Can someone give 5-10 illustrations of identifying oxidation and reduction species in redox reactions.

Oxidation - involves the loss of electrons or hydrogen OR gain of oxygen OR increase in oxidation state
Reduction - involves the gain of electrons or hydrogen OR loss of oxygen OR decrease in oxidation state
1.H₂ + F₂ ? 2 HF

The overall reaction may be written as two half-reactions:

H₂ ? 2 H⁺ + 2 e^- (the oxidation reaction)

F₂ + 2 e^- ? 2 F^- (the reduction reaction)

There is no net change in charge in a redox reaction so the excess electrons in the oxidation reaction must equal the number of electrons consumed by the reduction reaction. The ions combine to form hydrogen fluoride:

H₂ + F₂ ? 2 H⁺ + 2 F^- ? 2 HF

2. 2AgCl(s) + H₂(g) ? 2 H⁺(aq) + 2 Ag(s) + 2 Cl

AgCl

Ag has a +1 oxidation state

Cl has a -1 oxidation state

H₂ has an oxidation state of zero
H⁺ has a +1 oxidation state

• Ag has an oxidation state of zero.
• Cl^- has a -1 oxidation state.
• Ag went from +1 in AgCl to 0 in Ag. The silver atom gained an electron.H went from 0 in H₂ to +1 in H⁺. The hydrogen atom lost an electron.Cl kept its oxidation state constant at -1 throughout the reaction.Silver gained an electron. This means the silver was reduced.Hydrogen lost an electron. This means the hydrogen gas was oxidized.Its oxidation state was increased by one.. Chlorine's oxidation state was unchanged throughout the reaction.

3.2Na + Cl2 ------->2NaCl

Na –loss of 2 e-

Cl –gain of 2e-

4.2Na + O2-------->Na2O

Na-loss of 2 e-

O -gain 2 e-

5.2Na + S ------->Na2S

Na-loss of 2e-

S-gain of 2 e-

Thanks & Regard

Aarti Gupta

askiitians Faculty

Approved

Zn(s) + CuSO4(aq) ? ZnSO4(aq) + Cu(s)

Copperbegins in the +2 oxidation state and is reduced to a neutral state over the course of the reaction. Zinc begins neutral and is oxidized to the +2 state.

The oxidation half-reaction is:

Zn(s) ? Zn2++ 2e-

The reduction half-reaction is:

Cu2++ 2e-? Cu(s)

Sample 1. equation: 2 H2+ O2? 2 H2O

The sum of oxidation states in the reactants is equal to that in the products: 0 + 0 ? (2)(+1) + (-2)

In this equation both H2and O2are free elements and their oxidation states are 0. Theproductis H2O, in which the oxidation state of oxygen is -2 and that of each hydrogen is +1.

Sample 2. equation: 2 H2O ? 2 H2+ O2

Calculation: (2)(+1) + (-2) = 0 ? 0 + 0

In this equation the water is "decomposed" into hydrogen and oxygen, which are both neutral. Similar to the previous example, the H2O has a total oxidation state of 0, with each H taking on a +1 state and the O being -2. Thus, decomposition oxidizes oxygen from -2 to 0 and reduces hydrogen from +1 to 0.

Sample 3. equation: Cl2+ 2 NaBr ? 2 NaCl + Br2

Calculation: (0) + ((+1) + (-1) = 0) -> ((+1) + (-1) = (0) + 0

In this equation Br is replaced with Cl and Cl is reduced, while Br is oxidized.Sample 4. equation: Fe2O3+ 6 HCl ? 2 FeCl3+ 3 H2O

Explanation: In this equation Fe and H trade places and O and Cl trade places.

Reaction: 2H2O2(aq) ? 2H2O(l) + O2(g)

In the reactants H has an O has an O.S. of -1, which changes to -2 for the product, H2O (reduced) and 0 for the product, O2 (oxidized).

Thanks & Regards

Rinkoo Gupta

AskIITians Faculty