Balancing redox equations — ion-electron (half-reaction) method, oxidation-number method

1.3.7 · Chemistry › Chemical Reactions & Stoichiometry

Ordinary balancing methods yahan kyun fail karte hain?

$Mn O_{4}^{-} + F e^{2 +} \to M n^{2 +} + F e^{3 +}$ (acidic) jaisi reaction ke liye trial-and-error fail hoti hai kyunki:

Oxygen aur hydrogen solvent (water, $H^{+}$ , $O H^{-}$ ) se aate hain, skeleton mein likhe nahi hote.
Charge bhi balance hona chahiye, sirf atoms nahi.

Isliye hum ek systematic method chahte hain jo electrons ko explicitly track kare.

Method 1 — Ion–Electron (Half-Reaction) Method

KAISE KARO (acidic medium), step by step:

Skeleton ko ek oxidation half aur ek reduction half mein split karo.
O aur H ke alawa sab atoms balance karo.
O balance karo $H_{2} O$ add karke.
H balance karo $H^{+}$ add karke.
Charge balance karo $e^{-}$ more-positive side par add karke.
Har half ko multiply karo taaki electrons cancel ho jaayein; add karo aur simplify karo.

Basic medium: Pehle sab kuch acidic ki tarah karo, phir equal $O H^{-}$ dono sides mein add karo taaki har $H^{+}$ neutralize ho jaaye ( $H_{2} O$ bane), aur duplicate waters cancel kar do.

[!example] Worked Example 1 — $Mn O_{4}^{-}$ + $F e^{2 +}$ (acidic)

Reduction half: $Mn O_{4}^{-} \to M n^{2 +}$

O ko water se balance karo: $Mn O_{4}^{-} \to M n^{2 +} + 4 H_{2} O$ Kyun? Left side par 4 O hain; 4 waters right par add karo unhe supply karne ke liye.
H ko $H^{+}$ se balance karo: $Mn O_{4}^{-} + 8 H^{+} \to M n^{2 +} + 4 H_{2} O$ Kyun? Right par ab 8 H hain ( $4 H_{2} O$ se); left par 8 $H^{+}$ add karo.
Charge balance karo: Left $= - 1 + 8 = + 7$ , Right $= + 2$ . Left par $5 e^{-}$ add karo. $Mn O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O$ Kyun? $+ 7 + (- 5) = + 2$ . ✔ Mn $+ 7 \to + 2$ jaata hai, 5 electrons gain karta hai — bilkul match!

Oxidation half: $F e^{2 +} \to F e^{3 +} + e^{-}$ Kyun? Charge $+ 2 \to + 3$ matlab 1 electron ka loss.

Electrons equalize karo: 5 aur 1 ka LCM 5 hai → Fe half ko 5 se multiply karo. $Mn O_{4}^{-} + 8 H^{+} + 5F e^{2 +} \to M n^{2 +} + 4 H_{2} O + 5F e^{3 +}$

Check: charge left $= - 1 + 8 + 10 = + 17$ ; right $= + 2 + 15 = + 17$ . ✔

[!example] Worked Example 2 — Basic medium: $Mn O_{4}^{-} + I^{-} \to Mn O_{2} + I_{2}$

Reduction: $Mn O_{4}^{-} \to Mn O_{2}$

O: $Mn O_{4}^{-} \to Mn O_{2} + 2 H_{2} O$
H: $Mn O_{4}^{-} + 4 H^{+} \to Mn O_{2} + 2 H_{2} O$
charge: left $- 1 + 4 = + 3$ , right $0$ → left par $3 e^{-}$ add karo: $Mn O_{4}^{-} + 4 H^{+} + 3 e^{-} \to Mn O_{2} + 2 H_{2} O$

Oxidation: $2 I^{-} \to I_{2} + 2 e^{-}$

Equalize karo (LCM 6): reduction ×2, oxidation ×3: $2Mn O_{4}^{-} + 8 H^{+} + 6 I^{-} \to 2Mn O_{2} + 4 H_{2} O + 3 I_{2}$

Basic mein convert karo: dono sides mein $8 O H^{-}$ add karo. Left par $8 H^{+} + 8 O H^{-} = 8 H_{2} O$ ; right par 4 waters cancel karo: $2Mn O_{4}^{-} + 4 H_{2} O + 6 I^{-} \to 2Mn O_{2} + 8O H^{-} + 3 I_{2}$ OH⁻ kyun add karte hain? Base mein free $H^{+}$ nahi hota; use water mein neutralize karna ek physically real equation deta hai.

Method 2 — Oxidation-Number Method

KAISE KARO:

Har atom ko oxidation numbers assign karo.
Un atoms ko identify karo jo upar jaate hain (oxidized) aur niche jaate hain (reduced); har atom ka Δ calculate karo.
Species ko factors se multiply karo taaki total increase = total decrease ho.
Baaki atoms balance karo (O ko $H_{2} O$ se, H ko $H^{+}$ / $O H^{-}$ se), phir charge balance karo.

[!example] Worked Example 3 — $Cu + HN O_{3} \to Cu (N O_{3})_{2} + N O_{2} + H_{2} O$ (acidic)

Cu: $0 \to + 2$ , 2 electrons lose karta hai.
Kuch $HN O_{3}$ mein N: $+ 5 \to + 4$ ( $N O_{2}$ mein), 1 each gain karta hai. Sirf kuch kyun? $Cu (N O_{3})_{2}$ mein nitrate $+ 5$ par rehta hai (spectator); sirf wo N jo $N O_{2}$ banta hai wo reduce hota hai.
Electrons balance karo: 1 Cu (2 lose karta hai) ke liye 2 N reduce chahiye (1 each gain karein).

$Cu + 4HN O_{3} \to Cu (N O_{3})_{2} + 2N O_{2} + 2 H_{2} O$ Check: N: left 4 = right $2 + 2$ ✔; H: 4 = 4 ✔; O: 12 = $6 + 4 + 2$ ✔.

Figure — Balancing redox equations — ion-electron (half-reaction) method, oxidation-number method

Common mistake Common errors ko steel-man karke dekho

1. "Bas atoms balance karo, charge apne aap ho jaayega." Sahi lagta hai kyunki acid–base equations mein usually hota hai. Fix: ions charge carry karte hain; $e^{-}$ explicitly add karo aur dono sides check karo.

2. Base mein balance karne se pehle $O H^{-}$ add karna. Sahi lagta hai — "ye basic hai, $O H^{-}$ start se use karo." Fix: pehle acidic form mein balance karo (ye cleaner hai), phir end mein $H^{+}$ ko $O H^{-}$ se neutralize karo.

3. Example 3 mein har nitrate reduce karna. Sahi lagta hai — sab N ek jaisa dikhta hai. Fix: spectator ions ( $N O_{3}^{-}$ jo nitrate hi reh jaata hai) ka oxidation state nahi badalta.

4. LCM step bhool jaana. Sahi lagta hai — tune halves add kar diye. Fix: electrons poori tarah cancel hone chahiye; halves ko scale karo taaki $e^{-}$ lost = $e^{-}$ gained ho.

Recall Feynman: ek 12 saal ke bacche ko samjhao

Do bachchon ki kalpana karo jo marbles pass kar rahe hain. Ek baccha sirf marbles de sakta hai, doosra sirf le sakta hai. Jitni marbles di jaati hain, utni hi leni padti hain — koi floor par nahi gir sakti. Ek redox reaction mein ye "marbles" electrons hain: jo atom oxidize hota hai wo deta hai, jo reduce hota hai wo leta hai. Balance karna bas itna hai ki koi marble drop ya invent na ho, aur ye bhi ensure karo ki har tarah ka atom dono sides par equally dikhe.

Flashcards

Half-reaction method mein acidic medium mein O atoms kaise balance karte hain?

Oxygen ki kami wali side mein

H_{2} O

add karo (ek paani per missing O).

Water se O balance karne ke baad H kaise balance karte hain?

Hydrogen ki kami wali side mein

H^{+}

add karo.

Half-reaction mein charge kaise balance hota hai?

Jyada positive side par electrons (

e^{-}

) add karo jab tak net charges equal na ho jaayein.

Acid mein

Mn O_{4}^{-} \to M n^{2 +}

ki reduced half kya hai?

Mn O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O

Mn O_{4}^{-} \to M n^{2 +}

kitne electrons gain karta hai aur kyun?

5, kyunki Mn +7 se +2 jaata hai.

Ek balanced acidic equation ko basic mein convert karne ke liye kya karte hain?

Dono sides mein equal

O H^{-}

add karo taaki saare

H^{+}

water mein neutralize ho jaayein, phir duplicate waters cancel karo.

Redox balancing ka core conservation law kya hai?

Electrons lost = electrons gained (plus mass aur charge conservation).

Oxidation-number method mein kya equal hona chahiye?

Oxidation number mein total increase = oxidation number mein total decrease.

Cu + HN O_{3}

mein har

N O_{3}^{-}

reduce kyun nahi hota?

Cu (N O_{3})_{2}

mein nitrate ek spectator hai (+5 unchanged); sirf wo N jo

N O_{2}

banta hai wo reduce hota hai.

LEO GER ka matlab kya hai?

Lose Electrons = Oxidation; Gain Electrons = Reduction.

Connections

Oxidation states / oxidation number rules
Oxidizing and reducing agents
Electrochemical cells & standard electrode potentials
Stoichiometry & mole ratios
Acid–base neutralization (for the OH⁻ step)
Conservation of mass and charge