For the half reaction Bs - B2- - 2e- E1- - - 0-44 Vand B2- B3- - e- E2- - 1-3 VWhat is E- for the reaction- 3e- - B-3- Bs

The correct option is B 0.14 V B(s)→ B ^ 2 + 2e ^ E _ 1 ^ 0 = 0.44V #160; #160; #160; #160; i B ^ 2+ → B ^ 3+ + e ^ ...

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Redox Reactions and Couple

For the half ...

Question

For the half reaction
$B (s) \to B^{2 +} + 2 e^{-}$ $E_{1}^{\circ} = - 0.44 V$
and $B_{2} + \to B^{3 +} + e^{-}$ $E_{2}^{\circ} = 1.3 V$
What is $E^{\circ}$ for the reaction,
$3 e^{-} + B^{+ 3} \to B (s)$

0.86

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0.14

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- 0.14

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- 0.28

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Solution

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B (s)

\to

B^{2 +} + 2 e^{-}

E_{1}^{\circ} = - 0.44 V

B^{2 +}

\to

B^{3 +} + e^{-}

E_{2}^{\circ} = 1.3 V

E_{3}^{\circ}

3 e^{-} + B^{+ 3}

\to

A u B r_{4}^{-} (a q) + 3 e^{-} \to A u (s) + 4 B r^{-} (a q); E^{\circ} = - 0.86 V

E u^{3 +} (a q) + e^{-} \to E u^{2 +} (a q); E^{\circ} = - 0.43 V

S n^{2 +} (a q) + 2 e^{-} \to S n (s); E^{\circ} = - 0.14 V

I O^{-} (a q) + H_{2} O (l) + 2 e^{-} \to I^{-} (a q) + 2 O H^{-} (a q); E^{\circ} = + 0.49 V

E^{0}

S n^{4 +} (a q) + 4 e^{-} \to S n (s)

E^{0}

S n^{4 +} (a q) + 2 e^{-} \to S n^{2 +} (a q); E^{0} = 0.15 V

S n^{2 +} (a q) + 2 e^{-} \to S n (s); E^{0} = - 0.14 V

P b + S n^{2 +} \to P b^{2 +} + S n

P b \to P b^{2 +}, E^{o} = 0.13 V

S n^{2 +} + 2 e^{-} \to S n; E^{o} = - 0.14 V

E^{0}

F e^{3 +} (a q) + 3 e^{-} \to F e (s)

E^{0}

F e^{3 +} (a q) + e^{-} \to F e^{2 +} (a q); E^{0} = 0.77 V

F e^{2 +} (a q) + 2 e^{-} \to F e (s); E^{0} = - 0.44 V

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