Using the standard electrode potentials given in the table, predict if the reaction between the following is feasible:
$F e_{(a q)}^{3 +}$ and $I_{(a q)}^{-}$

$A g^{+} (a q)$ and $C u (s)$

$F e^{3 +} (a q)$ and $C u (s)$

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

$B r_{2} (a q)$ and $F e^{2 +} (a q)$

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Solution

The possible reaction between $F e_{(a q)}^{3 +} + I_{(a q)}^{-}$
$2 F e_{(a q)}^{3 +} + 2 I_{(a q)}^{-} ⟶ F e_{(a q)}^{2 +} + I_{2 (s)}$ is given by,
Oxidation half equation: $2 I_{(a q)}^{-} ⟶ I_{2 (s)} + 2 e^{-} : E^{\circ} = - 0.54 V$
Reduction half equation: $[F e_{(a q)}^{3 +} + e^{-} ⟶ F e_{(a q)}^{2 +}] \times 2; E^{\circ} = + 0.77 V$
$¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯$
$2 F e_{(a q)}^{3 +} + 2 I_{(a q)}^{-} ⟶ 2 F e_{(a q)}^{2 +} + I_{2 (s)}; E^{\circ} = + 0.23 V$
$E^{\circ}$ for the overall reaction is positive. Thus, the reaction between $F e_{(a q)}^{3 +}$ and $I_{(a q)}^{-}$ is feasible.

The possible reaction between $A g_{(a q)}^{+} + C u_{(s)}$ is given by,
$2 A g_{(a q)}^{+} + C u_{(s)} ⟶ A g_{(s)} + C u_{(a q)}^{2 +}$
Oxidation half equation: $C u_{(s)} ⟶ C u_{(a q)}^{2 +} + 2 e^{-} : E^{\circ} = - 0.34 V$
Reduction half equation: $[A l_{(a q)}^{+} + e^{-} ⟶ A g_{(s)}] \times 2; E^{\circ} = + 0.80 V$
$¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯$
$2 A g_{(a q)}^{+} + C u_{(s)} ⟶ 2 A g_{(s)} + C u^{2 +}; E^{\circ} = + 0.46 V$
$E^{\circ}$ positive for the overall reaction is positive. Hence, the reaction between $A g_{(a q)}^{+}$ and $C u_{(s)}$ is feasible.

The possible reaction between $F e_{(a q)}^{3 +}$ and $C u_{(s)}$ is given by,
$2 F e_{(a q)}^{3 +} + C u_{(s)} ⟶ 2 F e_{(s)}^{2 +} + C u_{(a q)}^{2 +}$
Oxidation half equation: $C u_{(s)} ⟶ C u_{(a q)}^{2 +} + 2 e^{-} : E^{\circ} = - 0.34 V$
Reduction half equation: $[F e_{(a q)}^{3 +} + e^{-} ⟶ F e_{(s)}^{2 +}] \times 2; E^{\circ} = + 0.77 V$
$¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯$
$2 F e_{(a q)}^{3 +} + C u_{(s)} ⟶ 2 F e_{(s)}^{2 +} + C u_{(a q)}^{2 +}; E^{\circ} = + 0.43 V$
$E^{\circ}$ positive for the overall reaction is positive. Hence, the reaction between $F e_{(a q)}^{3 +}$ and $C u_{(s)}$ is feasible.

The possible reaction between $A g_{(s)}$ and $F e_{(a q)}^{3 +}$
$A g_{(s)} + 2 F e_{(a q)}^{3 +} ⟶ A g_{(a q)}^{+} + F e_{(a q)}^{2 +}$
Oxidation half equation: $A g_{(s)} ⟶ A g_{(a q)}^{+} + e^{-} : E^{\circ} = - 0.80 V$
Reduction half equation: $F e_{(a q)}^{3 +} + e^{-} ⟶ F e_{(a q)}^{2 +}; E^{\circ} = + 0.77 V$
$¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯$
$A g_{(s)} + F e_{(a q)}^{3 +} ⟶ A g_{(a q)}^{+} + F e_{(a q)}^{2 +}; E^{\circ} = - 0.03 V$
Here, $E^{\circ}$ for the overall reaction is negative. Hence, the reaction between $A g_{(s)}$ and $F e_{(a q)}^{3 +}$ is not feasible.

The possible reaction between $B r_{2 (a q)}$ and $F e_{(a q)}^{2 +}$ is given by,
$B r_{2 (s)} + 2 F e_{a q}^{2 +} ⟶ 2 B r_{(a q)}^{-} + 2 F e_{(a q)}^{3 +}$
Oxidation half equation: $F e_{(a q)}^{2 +} ⟶ F e_{(a q)}^{3 +} + e^{-} \times 2 : E^{\circ} = - 0.77 V$
Reduction half equation: $B r_{2 (a q)} + 2 e^{-} ⟶ 2 B r_{(a q)}^{-}; E^{\circ} = + 1.09 V$
$¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯$
$B r_{2 (a q)} + 2 F e_{(a q)}^{2 +} ⟶ 2 B r_{(a q)}^{-} + 2 F e_{(a q)}^{3 +}; E^{\circ} = - 0.32 V$
Here, $E^{\circ}$ for the overall reaction is positive. Hence, the reaction between $B r_{2 (a q)}$ and $F e_{(a q)}^{2 +}$ is feasible.

Suggest Corrections

Similar questions

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 potentials given in the table, predict if the reaction between the following is possible.

F e^{3 +} (a q)

and

I^{-} (a q)

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 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

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