Question

balance the following redox reaction by ion electron method in basic medium. Cr(oOH) 3 +Io3- = Cr3+ + I-​

Answer 1

$Oxidation number of chromium in reactant side of reaction is +3, whereas in product side it has +6 oxidation state. Hence, we can conclude that Cr(OH)3 is oxidized.</p><p>I has oxidation +5  in reactant side of reaction and −1 on the product side, so it has reduced in the reaction. By this we can say that IO3− is the oxidising agent</p><p>Electron gain = (initial oxidation state of I − final oxidation state of I) = 6.</p><p></p><p>$

Answer 2

Answer:

In this question we will balance Cr(OH) WITH Cro4 2- and Io3 - with I-

Cr(OH)3 GIVES Cr+3 + 3OH-

Hear OH shows that the equation should be balanced in basic medium

Cr+3----->(Cro4)2-

Cr+3----->(Cro4)2- +4H2O

8OH- + Cr+3---->(Cro4)2- + 4H2O

8OH- + cr+3--->4H2O+(Cro4)2-+ 3e- this is our first equation

Now

IO3 - ----->I-

IO3 - +3H2O ----> I- + 6OH-

IO3- + 3H2O + 6e- -----> I- +6OH- Tthis is our second equation

Multiplying first euation with three and second equation with six

We get our equation as

3IO3- + 9H2O+6Cr(OH)3------> 6(Cro4)2- +24H2O + 3I- + 18OH-

Then by simplifying we get he balanced equation as

IO3 - + 3H2O+2Cr(OH)3 -----> 2Cro4 2- 8 H2O +I- 6OH-

Explanation:

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