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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$ $2SO_4^{2-}$ $+$$10H_2O - Simplify: S_2O_3^{2-} - 8e^- + 10 \;OH^- \longrightarrow 2SO_4^{2-}$$+$ $5H_2O$