)

Which of the following is NOT a step you would take to balance this reaction?

A. Add two electrons to one side of a half-reaction

B. Multiply one half-reaction by 3

C. Add 10OH–(aq) to one side of a half-reaction.

D. Do not change any of the numbers to balance the redox atoms in the two half-reactions

2007-02-28 02:56:34 · 4 answers · asked by rnjes21 2 in Science & Mathematics Chemistry

4 answers

2 electrons are transferred by Sulfur and oxidation state of Chromium goes from +6 to +3 so B is the only logical choice.

Answer B)

2007-02-28 03:07:03 · answer #1 · answered by Dr Dave P 7 · 0 0

Split it into two half-reactions.

Oxidation: S2- ---> S + 2 e-
Reduction: CrO42- + 3e- + 4 H2O---> Cr(OH)3 + 5OH-

Overall: 3S2- + 2CrO42- + 8 H2O ---> 3S + 2 Cr(OH)3 + 10OH-

The answers are worded a little weird. Technically, you're not adding 10OH- to a half-reaction, but you end up there because you have to cancel the electrons out in the final reaction. And you're not changing any of the numbers in the half-reactions amongst the atoms actually being oxidized and reduced, but you do have to in the final equation. But you do have to multiply one half-reaction by 3, and you also have to add two electrons to one side of a half-reaction.

2007-02-28 03:11:20 · answer #2 · answered by TheOnlyBeldin 7 · 1 0

First, develop into conscious of the oxidation states of the species being oxidised and decreased: Cl in Cl2O7 is +7 O in H2O2 is -a million Cl in ClO2- is +3 O in O2 is 0 The Cl decreases by 4, the O will boost by a million - so for each Cl decreased there must be 4 O oxidised interior the balanced equation. stability the Cl interior the equation first, then in simple terms be positive you've 4 circumstances as many O being oxidised from H2O2 to O2. upon getting performed this, upload up the costs on the left and evaluate with the costs on the right. cause them to equivalent by including an appropriate type of OH- to the left. To finally stability the equation, upload an appropriate type of water molecules (H2O) on the right to stability the H. hey presto, each thing is now balanced. there is sufficient suggestion the following for you to do the problem your self without me telling you the answer.

2016-12-05 01:40:49 · answer #3 · answered by bartow 4 · 0 0

B is the answer

2007-02-28 05:19:42 · answer #4 · answered by genius'n'training 2 · 0 0

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