The redox reaction involving the reducing power of hydrogen sulphide is- S - 2H- - 2e- - H2S- E-S-H2S - - 0-14VTwo other half equations are- Fe3- - e- - Fe2- E-Fe3-Fe2- - -0-77 V Br2 - 2e- - 2Br- E-Br2-Br- - 1-07 VUnder standard conditions- hydrogen sulphide reacts with iron III ions- If this statement is true enter 1- else enter 0-

Yes, Fe^3+ ions will oxidise H_2S to S 2Fe^3+ + H_2S → 2Fe^2+ + 2H^+ + S Since , E^∘ = 0.77 0.14 = 0.63 V and thus G^∘ = nFE^ ...

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Equilibrium Constant from Nernst Equation

The redox rea...

Question

The redox reaction involving the reducing power of hydrogen sulphide is:
$S + 2 H^{+} + 2 e^{-} \to H_{2} S, E_{S / H_{2} S}^{\circ} = + 0.14 V$
Two other half equations are:
$F e^{3 +} + e^{-} \to F e^{2 +} E_{F e^{3 +} / F e^{2 +}}^{\circ} = + 0.77 V$
$B r_{2} + 2 e^{-} \to 2 B r^{-} E_{B r_{2} / B r^{-}}^{\circ} = 1.07 V$
Under standard conditions, hydrogen sulphide reacts with iron $(I I I)$ ions.
If this statement is true enter 1, else enter 0.

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F e^{2 +} (a q) + 2 e^{-} \to F e (s)

E^{⊖} = - 0.44 V

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

E^{⊖} = - 1.66 V

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F e^{3 +} + e^{-} \to F e^{2 +}; (0.77 V

F e^{2 +} ⇌ F e; E^{\circ} = - 0.44 V

F e^{3 +} ⇌ F e^{2 +}; E^{\circ} = + 0.77 V

F e^{2 +}, F e^{3 +}

F e

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

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(E^{o})

F e^{3 +} \to F e

F e^{2 +} + 2 e^{-} \to F e; E_{F e^{2 +} / F e}^{o} = - 0.47 V

F e^{3 +} + e^{-} \to F e^{2 +}; E_{F e^{3} / F e^{2 +}}^{o} = + 0.77 V

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