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Standard XII

Chemistry

Introduction

Using the sta...

Question

Using the standard electrode potentials given in the Table, predict if the reaction between the following is feasible.
$A g (s)$ and $F e^{3 +} (a q)$

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Solution

Oxidation half reaction is $A g (s) \to A g^{+} (a q) + e^{-}; E^{0} = - 0.80 V$ .
Reduction half reaction is $F e^{3 +} (a q) + e^{-} \to F e^{2 +} (a q); E^{0} = + 0.77 V$ .
The net cell reaction is $2 F e^{3 +} (a q) + A g (s) \to 2 F e^{2 +} (a q) + A g^{+}; E^{0} = - 0.03 V$ .
Since, the cell potential is negative, the reaction is not feasible.

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Similar questions

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 possible.

F e^{3 +} (a q)

and

I^{-} (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. Using the standard electrode potential , find out the pair between which redox reaction is not feasible .

E^{0}

values :

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

= +0.77;

I_{2} / I^{-}

= + 0.54;

C u^{2 +} / C u

= +0.34;

A g^{+} / A g

= +0.80V

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Using the standard electrode potentials given in the Table, predict if the reaction between the following is feasible.Ag(s) and Fe3+(aq)

Oxidation half reaction is Ag(s)→Ag+(aq)+e−;E0=−0.80V.Reduction half reaction is Fe3+(aq)+e−→Fe2+(aq);E0=+0.77V.The net cell reaction is 2Fe3+(aq)+Ag(s)→2Fe2+(aq)+Ag+;E0=−0.03V.Since, the cell potential is negative, the reaction is not feasible.

Using the standard electrode potentials given in the Table, predict if the reaction between the following is feasible.
$A g (s)$ and $F e^{3 +} (a q)$