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

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

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Solution

(i) A reaction is feasible, if its $E_{c e l l}^{c e l l}$ value is positive.

$F e^{3 +} (a q) + I^{-} (a q) \to F e^{2 +} (a q) + \frac{1}{2} I_{2} (g)$

$E_{c e l l}^{\circ} = E_{c e l l}^{\circ} (\frac{F e^{3 +}}{F e^{26}}) - E_{c e l l}^{\circ} \frac{\frac{1}{2} I_{2}}{I^{-}} = 0.77 - 0.54 = 0.23 V$
The reaction is feasible.

(ii) A reaction is feasible, if its $E_{c e l l}^{c e l l}$ value is positive.

$C u (s) + 2 A g^{+} (a q) \to 2 A g (s) + C u^{2 +} (a q)$
$E_{c e l l}^{\circ} = E_{c e l l}^{\circ} (\frac{A g^{+}}{A g^{0}}) - E_{c e l l}^{\circ} (\frac{C u^{2 +}}{C u}) = 0.80 - 0.34 = 0.46 V$
The reaction is feasible

(iii) A reaction is feasible, if its $E_{c e l l}^{c e l l}$ value is positive.

$F e^{3 +} (a q) + B r^{-} (a q) \to F e^{2 +} (a q) + \frac{1}{2} B r_{2} (g)$

$E_{c e l l}^{\circ} = E_{c e l l}^{\circ} (\frac{F e^{3 +}}{F e^{2 +}}) - E_{c e l l}^{\circ} (\frac{\frac{1}{2} B r_{2}}{B r^{-}}) = 0.77 - 1.09 = - 0.32 V$

The reaction is not feasible

(iv) A reaction is feasible, if its $E_{c e l l}^{c e l l}$ value is positive.

$A g (s) + F e^{3 +} \to A g^{+} (a q) + F e^{2 +} (a q)$

$E_{c e l l}^{\circ} = E_{\frac{F e^{3 +}}{F e^{2 +}}}^{\circ} - E_{\frac{A g^{+}}{A g}}^{\circ} = 0.77 - 0.80 = - 0.03 V$
The reaction is not feasible

(v) A reaction is feasible, if its $E_{c e l l}^{c e l l}$ value is positive.

$\frac{1}{2} B r_{2} (a q) + F e^{2 +} (a q) \to B r^{-} (a q) + F e^{3 +} (a q)$

$E_{c e l l}^{\circ} = E_{\frac{1}{2} \frac{B r_{2}}{B r^{-}}}^{\circ} - E_{\frac{F e^{3 +}}{F e^{2 +}}}^{\circ} = 1.09 - 0.77 = 0.32 V$

The reaction is feasible.

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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 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 potential, find out the pair between which redox reaction is not feasible.

$E ⊖$ values :

$\frac{F e^{3 +}}{F e^{2 +}} = + 0.77 \frac{I_{2}}{I^{-}} = + 0.54; \frac{C u^{2 +}}{C u} = + 0.34 \frac{A g^{+}}{A g} = + 0.80 V$

(a) $F e^{3 +}$ and $I^{-}$
(b) $A g^{+}$ and Cu
(c) $F e^{3 +}$ and Cu
(d) Ag and $F e^{3 +}$

A g (s)

F e^{3 +} (a q)

F e^{3 +}

C u (s)

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