For a particular redox reaction Cr is oxidized to CrO42 Solution Let\'s first ba

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Let's first balance the one that involves an acidic solution. CrO4(2-) + Fe(2+) -> Cr(3+) + Fe(3+) In an acidic solution, we put H+ to balance the hydrogen atoms while we put H2O to balance the O atoms. First, we write the half-reactions. Reduction: CrO4(2-) -> Cr(3+) Oxidation: Fe(2+) -> Fe(3+) Let's first balance the oxidation half-reaction. The atoms are balanced but the charges are not. The left side has a charge of +2 while the right has a charge of +3. To compensate for the charge difference, we need to add an electron on the right side. Fe(2+) -> Fe(3+) + e- The charges and number of atoms are now balanced. Let's now balance the reduction half-reaction. The O atoms are not balanced. there's 4 on the left and none on the right. To solve this problem, we need to add 4 H2O molecules on the right side. CrO4(2-) -> Cr(3+) + 4H2O Now, there are 8 H atoms on the right while none on the left side. To compensate, we need to add 8 H+ ions to balance the H atoms. CrO4(2-) + 8H+ -> Cr(3+) + 4H2O Now, the thing is balanced in terms of number of atoms. The charges, however, aren't. The net charge on the left is +6 while on the right, +3. To balance the charge, we need to add 3 electrons. CrO4(2-) + 8H+ + 3e- -> Cr(3+) + 4H2O Let's now multiply both equations by a number that will equalize the number of electrons on both. That way, electrons lost will be equal to electrons gained. 3[Fe(2+) -> Fe(3+) + e-] 1[CrO4(2-) + 8H+ + 3e- -> Cr(3+) + 4H2O] We now have the equations 3Fe(2+) --> 3Fe(3+) + 3e- CrO4(2-) + 8H+ + 3e- -> Cr(3+) + 4H2O Add together the half-reactions to produce one equation: 3Fe(2+) + CrO4(2-) + 8H+ + 3e- ->3Fe(3+) + 3e- + Cr(3+) + 4H2O Combining and/or cancelling like terms gives 3Fe(2+) + CrO4(2-) + 8H+ ->3Fe(3+) + Cr(3+) + 4H2O

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