Question

balancing the chemical reaction by ion electron method.

equation upar hai​

Answer 1

$\huge\underline\purple{\sf Answer:-}$

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Step 1 :- Write both the half reaction .

$\large{\sf MnO_{4}^{-}→Mn^{+2}}$ (Reduction half)

$\large{\sf C_{2}O_{4}^{-2}→CO_{2}}$ (Oxidation)

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Step 2 :- Atoms other than H and O are Balanced .

$\large{\sf MnO_{4}^{-}→Mn^{+2}}$

$\large{\sf C_{2}O_{4}^{-2}→2CO_{2}}$

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Step 3 :- Balance O-atom by addition of H2O to another side .

$\large{\sf MnO_{4}^{-}→Mn^{+2}+4H_{2}O}$

$\large{\sf C_{2}O_{4}^{-2}→2CO_{2}}$

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Step 4 :- Balance H-atoms by addition to another side .

$\large{\sf MnO_{4}^{-}+8H^{+}→Mn^{+2}+4H_{2}O}$

$\large{\sf C_{2}O_{4}^{-2}→2CO_{2}}$

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Step 5 :- Now balance the charge by addition of Electron.

$\large{\sf MnO_{4}^{-}+8H^{+}+5e^{-}→Mn^{+2}+4H_{2}O ×2}$

$\large{\sf C_{2}O_{4}^{-2}+8H^{+}+5e^{-}→2CO_{2} ×5}$

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$\large{\sf 2MnO_{4}^{-}+16H^{+}+10e^{-}→2Mn^{+2}+8H_{2}O}$

$\large{\sf 5C_{2}O_{4}^{-2}→10CO_{2}+10e^{-}}$

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So final balanced reaction is :-

$\large\red{\sf 2MnO_{4}^{-}+5C_{2}O_{4}^{-2}+16H^{+}→10CO_{2}+2Mn^{+2}+8H_{2}O}$