Question

Complete the following ionic equation:

\[\ce{Cr2O^2-_7 + 14H+ + 6e- ->}\]

Answer 1

\[\ce{Cr2O^2-_7 + 14H+ + 6e- -> 2Cr^3+ + 7H2O}\]

Write an unbalanced reaction: 

\[\ce{Cr2O^2-_7 + 14H+ + 6e- -> Cr}\]

Balance the atoms other than hydrogen and oxygen:

\[\ce{Cr2O^2-_7 + 14H+ + 6e- -> 2Cr}\]

Balance the oxygen by adding water to the side that needs oxygen:

\[\ce{Cr2O^2-_7 + 14H+ + 6e- -> 2Cr + 7H2O}\]

Balance the charge by adding electrons or charge:

\[\ce{Cr2O^2-_7 + 14H+ + 6e- -> 2Cr^3+}\]