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Standard XII

Chemistry

Electron Gain Enthalpy

Using the sta...

Question

Using the standard electrode potentials given in the table, predict if the reaction between the following is possible.
$F e^{3 +}$ (aq) and $C u (s)$

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Solution

Oxidation half reaction is $C u (s) \to C u^{2 +} (a q) + 2 e^{-}; E^{0} = - 0.34 V$ .
Reduction half reaction is $[F e^{3 +} (a q) + e^{-} \to F e^{2 +} (a q)] \times 2; E^{0} = + 0.77 V$ .
The net cell reaction is $2 F e^{3 +} (a q) + C u (s) \to 2 F e^{2 +} (a q) + C u^{2 +}; E^{0} = + 0.43 V$ .
Since, the cell potential is positive, the reaction is feasible.

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Similar questions

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

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

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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)

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Q. Using the standard electrode potentials given in Table

3.1

, predict if the reaction between the following is feasible:
(i)

F e^{3 +} (a q)

and

I^{-} (a q)

(ii)

A g^{+} (a q)

and

C u (s)

(iii)

F e^{3 +} (a q)

and

B r^{-} (a q)

(iv)

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F e^{3 +} (a q)

(v)

B r_{2} (a q)

and

F e^{2 +} (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

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

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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 Cu(s)

Oxidation half reaction is Cu(s)→Cu2+(aq)+2e−;E0=−0.34V.Reduction half reaction is [Fe3+(aq)+e−→Fe2+(aq)]×2;E0=+0.77V.The net cell reaction is 2Fe3+(aq)+Cu(s)→2Fe2+(aq)+Cu2+;E0=+0.43V.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.
$F e^{3 +}$ (aq) and $C u (s)$