cr2o72-+ I-=cr3++i2 basa setengah reaksi
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Pertanyaan
cr2o72-+ I-=cr3++i2 basa setengah reaksi
1 Jawaban
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1. Jawaban Ramsyahnovaldy
Break up the reaction into two half-reactions -- one for Cr and one for I.
Cr2O72- -----> 2Cr3+
I- ------> IO3-
After balancing the Cr and I, balance the O by adding H2O.
Cr2O72- -----> 2Cr3+ + 7H2O
I- 3H2O ------> IO3-
Next, balance the H by adding H+
Cr2O72- + 14H+ -----> 2Cr3+ + 7H2O
I- + 3H2O ------> IO3- + 6H+
Balance the charges by adding e- to the more positive side of each half-reaction.
Cr2O72- + 14H+ + 6e- -----> 2Cr3+ + 7H2O (reduction)
I- + 3H2O ------> IO3- + 6H+ + 6e- (oxidation)
Since each half-reaction has 6e- involved, just add the half-reactions together. Simplify the water and H+.
Cr2O72- + 14H+ + 6e- -----> 2Cr3+ + 7H2O
I- + 3H2O ------> IO3- + 6H+ + 6e-
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I- + Cr2O72- + 8H+ -----> 2Cr3+ + 4H2O
Semoga dapat membantu yah