Correct answer: (B) 0.535 V. BrO3-+5→BrO^(-)+1→0.45 V 1/2Br_20→1.07 V Br^(-) For BrO3-+5→BrO^(-)+1→0.45 V 1/2Br_20→1.07 V Br^(-)-1For BrO3- to BrO ^(- )∆G =-nFE°∆G =-4FE1°For BrO- to 1/2 Br_(2 )∆G =-0.45 FE°For 1/2Br_(2 )to Br^(- )∆G =-1.07FE°For BrO3- to Br^(-)∆G =-6FE°Therefore, -6FE° = -4FE1°-0.45FE°-1.07FE°E1°=0.535 V

Q25 / 160 · TS EAMCET 2020 (09 Sep Shift 1) · Chemistry

The value of $E_{1}^{o}$ is

A
$0.76 V$
Correct answer
B
$0.535 V$
C
$0.428 V$
D
$1.12 V$

Solution

\overset{+ 5}{B r O_{3}^{-}} \to \overset{+ 1}{B r O^{-}} \overset{0.45 V}{\to} \frac{1}{2} \overset{0}{B r_{2}} \overset{1.07 V}{\to} \overset{}{B r^{-}}

For

\overset{+ 5}{B r O_{3}^{-}} \to \overset{+ 1}{B r O^{-}} \overset{0.45 V}{\to} \frac{1}{2} \overset{0}{B r_{2}} \overset{1.07 V}{\to} \overset{- 1}{B r^{-}} F o r B r O_{3}^{-} t o B r O^{-} ∆ G = - n F E ° ∆ G = - 4 F E_{1}^{°} F o r B r O_{}^{-} t o \frac{1}{2} B r_{2} ∆ G = - 0.45 F E ° F o r \frac{1}{2} B r_{2} t o B r^{-} ∆ G = - 1.07 F E ° F o r B r O_{3}^{-} t o B r^{-} ∆ G = - 6 F E ° T h e r e f o r e, - 6 F E ° = - 4 F E_{1}^{°} - 0.45 F E ° - 1.07 F E ° E_{1}^{°} = 0.535 V

Hence, the correct answer is option: 'B'.

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The value of E1o is

Solution

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The value of $E_{1}^{o}$ is