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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