Đáp án đúng là: B
K I − 1 K\mathop I\limits^{ - 1} K I − 1 + H 2 S + 6 O 4 {H_2}\mathop S\limits^{ + 6} {O_4} H 2 S + 6 O 4 ⟶ I 2 0 {\mathop I\limits^0 _2} 2 I 0 + H 2 S − 2 {H_2}\mathop S\limits^{ - 2} H 2 S − 2 + K2 _{2} 2 SO4 _{4} 4 + H2 _{2} 2 O
4 × 1 × ∣ 2 I − 1 → I 2 0 + 2 e S + 6 + 8 e → S − 2 \begin{array}{cccccccccccccccccccc}{4\,\, \times }\\{1\,\, \times }\end{array}\,\left| \begin{array}{l}2\mathop I\limits^{ - 1} \,\,\, \to \,\,\,{\mathop I\limits^0 _2}\, + 2e\,\\\mathop S\limits^{ + 6} \,\,\,\, + 8e \to \,\,\,\mathop S\limits^{ - 2} \,\,\,\end{array} \right. 4 × 1 × 2 I − 1 → 2 I 0 + 2 e S + 6 + 8 e → S − 2
⇒ 8K I − 1 K\mathop I\limits^{ - 1} K I − 1 + 5H 2 S + 6 O 4 {H_2}\mathop S\limits^{ + 6} {O_4} H 2 S + 6 O 4 ⟶ 4I 2 0 {\mathop I\limits^0 _2} 2 I 0 + H 2 S − 2 {H_2}\mathop S\limits^{ - 2} H 2 S − 2 + 4K2 _{2} 2 SO4 _{4} 4 + 4H2 _{2} 2 O