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2.5 points Save Answer his half-reaction occurs in acid. How many waters are needed? ood) OA. 7 B.1 D. 14
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Answer #1

Lets calculate Oxidation state of Cr in Cr2O7-2

lets the oxidation number of Cr be x

use:

7* oxidation number (O) + 2* oxidation number (Cr) = net charge

7*(-2)+2* x = -2

-14 + 2 * x = -2

x = 6

So oxidation number of Cr = +6

Oxidation state of Cr changes from 0 to +6

The reaction is:

Cr (s) —> Cr2O7-2 (aq)

Balance Cr:

2 Cr (s) —> Cr2O7-2 (aq)

Balance electron:

2 Cr (s) —> Cr2O7-2 (aq) + 12 e-

Balance O:

2 Cr (s) + 7 H2O —> Cr2O7-2 (aq) + 12 e-

Balance H:

2 Cr (s) + 7 H2O —> Cr2O7-2 (aq) + 12 e- + 18 H+

The equation is now balanced

There are 7 H2O needed

Answer: A

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