The equilibrium constant for the reaction- Sn s - 2H - - Sn2 - - H2 g at 25- C is- E- Sn2 - -Sn - 0-14

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Equilibrium Constant from Nernst Equation

The equilibri...

Question

The equilibrium constant for the reaction, $S n (s) + 2 H^{+} \to S n^{2 +} + H_{2} (g) a t 25^{\circ} C$ is:
$E^{\circ} (S n^{2 +} | S n) = 0.14$

5.5 \times 10^{4}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2.3 \times 10^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1.1 \times 10^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1.009

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

Suggest Corrections

Similar questions

3 S n (s) + 2 C r_{2} O_{7}^{2 -} + 28 H^{+} \to 3 S n^{+ 4} + 4 C r^{3 +} + 14 H_{2} O

E^{\circ}

S n / S n^{2 +} = 0.136 V E^{\circ}

S n^{2 +} / S n^{4} = - 0.154 V

E^{\circ}

C r_{2} O_{7}^{2 -} / C r^{3 +} = 1.33 V

K = 10^{c}

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

\frac{[P b^{2 +}]}{[S n^{2 +}]}

E_{c e l l} = 0

E_{P b / P b^{2 +}}^{\circ} = 0.13 v o l t

E_{S n^{2 +} / S n}^{\circ} = - 0.14 v o l t

S n (s) | S n^{2 +} (a q ., 1 M) | | P b^{2 +} (a q ., 1 M) | P b (s)

\frac{[S n^{2 +}]}{[P b^{2 +}]}

E_{\frac{S n^{2 +}}{S n}}^{\circ} = - 0.14 V; E_{\frac{P b^{2 +}}{P b}}^{\circ} = - 0.13 V, \frac{2.303 R T}{F} = 0.06 V)

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 +}

2 B r^{-} (a q) + S n^{2 +} (a q) \to B r_{2} (1) + S n (s)

E_{S n}^{\circ} = - 0.114 V, E_{B r_{2}}^{\circ} = + 1.09 V

Join BYJU'S Learning Program