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Method for equilibrating an oxidation or reduction half reaction

In acid or neutral medium

Example: Equilibrate in acid medium: $..Cr_2O_7^{2-}$ $..$ $\longrightarrow$ $.. Cr^{3+}$ $..$

Method: - Equilibrate in relation to the central atom: $Cr_2$$O_7^{2-}$ $..$ $\longrightarrow$$ 2$$Cr^{3+}$ - Find the variation of the oxidation number(s) of the central atom: $\Delta o.n.(Cr)$ $=$ $6$ $-$ $12$ $=$ $-6$ - Deduce the number of exchanged electrons: $Cr_2O_7^{2-}$$+$ $6e^-$$+$ $..$ $\longrightarrow$ $2Cr^{3+}$ $+$ $..$ - Equilibrate eventually with $H^+$ and $H_2O$: $Cr_2O_7^{2-}$ $+$ $6e^-$ $+$$14H^+$$\longrightarrow$ $2Cr^{3+}$ $+$$7H_2O$

In basic medium

Example: Equilibrate in basic medium: $..S_2O_3^{2-}$ $..$ $\longrightarrow$ $.. SO_4^{2-}$ $..$

Method: - Equilibrate first in acid medium: $S_2O_3^{2-}$ $-$ $8e^-$ $+$ $5H_2O$ $\longrightarrow$ $2SO_4^{2-}$ $+$ $10H^+$ - Neutralise $H^+$ ions by $OH^-$ ions introduced on both sides of the reaction arrow. $S_2O_3^{2-}$ $-$ $8e^-$ $+$ $5H_2O$$+$ $10\; OH^-$$\longrightarrow$ $2SO_4^{2-}$ $+$$10H_2O$ - Simplify: $S_2O_3^{2-}$ $-$ $8e^-$ $+$ $10 \;OH^-$ $\longrightarrow$ $2SO_4^{2-}$$+$ $5H_2O$