Concord High Chemistry Honors
Balancing Redox Equations Practice
Balancing Redox Equations Practice Problem #3:
SO2 (aq) + Cr2O72- (aq) + H+ (aq) — > Cr3+ (aq) + SO42- (aq) + H2O (l)
- Assign oxidation numbers to each element in the equation. Figure out which elements have their oxidation numbers change in the reaction.
S = oxidation number increases from +4 [in SO2] to +6 [in SO42-]
Cr = oxidation number decreases from +6 [in Cr2O72-] to +3 [in Cr3+]
- Figure out the oxidation number increase or decrease of the elements.
S = increases by 2
Cr = decreases by 3
- Find the lowest common multiple of the increase in oxidation number and the decrease in oxidation number. Figure out what number you need to multiply the increase/decrease in oxidation number by to get the lowest common multiple. This number is the number of atoms of each particular element needed on both sides of the equation. Fix the coefficients of the affected elements.
Lowest common multiple is 6.
S = 2 X 3 = 6 You need 3 atoms of S on each side.
Cr = 3 X 2 = 6 You need 2 atoms of Cr on each side.
3SO2 (aq) + 1Cr2O72- (aq) + H+ (aq) — > 2Cr3+ (aq) + 3SO42- (aq) + H2O (l)
- Balance the rest of the equation like regular.
• O and H on both sides are unequal. You can only change the coefficients of the H+ and the H2O because all the other species are “locked in.”
Use a system of equations like in algebra to find coefficients of H+ (x) and H2O (y)
atoms of H: 0 + 0 + x(1) = 0 + 0 + y(2)
atoms of O: 2 + 7 + x(0) = 0 + 12 + y(1)
Solve the system of equations to get x=(2) and y =(1).
Plug them into the redox equation in front of H+ and H2O.
3SO2 (aq) + Cr2O72- (aq) + 2H+ (aq) — > 2Cr3+ (aq) + 3SO42- (aq) + 1H2O (l)
- You may wish to double-check your balanced redox equation by checking that the total number of charges on either side is equal.
Double-check on your own. It works.
3SO2 (aq) + Cr2O72- (aq) + 2H+ (aq) — > 2Cr3+ (aq) + 3SO42- (aq) + H2O (l)
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