Consider the following equation: Cr\(_2\)O\(_7\)\(^{2-}\)(aq) + 14H\(^+\)(aq) + 6e\(^-\) → 2Cr\(^{3+}\)(ag) + 7H\(_2\)O(1). The oxidation number of chromium changes from

badformular

10 years ago

yes becos oxygen has a valency of -2,7*-2 =-2(the charge oxygen carries)=-14=-2.....carring 14 to the other side it will be +14-2=12...now the oxidation number of oxygen is 12 buh we luking for cr2....just divide 12 by the 2 in cr...=6.then to the product of the reaction its just simple cos the charge now is the oxidation number

ss3ascience

1 year ago

why no explanation