Given the unbalanced chemical reaction
[tex]Cr_2O_7^{2-}+CH_3OH→HCO_2H+Cr^{3+}[/tex]Separate into half reactions
[tex]\begin{gathered} Cr_2O_7^{2-}\rightarrow Cr^{3+}(Reduction) \\ CH_3OH\rightarrow HCO_2H(Oxidation) \end{gathered}[/tex]Balance all other atoms except hydrogen and oxygen to have:
[tex]\begin{gathered} CH_3OH\rightarrow HCO_2H \\ Cr_2O_7^{2-}\rightarrow2Cr^{3+} \end{gathered}[/tex]Balance the oxygen atoms by adding water
[tex]\begin{gathered} CH_3OH+H_2O\rightarrow HCO_2H \\ Cr_2O_7^{2-}\operatorname{\rightarrow}2Cr^{3+}+7H_2O \end{gathered}[/tex]Balance the hydrogen atoms by adding hydrogen ion
[tex]\begin{gathered} CH_3OH+H_2O\operatorname{\rightarrow}HCO_2H+4H^+ \\ Cr_2O_7^{2-}+14H^+\operatorname{\rightarrow}2Cr^{3+}+7H_2O \end{gathered}[/tex]Balance the charges
[tex]\begin{gathered} CH_3OH+H_2O\operatorname{\rightarrow}HCO_2H+4H^++4e^-\text{ * 3} \\ Cr_2O_7^{2-}+14H^++6e^-\operatorname{\rightarrow}2Cr^{3+}+7H_2O\text{ * 2} \\ _{------------------------------------} \\ 3CH_3OH+3H_2O\operatorname{\rightarrow}3HCO_2H+12H^++12e^- \\ 2Cr_2O_7^{2-}+28H^++12e^-\operatorname{\rightarrow}4Cr^{3+}+14H_2O \end{gathered}[/tex]Cancel the charges and add the resulting equation to have:
[tex]3CH_3OH+2Cr_2O_7^{2-}+16H^+\rightarrow3HCO_2H+4Cr^{3+}+11H_2O[/tex]