How to balance the following equation acidic solution: Cr2O7-2 + Mg----Cr+3 +Mg+

Question

How to balance the following equation acidic solution: Cr2O7-2 + Mg----Cr+3 +Mg+2 each step would be helpful even how to do oxidation numbersand how to get charges. How to balance the following equation acidic solution: Cr2O7-2 + Mg----Cr+3 +Mg+2 each step would be helpful even how to do oxidation numbersand how to get charges.

Explanation / Answer

Cr2O72- =>  2Cr3+ O has -2 charge, but Cr2O72- hasoverall charge of -2. So 7*-2 + (oxidation state of Cr2 indichromate ) = -2 Each Cr is +6. 7*-2 + 2*6 = -2 On the right Cr3+, so 3 electrons are used to reduce each Cr inCr2O7 Cr2O72- + 6e- =>   2Cr3+ Now add H2O on the right to balance O's. Add 7H2O Cr2O72- + 6e- =>   2Cr3+ + 7H2O Add 14 H+ on the left to balance H2O Cr2O72- + 14H+ +6e- =>   2Cr3+ +7H2O                 Eq.1 Now move onto the oxdiation reaction Mg => Mg2+ Mg = > Mg2+ + 2e-               Eq.2        Magnesium is oxidized to2+ Now add Eq.1 and Eq.2 such that you eliminatee-'s. So multiply equation 2 by 3 and add to 1 Cr2O72- 14H+ +6e- =>   2Cr3+ + 7H2O                         3*(Mg = > Mg2+ + 2e-) ------------------------------------------------------------ Cr2O72- 14H+ +6e- + 3 Mg =>   2Cr3+ + 7H2O +6e- + 3Mg2+ Eliminate like terms on both sides (electrons) Cr2O72- 14H+   + 3Mg =>   2Cr3+ + 7H2O + 3Mg2+

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