Given: E°Au3+/Au = 1.4 V , E°O2/H2O = 1.23 V, E°Br2/Br- = 1.09 V , E°S2O82- /SO4-2 = 2.08 V Which will be best oxidising agent ?

Question

Given: E°Au3+/Au = 1.4 V , E°O2/H2O = 1.23 V, E°Br2/Br- = 1.09 V , E°S2O82- /SO4-2 = 2.08 V Which will be best oxidising agent ? (A) Au3+ (B) O2 (C) Br2 (D) S2 O8-2

Tardigrade Admin · Accepted Answer

The species which has highest value of reduction potential will be best oxidising agent. Hence S2 O8-2

Q. Given : $E_{A u^{3 +} / A u}^{\circ} = 1.4 V, E_{O_{2} / H_{2} O}^{\circ} = 1.23 V, E_{B r_{2} / B r^{-}}^{\circ} = 1.09 V, E_{S_{2} O_{8}^{2 -} / S O_{4}^{- 2}}^{\circ} = 2.08 V$
Which will be best oxidising agent ?

$Au X^{3 +}$

$O_{2}$

$Br X_{2}$

$S X_{2} O X_{8} X^{- 2}$

The species which has highest value of reduction potential will be best oxidising agent. Hence $S X_{2} O X_{8} X^{- 2}$

Q. Given : EAu3+/Au∘​=1.4V,EO2​/H2​O∘​=1.23V,EBr2​/Br−∘​=1.09V,ES2​O82−​/SO4−2​∘​=2.08V Which will be best oxidising agent ?

AuX3+

O2​

BrX2​

SX2​ OX8​X−2