The reaction- 2Br - aq - Sn2 - aq - Br2 1 - Sn s with the standard potentials- E- Sn - - 0-114V- E- Br2 - - 1-09V- is-

The correct option is A spontaneous in reverse direction 2Br^ +Sn^2+⟶ Br_2+Sn(s) E_cell=E^o_Sn/Sn^2+ E^o_Br_2/Br^ = 0.114 0.109= 0.223 V Hence, ...

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Gibb's Energy and Nernst Equation

The reaction,...

Question

The reaction, $2 B r^{-} (a q) + S n^{2 +} (a q) \to B r_{2} (1) + S n (s)$ with the standard potentials, $E_{S n}^{\circ} = - 0.114 V, E_{B r_{2}}^{\circ} = + 1.09 V$ , is:

spontaneous in reverse direction

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spontaneous in forward direction

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at equilibrium

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non-spontaneous in reverse direction

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Solution

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Similar questions

E_{F e^{3 +} / F e^{2 +}}^{\circ} = + 0.77 V; E_{S n^{2 +} / S n}^{\circ} = - 0.14 V

S n_{(s)} + 2 F e_{(a q)}^{3 +} \to 2 F e_{(a q)}^{2 +} + S n_{(a q)}^{2 +}

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

A g^{+} (a q .) + e^{-} \to A g (s)

S n^{2 +} (a q .) + 2 e^{-} \to S n (s)

25^{o} C

0.80 v o l t

- 0.14 v o l t

e m f

S n | S n^{2 +} (1 M) | | A g^{+} (1 M) | A g

C l_{2} (g) + 2 e^{-} \to 2 C l^{-} (a q) + 1.36 V

B r_{2} (g) + 2 e^{-} \to 2 B r^{-} (a q) + 1.09 V

S n_{2 +} + 2 e^{-} \to S n - 0.14 V

P b^{2 +}

S n^{2 +}

S n (s) + P b^{2} (a q .) \to S n^{2 +} (a q .) + P b (s)

E_{S n^{2 +} / S n}^{\circ} = - 0.136 v o l t

E_{P b^{2 +} / P b}^{\circ} = - 0.126 v o l t

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