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Lana71 [14]
3 years ago
5

Identify the atoms that are oxidized and reduced, the change in oxidation state for each, and the oxidizing and reducing agents

in each of the following equations: (a) Mg(s)+NiCl2(aq)⟶MgCl2(aq)+Ni(s) (b) PCl3(l)+Cl2(g)⟶PCl5(s) (c) C2H4(g)+3O2(g)⟶2CO2(g)+2H2O(g) (d) Zn(s)+H2SO4(aq)⟶ZnSO4(aq)+H2(g) (e) 2K2S2O3(s)+I2(s)⟶K2S4O6(s)+2KI(s) (f) 3Cu(s)+8HNO3(aq)⟶3Cu(NO3)2(aq)+2NO(g)+4H2O(l)
Chemistry
2 answers:
Dominik [7]3 years ago
7 0

Answer:

<u>An oxidizing agent</u><u>,</u> <u><em>gains electrons and is reduced</em></u><u> </u>in a chemical reaction. It is usually in the highest oxidation state as it gains (accepts) electrons and is reduced.

<u>A reducing agent</u> <u><em>loses electrons and is oxidized</em></u><em> </em>in a chemical reaction. A reducing agent is typically in one of i<em>ts lower possible oxidation states</em>, because it loses electrons (<em>electron donor</em>).

Having noted this, the following are the oxidation states of the substances in each equation.

Explanation:

Mg(s )+ NiCl2(aq)⟶MgCl2(aq)+ Ni(s) :

  • <em>Mg is oxidized </em><em>because it has lost electrons</em><em> </em>(oxidation number 0 to +2);
  • <em>Ni is reduced </em>as it has gained electrons (oxidation number 2+ to 0)

PCl3(l )+ Cl2(g)⟶ PCl5(s)

  • <em>P is oxidized as it haslost electrons </em>(oxidation number 0 to +2);
  • <em>Ni is reduced </em>as it has gained electrons (oxidation number 2+ to 0)

C2H4(g) + 3O2(g)⟶2CO2(g) + 2H2O(g)

  • <em> H2 is oxidized </em><em>as it has lost electrons</em><em> </em>(oxidation number +2 to +1);
  • <em>C is reduced </em>as it has gained electrons (oxidation number -2 to +4)

Zn(s) + H2SO4(aq) ⟶ZnSO4(aq) + H2(g)

  • <em> Zn is oxidized </em><em>as it has lost electrons</em><em> </em>(oxidation number 0 to +2);
  • <em>H2 is reduced </em>as it has gained electrons (oxidation number +2 to 0)

2K2S2O3(s) + I2(s)⟶ K2S4O6(s) +2KI(s)

  • <em> H2 is oxidized </em><em>as it has lost electrons</em><em> </em>(oxidation number +2 to +1);
  • <em>C is reduced </em>as it has gained electrons (oxidation number -2 to +4)

3Cu(s) + 8HNO3(aq) ⟶ 3Cu(NO3)2(aq) + 2NO(g) + 4H2O(l)

  • <em>Cu is oxidized </em><em>as it has lost electrons</em><em> </em>(oxidation number 0 to +2);
  • <em>Nitrogen is reduced </em>as it has gained electrons (oxidation number +5 to +4)
SashulF [63]3 years ago
5 0

Answer:

Explanation:

Step 1. Assign oxidation numbers to all

elements

PbS(s) + O2

(g) PbO(s) + SO2

(g)

• Step 2. Identify oxidized and reduced species

– PbS was oxidized (O.N. of S: -2 -> +4)

– O2 was reduced (O.N. of O: 0 -> -2)

• Step 3. Compute e-

lost and e- gained

– In the oxidation: 6e- were lost from S

– In the reduction: 2e- were gained by each O

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Alright sorry you're getting the answer hours later, but i can help with this.
so you're looking for specific heat, the equation for it is <span>macaΔTa = - mbcbΔTb with object a and object b. that's mass of a times specific heat of a times final minus initial temperature of a equals -(mass of b times specific heat of b times final minus initial temperature of b)
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