Core
Reactions: Balancing Redox Equations
Slide 9 of 29
Redox reactions in acidic medium: Step 5
–The charge on the left side of the reduction half-equation is -2 + 14 = 12, while on the
right side it is 2 (3+) = 6. So the left side needs 6 electrons to bring the charge
down to 6+.
Reduction half-equation: Cr2O72- + 14H+ + 6e- ® 2Cr3+ + 7H2O [Gain of e-]
Reduction half-equation: Cr2O72- + 14H+ + 6e- ® 2Cr3+ + 7H2O [Gain of e-]
Step 5. Now that we have the two balanced half-reactions,
the two can be combined after ensuring that the extent of
oxidation (number of electrons lost) is matched by the
extent of reduction (number of electrons gained)
–i) Since number of electrons lost is only 2 while that gained is
6, multiplying the oxidation half-equation by 3 balances it
out
Oxidation
half-equation: 6Cl- ® 3Cl2 + 6e-
Reduction half-equation: Cr2O72- + 14H+ + 6e- ® 2Cr3+ + 7H2O