Balance the redox equations and give the half-cell reactions and identify the oxidizing agent, species oxidized, reducing agent, and species reduced.
In an acidic solution: CrO4^2-(aq)+ Cl-(aq)--> Cr^3+(aq) +HClO2(aq)
In an acidic solution: Fe^2+(aq) + MnO4-(aq) --> Fe^3+(aq)+ Mn^2+(aq)
In acidic solution: NH4+(aq) + NO3-(aq)--> N2O(g)
In basic solution: N2H4(aq) + Cu(OH)2(s) ---> N2(g) + Cu(s)
In basic solution: Ca(OH)2(aq)+ C(s) + ClO2(g)--> CaCO3(s) +ClO2-(aq)
For these questions, you have to use the EOHC technique....
E -everything besides Oxygens and hydrogens
O -Oxygens by adding waters
H -Hydrogens by adding protons H+
C -charge by adding electrons e-
First thing to do is to split into half reactions (sometimes not so easy)
TIP: Figure out which elements are changing oxidation state first, and try to ignore elements that are not changing in oxidation state.
For the first question:
Cr04^-2 --> Cr^3+ is the one half reaction to balance
Cl^- --> HClO2 is the second half reaction to balance
Balancing the first....
CrO4^-2 --> Cr^3+
Everything else is simply the Cr......already balanced
Add 4 waters to the RHS (right hand side)
CrO4^-2 --> Cr^3+ + 4H2O
Add 8 protons to ...
The solution is comprised of a detailed explanation of how to balance redox reactions, and identify oxidizing agents, reducing agents.