Hoàn thành phương trình hóa học sau bằng phương pháp thăng bằng electron: a) FeCl3 + Cu —⟩ FeCl2 + CuCl2 b) Ag + HNO3 loãng —⟩ AgNO3 + NO + H2O

a) 

$\mathop{Fe}\limits^{+3}Cl_3+Cu→\mathop{Fe}\limits^{+2}Cl_2+\mathop{Cu}\limits^{+2}Cl_2$

$\mathop{Fe}\limits^{+3}+1e→\mathop{Fe}\limits^{+2}$  || $\times2$

$\mathop{Cu}\limits^{0}→\mathop{Cu}\limits^{+2}+2e$   || $\times1$

⇒ $2\mathop{Fe}\limits^{+3}Cl_3+Cu→2\mathop{Fe}\limits^{+2}Cl_2+\mathop{Cu}\limits^{+2}Cl_2$

b)

$\mathop{Ag}\limits^{o}+H\mathop{N}\limits^{+5}O_{3(Loãng)}→\mathop{Ag}\limits^{+1}NO_3+\mathop{N}\limits^{+2}O+H_2O$

$\mathop{Ag}\limits^{0}→\mathop{Ag}\limits^{+1}+1e$   || $\times3$

$\mathop{N}\limits^{+5}+3e→\mathop{N}\limits^{+2}$    ||  $\times1$

⇒ $3\mathop{Ag}\limits^{o}+4H\mathop{N}\limits^{+5}O_{3(Loãng)}→3\mathop{Ag}\limits^{+1}NO_3+\mathop{N}\limits^{+2}O+2H_2O$