The oxidation of chromium in Cr\(_2\)O\(_7\)\(^{2-}\) ?
Cr\(_2\)O\(_7\)\(^{2-}\)
Cr\(_2\)O\(_7\) = - 2
Cr\(_2\) + O\(_7\) = - 2
2Cr + 7(-2) = - 2
2Cr - 14 = - 2
2Cr = -2 + 14
2Cr = 12
Cr = \(\frac{12}{2}\)
Cr = + 6
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