Using the standard electrode potentials given in the table, predict if the reaction between the following is possible.
$F e^{3 +} (a q)$ and $I^{-} (a q)$

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Solution

The oxidation half reaction is $2 I^{-} (a q) \to I_{2} (s) + 2 e^{-}; E^{0} = - 0.54 V$ .
The reduction half reaction is $[F e^{3 +} (a q) + e^{-} \to] \times 2; E^{0} = + 0.77 \times 2 V = + 0.77 V$ .
The net cell reaction is $2 F e^{3 +} (a q) + 2 I^{-} (a q) \to 2 F e^{2 +} (a q) + I_{2} (s); E^{0} = + 23 V$ .
Since, the cell potential is positive, the reaction is feasible.

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Using the standard electrode potentials given in the Table 8.1, predict if the reaction between the following is feasible:

(a) Fe³⁺(aq) and I^–(aq)

(b) Ag⁺(aq) and Cu(s)

(d) Ag(s) and Fe³⁺(aq)

(e) Br₂(aq) and Fe²⁺(aq)

Using the standard electrode potentials given in the Table, predict if the reaction between the following is feasible.

A g (s)

and

F e^{3 +} (a q)

Using the standard electrode potentials given in Table 3.1, predict if the reaction between the following is feasible:

(i) Fe³⁺(aq) and I⁻(aq)

(ii) Ag⁺ (aq) and Cu(s)

(iii) Fe³⁺ (aq) and Br⁻(aq)

(iv) Ag(s) and Fe³⁺ (aq)

(v) Br₂(aq) and Fe²⁺(aq).

Using the standard electrode potential valuesl predict if the reaction between the following is feasible :

(i) $F e^{3 +} (a q) a n d I^{-} (a q)$

(ii) $A g^{+} (a q) a n d C u (s)$

(iii) $F e^{3 +} (a q) a n d B r^{-} (a q)$

(iv) $A g (s)$ and $F e^{3 +} (a q)$

(v) $B r_{2} (a q) a n d F e^{2 +} (a q)$

Given standard electrode potentials

$E_{\frac{1}{2} \frac{t_{2}}{1^{-}}}^{\circ} = + 0.541 V E_{(\frac{C u^{2 +}}{C u})}^{\circ} = + 0.34 V E_{(\frac{1}{2} \frac{B r_{2}}{B r^{-}})}^{\circ} = + 1.09 V E_{(\frac{A g^{+}}{A g})}^{\circ} = + 0.80 V E_{(\frac{F e^{3 +}}{F e^{2 +}})}^{\circ} = + 0.77 V$

Q. The standard emf of the following electrodes are;

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

under standard conditions, the potential for the reaction,

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

is:

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Using the standard electrode potentials given in the table, predict if the reaction between the following is possible.Fe3+(aq) and I−(aq)

The oxidation half reaction is 2I−(aq)→I2(s)+2e−;E0=−0.54V.The reduction half reaction is [Fe3+(aq)+e−→]×2;E0=+0.77×2 V=+0.77V.The net cell reaction is 2Fe3+(aq)+2I−(aq)→2Fe2+(aq)+I2(s);E0=+23V.Since, the cell potential is positive, the reaction is feasible.

Using the standard electrode potentials given in the table, predict if the reaction between the following is possible.
$F e^{3 +} (a q)$ and $I^{-} (a q)$