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Question

Two half reactions are given as follows:
2e+H++H5IO6IO3+3H2O
CrCr3++3e
Final balanced reaction is:

A
3H++3H5IO6+2Cr2Cr3++3IO3+9H2O
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B
5H++3H5IO6+2Cr2Cr3++3IO3+10H2O
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C
3H++3H5IO6+4Cr4Cr3++3IO3+9H2O
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D
3H++3H5IO6+3Cr3Cr3++3IO3+9H2O
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Solution

The correct option is A 3H++3H5IO6+2Cr2Cr3++3IO3+9H2O
Multiply each given half reactions by a number to balance electrons on both sides.
[2e+H++H5IO6IO3+3H2O]×3(1)
[CrCr+3+3e]×2(2)
Now adding equation (1) and (2) we get balanced equation
3H++3H5IO6+2Cr2Cr+3+3IO3+9H2O .

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