The unbalanced redox reaction is \\\\(\\displaystyle MnO_{4}^{-} + C_{2}O_{4}^{2-} + H^{+}\\rightarrow Mn^{2+} + CO_{2} + H_{2}O\\\\)Balance C atoms.\\\\(\\displaystyle MnO_{4}^{-} + C_{2}O_{4}^{2-} + H^{+}\\rightarrow Mn^{2+} + 2CO_{2} + H_{2}O\\\\)The oxidation number of Mn decreases from \\\\(\\displaystyle +7\\\\) to \\\\(\\displaystyle +2\\\\).The decrease in the oxidation number is \\\\(\\displaystyle 7-2=5\\\\)The oxidation number of C increases from \\\\(\\displaystyle +3\\\\) to \\\\(\\displaystyle +4\\\\).The increase in the oxidation number for 1 C atom is \\\\(\\displaystyle 4-3=1\\\\)The increase in the oxidation number for 2 C atom is \\\\(\\displaystyle 2 \ imes 1=2\\\\)To balance the increase in the oxidation number with decrease in the oxidation number, multiply Mn containing species with 2 and C containing species with 5.\\\\(\\displaystyle 2MnO_{4}^{-} + 5C_{2}O_{4}^{2-} \\rightarrow 2Mn^{2+} + 10CO_{2} \\\\) Balance O atoms by adding 8 water molecules to the products side.\\\\(\\displaystyle 2MnO_{4}^{-} + 5C_{2}O_{4}^{2-} \\rightarrow 2Mn^{2+} + 10CO_{2} + 8H_{2}O\\\\) Balance H atoms by adding 16 \\\\(\\displaystyle H^+\\\\) ions to the reactants side.\\\\(\\displaystyle 2MnO_{4}^{-} + 5C_{2}O_{4}^{2-} + 16H^{+}\\rightarrow 2Mn^{2+} + 10CO_{2} + 8H_{2}O\\\\) This is the balanced equation.