Calculate the equilibrium constant for the following reaction- 3Sn - 2Cr2O72- - 28 H- - 3Sn4- - 4Cr3- - 14 H2O EoSn-Sn4- - 0-009 V EoCr2O72- - Cr3- - 1-33 V

The correct option is C 10^272 3Sn^++2Cr_2O_7^2 +8H^+ ⟶ 3Sn^4++4Cr^3++14H_2O E^o_cell=E^o_cathode E^o_anode (Reduction) as Cr^+6⟶ Cr^+3⟶ I ...

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Characteristics of Equilibrium Constant

Calculate the...

Question

Calculate the equilibrium constant for the following reaction:
$3 S n + 2 C r_{2} O_{7}^{2 -} + 28 H^{+} \to 3 S n^{4 +} + 4 C r^{3 -} + 14 H_{2} O$
$E_{S n / S n^{4 +}}^{o} = 0.009 V$
$E_{C r_{2} O_{7}^{2 -} / C r^{3 -}}^{o} = 1.33 V$

10^{413}

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10^{1.8}

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10^{272}

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10^{317}

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Solution

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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}

C r_{2} O_{7}^{2 -} + I^{-} \to I_{2} + C r^{3 +} E_{c e l l}^{o} = 0.79; E_{C r_{2} O_{7}^{2 -}}^{o} = 1.33 V, E_{t_{2}}^{o}

E_{C l_{2} / C l^{-}}^{o} = 1.36 V, E_{C r^{3 +} / C r}^{\circ} = - 0.74 V

E_{C r_{2} O_{7}^{2 -} / C r^{3 +}}^{o} = 1.33 V, E_{M n O_{4}^{-} / M n^{2 +}}^{o} = 1.51 V

298 K

2 F e^{3 +} + S n^{2 +} \to 2 F e^{2 +} + S n^{4 +}

E_{S n^{4 +} / S n^{2 +}}^{\circ} = 0.15 v o l t; E_{F e^{3 +} / F e^{2 +}}^{\circ} = 0.771 v o l t

E_{C r^{3 +} / C r}^{o} = - 0.74 V; E_{M n O_{4}^{-} / M n^{2 +}}^{o} = 1.51 V

E_{C r_{2} O_{7}^{2 +} / C r^{3 +}}^{o} = 1.33 V; E_{C l / C l^{-}}^{o} = 1.36 V

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