For the half reaction Bs - B2- - 2e- E 1 -0-44Vand B2- - B3- - e- E 2 -1-3VWhat is E3- for the reaction 3e- - B-3 - Bs-

The correct option is A 0.86 V #160; #160; B( s ) → B ^ 2+ + 2e ^ E _ 1 ^ 0 = 0.44V + #160; #160; B ^ 2+ → B ^ 3+ + e ^ ...

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Ion Electron Method

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 $E_{3}^{\circ}$ for the reaction
$3 e^{-} + B^{+ 3}$ $\to$ B(s):

0.86 V

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0.14 V

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- 0.14 V

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- 0.28 V

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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^{\circ}

3 e^{-} + B^{+ 3} \to B (s)

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_{c e l l}^{\circ}

2 C r (s) + 3 F e^{2 +} (0.01 M) \to 2 C r^{3 +} (0.01 M) + 3 F e (s)

E_{c e l l} = 0.261 V

E^{\circ}

[E^{o} (F e^{2 +} / F e) = - 0.44 V]

E^{\circ} (A^{2 +} / A) = - 2.37 V

E^{\circ} (B^{2 +} / B) = - 0.14 V

E_{c e l l}^{\circ}

2 C r (s) + 3 F e^{2 +} (0.01 M) \to 2 C r^{3 +} (0.01 M) + 3 F e (s)

E_{c e l l} = 0.261 V

E^{\circ}

[E^{o} (F e^{2 +} / F e) = - 0.44 V]

E^{\circ} (A^{2 +} / A) = - 2.37 V

E^{\circ} (B^{2 +} / B) = - 0.14 V

E_{1}

E_{2}

E_{3}

E_{3} = \frac{n_{1} E_{1} + n_{2} E_{2}}{n_{3}}

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