2NaCro2 + 2NaOH NalO3 . 2Na2CrO4 Nal + H2O --> the change in oxidation number for I (iodine) is equivalent to:

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Chapter18: Electrochemistry
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Problem 36E: Consider the following galvanic cell: a. Label the reducing agent and the oxidizing agent, and...
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According to the following balanced equation:
2NaCrO2 +
2NaOH
NalO3 .
--> 2Na2CrO4 +
Nal + H20
the change in oxidation number for I (iodine) is equivalent to:
a) 4 electrons lost.
O b) 4 electrons gained.
c) 6 electrons lost.
d) 6 electrons gained.
Transcribed Image Text:According to the following balanced equation: 2NaCrO2 + 2NaOH NalO3 . --> 2Na2CrO4 + Nal + H20 the change in oxidation number for I (iodine) is equivalent to: a) 4 electrons lost. O b) 4 electrons gained. c) 6 electrons lost. d) 6 electrons gained.
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