Tìm 30 phản ứng oxi hóa khử , và cân bằng ch 13/08/2021 Bởi Ruby Tìm 30 phản ứng oxi hóa khử , và cân bằng ch
Đáp án: Bạn tham khảo lời giải ở dưới nhé!!! Giải thích các bước giải: \(\begin{array}{l}CrS + 6HN{O_3} \to Cr{(N{O_3})_3} + 3N{O_2} + S + 3{H_2}O\\C{r^{2 + }} \to C{r^{3 + }} + 1e\\{S^{2 – }} \to S + 2e\\{N^{5 + }} + 1e \to {N^{4 + }}\\2NaCr{O_2} + 3B{r_2} + 8NaOH \to 2N{a_2}Cr{O_4} + 6NaBr + 4{H_2}O\\Cr{O_2}^ – + 4O{H^ – } \to Cr{O_4}^{2 – } + 2{H_2}O + 3e\\B{r_2} + 2e \to 2B{r^ – }\\2KMn{O_4} + 3{K_2}S{O_3} + {H_2}O \to 2Mn{O_2} + 3{K_2}S{O_4} + 2KOH\\Mn{O_4}^ – + 2{H_2}O + 3e \to Mn{O_2} + 4O{H^ – }\\S{O_3}^{2 – } + {H_2}O \to S{O_4}^{2 – } + 2{H^ + } + 2e\\6FeS{O_4} + {K_2}C{r_2}{O_7} + 7{H_2}S{O_4} \to 3F{e_2}{(S{O_4})_3} + {K_2}S{O_4} + C{r_2}{(S{O_4})_2} + 7{H_2}O\\F{e^{2 + }} \to F{e^{3 + }} + 1e\\2C{r^{6 + }} + 6e \to 2C{r^{3 + }}\\3N{a_2}S{O_3} + 2KMn{O_4} + {H_2}O \to 3N{a_2}S{O_4} + 2Mn{O_2} + 2KOH\\{S^{4 + }} \to {S^{6 + }} + 2e\\M{n^{7 + }} + 3e \to M{n^{4 + }}\\3F{e_3}{O_4} + 28HN{O_3} \to 9Fe{(N{O_3})_3} + NO + 14{H_2}O\\3F{e^{\dfrac{8}{3} + }} \to 3F{e^{3 + }} + 1e\\{N^{5 + }} + 3e \to {N^{2 + }}\\3A{s_2}{S_3} + 28HN{O_3} + 4{H_2}O \to 6{H_3}As{O_4} + 28NO + 9{H_2}S{O_4}\\A{s^{3 + }} \to A{s^{5 + }} + 2e\\{S^{ – 2}} \to {S^{6 + }} + 8e\\3F{e_x}{O_y} + (12x – 6y)HN{O_3} \to 3xFe{(N{O_3})_3} + (3x – 2y)NO + (6x – 3y){H_2}O\\xF{e^{\dfrac{{2y}}{x} + }} \to xF{e^{3 + }} + (3x – 2y)\\{N^{5 + }} + 3e \to {N^{2 + }}\end{array}\) \(\begin{array}{l}{K_2}C{r_2}{O_7} + 14HCl \to 2KCl + 2CrC{l_3} + 3C{l_2} + 7{H_2}O\\C{r_2}^{6 + } + 3e \to 2C{r^{3 + }}\\2C{l^ – } \to C{l_2} + 2e\\10FeS{O_4} + 2KMn{O_4} + 8{H_2}S{O_4} \to 5F{e_2}{(S{O_4})_3} + {K_2}S{O_4} + 2MnS{O_4} + 8{H_2}O\\F{e^{2 + }} \to F{e^{3 + }} + 1e\\M{n^{7 + }} + 5e \to M{n^{2 + }}\\6FeS{O_4} + {K_2}C{r_2}{O_7} + 7{H_2}S{O_4} \to 3F{e_2}{(S{O_4})_3} + C{r_2}{(S{O_4})_3} + {K_2}S{O_4} + 7{H_2}O\\F{e^{2 + }} \to F{e^{3 + }} + 1e\\C{r_2}^{6 + } + 3e \to 2C{r^{3 + }}\\3Cu + 2KN{O_3} + 4{H_2}S{O_4} \to 3CuS{O_4} + {K_2}S{O_4} + 2NO + 4{H_2}O\\Cu \to C{u^{2 + }} + 2e\\{N^{5 + }} + 3e \to {N^{2 + }}\\P + 5HN{O_3} \to {H_3}P{O_4} + 5N{O_2} + {H_2}O\\P \to {P^{5 + }} + 5e\\{N^{5 + }} + 1e \to {N^{4 + }}\\S + 6HN{O_3} \to {H_2}S{O_4} + 6N{O_2} + 2{H_2}O\\S \to {S^{6 + }} + 6e\\{N^{5 + }} + 1e \to {N^{4 + }}\\C + 2{H_2}S{O_4} \to C{O_2} + 2S{O_2} + 2{H_2}O\\C \to {C^{4 + }} + 4e\\{S^{6 + }} + 2e \to {S^{4 + }}\\{H_2}S + 2{H_2}S{O_4} \to 3S{O_2} + 2{H_2}O\\{S^{2 – }} \to {S^{4 + }} + 6e\\{S^{6 + }} + 2e \to {S^{4 + }}\\S + 2{H_2}S{O_4} \to 3S{O_2} + 2{H_2}O\\S \to {S^{4 + }} + 4e\\{S^{6 + }} + 2e \to {S^{4 + }}\end{array}\) Bình luận
Đáp án:
Bạn tham khảo lời giải ở dưới nhé!!!
Giải thích các bước giải:
\(\begin{array}{l}
CrS + 6HN{O_3} \to Cr{(N{O_3})_3} + 3N{O_2} + S + 3{H_2}O\\
C{r^{2 + }} \to C{r^{3 + }} + 1e\\
{S^{2 – }} \to S + 2e\\
{N^{5 + }} + 1e \to {N^{4 + }}\\
2NaCr{O_2} + 3B{r_2} + 8NaOH \to 2N{a_2}Cr{O_4} + 6NaBr + 4{H_2}O\\
Cr{O_2}^ – + 4O{H^ – } \to Cr{O_4}^{2 – } + 2{H_2}O + 3e\\
B{r_2} + 2e \to 2B{r^ – }\\
2KMn{O_4} + 3{K_2}S{O_3} + {H_2}O \to 2Mn{O_2} + 3{K_2}S{O_4} + 2KOH\\
Mn{O_4}^ – + 2{H_2}O + 3e \to Mn{O_2} + 4O{H^ – }\\
S{O_3}^{2 – } + {H_2}O \to S{O_4}^{2 – } + 2{H^ + } + 2e\\
6FeS{O_4} + {K_2}C{r_2}{O_7} + 7{H_2}S{O_4} \to 3F{e_2}{(S{O_4})_3} + {K_2}S{O_4} + C{r_2}{(S{O_4})_2} + 7{H_2}O\\
F{e^{2 + }} \to F{e^{3 + }} + 1e\\
2C{r^{6 + }} + 6e \to 2C{r^{3 + }}\\
3N{a_2}S{O_3} + 2KMn{O_4} + {H_2}O \to 3N{a_2}S{O_4} + 2Mn{O_2} + 2KOH\\
{S^{4 + }} \to {S^{6 + }} + 2e\\
M{n^{7 + }} + 3e \to M{n^{4 + }}\\
3F{e_3}{O_4} + 28HN{O_3} \to 9Fe{(N{O_3})_3} + NO + 14{H_2}O\\
3F{e^{\dfrac{8}{3} + }} \to 3F{e^{3 + }} + 1e\\
{N^{5 + }} + 3e \to {N^{2 + }}\\
3A{s_2}{S_3} + 28HN{O_3} + 4{H_2}O \to 6{H_3}As{O_4} + 28NO + 9{H_2}S{O_4}\\
A{s^{3 + }} \to A{s^{5 + }} + 2e\\
{S^{ – 2}} \to {S^{6 + }} + 8e\\
3F{e_x}{O_y} + (12x – 6y)HN{O_3} \to 3xFe{(N{O_3})_3} + (3x – 2y)NO + (6x – 3y){H_2}O\\
xF{e^{\dfrac{{2y}}{x} + }} \to xF{e^{3 + }} + (3x – 2y)\\
{N^{5 + }} + 3e \to {N^{2 + }}
\end{array}\)
\(\begin{array}{l}
{K_2}C{r_2}{O_7} + 14HCl \to 2KCl + 2CrC{l_3} + 3C{l_2} + 7{H_2}O\\
C{r_2}^{6 + } + 3e \to 2C{r^{3 + }}\\
2C{l^ – } \to C{l_2} + 2e\\
10FeS{O_4} + 2KMn{O_4} + 8{H_2}S{O_4} \to 5F{e_2}{(S{O_4})_3} + {K_2}S{O_4} + 2MnS{O_4} + 8{H_2}O\\
F{e^{2 + }} \to F{e^{3 + }} + 1e\\
M{n^{7 + }} + 5e \to M{n^{2 + }}\\
6FeS{O_4} + {K_2}C{r_2}{O_7} + 7{H_2}S{O_4} \to 3F{e_2}{(S{O_4})_3} + C{r_2}{(S{O_4})_3} + {K_2}S{O_4} + 7{H_2}O\\
F{e^{2 + }} \to F{e^{3 + }} + 1e\\
C{r_2}^{6 + } + 3e \to 2C{r^{3 + }}\\
3Cu + 2KN{O_3} + 4{H_2}S{O_4} \to 3CuS{O_4} + {K_2}S{O_4} + 2NO + 4{H_2}O\\
Cu \to C{u^{2 + }} + 2e\\
{N^{5 + }} + 3e \to {N^{2 + }}\\
P + 5HN{O_3} \to {H_3}P{O_4} + 5N{O_2} + {H_2}O\\
P \to {P^{5 + }} + 5e\\
{N^{5 + }} + 1e \to {N^{4 + }}\\
S + 6HN{O_3} \to {H_2}S{O_4} + 6N{O_2} + 2{H_2}O\\
S \to {S^{6 + }} + 6e\\
{N^{5 + }} + 1e \to {N^{4 + }}\\
C + 2{H_2}S{O_4} \to C{O_2} + 2S{O_2} + 2{H_2}O\\
C \to {C^{4 + }} + 4e\\
{S^{6 + }} + 2e \to {S^{4 + }}\\
{H_2}S + 2{H_2}S{O_4} \to 3S{O_2} + 2{H_2}O\\
{S^{2 – }} \to {S^{4 + }} + 6e\\
{S^{6 + }} + 2e \to {S^{4 + }}\\
S + 2{H_2}S{O_4} \to 3S{O_2} + 2{H_2}O\\
S \to {S^{4 + }} + 4e\\
{S^{6 + }} + 2e \to {S^{4 + }}
\end{array}\)