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# Tutor profile: Taylor B.

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Taylor B.
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## Questions

### Subject:Chemistry

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Question:

Consider the following oxidation reduction reaction: Fe2O3 (s) + 3 CO (g) --> 2 Fe (s) + 3 CO2 (g) Show and balance the two half reactions

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Taylor B.

The first step in balancing half reactions is determining the oxidation numbers of each atom in the reaction. The oxidation numbers are as follows in parentheses for each atom: Fe2(+3)O3(-2) (s) + 3 C(+2)O(-2) (g) --> 2 Fe(0) (s) + 3 C(+4)O2(-2) (g) This allows us to separate the half reactions based on whether they are oxidation or reduction. The oxidation reaction is the half reaction that loses electrons so the oxidation reaction before balancing is as follows: 3 CO (g) --> 3 CO2 (g) This is because the oxidation number of the carbon atom rises from 2+ to 4+, and since electrons are negative charges, a higher positive charge indicates the loss of electrons. The unbalanced reduction reaction is the remaining half of the reaction: Fe2O3 (s) --> 2 Fe (s) This is the reduction half of the reaction because it gains electrons, changing the charge of the iron atom from +3 to 0. So, our unbalanced reactions are: O: 3 CO (g) --> 3 CO2 (g) R: Fe2O3 (s) --> 2 Fe (s) To balance these reactions, first we balance any atoms other than oxygen and hydrogen because they have special rules. As both half reactions here are already balanced without the oxygen and hydrogen atoms, we can now balance for these atoms, For oxygen, any needed oxygen atoms are added by adding H2O molecules so when we balance these equations for oxygen we get these half reactions: O: H2O + 3 CO (g) --> 3 CO2 (g) R: Fe2O3 (s) --> 2 Fe (s) + 3 H2O When balancing hydrogen atoms, we add protons (H+) for each needed hydrogen atom and get the resulting reactions: O: 3 H2O (l) + 3 CO (g) --> 3 CO2 (g) + 6 H+ R: 6 H+ + Fe2O3 (s) --> 2 Fe (s) + 3 H2O (l) The final step in balancing half reactions is balancing charges. Charges for each side are listed in brackets after final ion: O: 3 H2O (l) + 3 CO (g) [0] --> 3 CO2 (g) + 6 H+ [+6] R: 6 H+ + Fe2O3 (s) [+6] --> 2 Fe (s) + 3 H2O (l) [0] Now that we have the charges for each side, we add electrons to balance the charges: O: 3 H2O (l) + 3 CO (g) [0] --> 3 CO2 (g) + 6 H+ + 6 e- [0] R: 6 H+ + Fe2O3 (s) + 6 e- [0] --> 2 Fe (s) + 3 H2O (l) [0] Our final balanced half reactions are as follows: O: 3 H2O (l) + 3 CO (g) --> 3 CO2 (g) + 6 H+ + 6 e- R: 6 H+ + Fe2O3 (s) + 6 e- --> 2 Fe (s) + 3 H2O (l)

### Subject:Psychology

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Question:

In Pavlov’s dog experiment the food was the ________ stimuli; while the metronome ticking was the _______ stimuli.

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Taylor B.

Unconditioned; neutral

### Subject:Anatomy

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Question:

Identify all the carpal bones

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Taylor B.

Scaphoid, lunate, triquetral, trapezium, pisiform, trapezoid, hamate, capitate

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