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Standard X

Chemistry

Electrochemical Series

Using the sta...

Question

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

F e^{3 +} a n d I^{-}

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A g^{+} a n d C u

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F e^{3 +} a n d C u

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A g

and

F e^{3 +}

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Solution

The correct option is D $A g$ and $F e^{3 +}$
Given $E^{0}$ 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
Thus the order of getting reduced is:
$A g^{+} > F e^{3 +} > I_{2} > C u^{2 +}$
Possible reactions are:
$I_{2} + C u \to I^{-} + C u^{2 +}$ :
$F e^{3 +} + I^{-} \to F e^{2 +} + I_{2}$
$F e^{3 +} + C u \to F e^{2 +} + C u^{2 +}$
$A g^{+} + F e^{2 +} \to A g + F e^{3 +}$
$A g^{+} + I^{-} \to A g + I_{2}$
$A g^{+} + C u \to A g + C u^{2 +}$

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Similar questions

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 $F e^{3 +}$

Q. The more positive the value of

E^{\circ}

, the greater is the tendency of the species to getreduced. Using the standard electrode potential of redox couples given below find outwhich of the following is the strongest oxidising agent.

E^{\circ}

values:

F e^{3 +} / F e^{2 +} = + 0.77; I_{2} (s) / I^{-} = + 0.54;

C u^{2 +} / C u = + 0.34; A g^{+} / A g = + 0.80 V

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 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)

A g (s)

and

F e^{3 +} (a q)

(v)

B r_{2} (a q)

and

F e^{2 +} (a q)

The more positive the value of $E^{⊖}$ , the greater is the tendency of the species to get reduced. Using the standard electrode potential of redox couples given below find out which of the following is the strongest oxidising agent.

E\ominus values :

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

(a) $F e^{3 +}$
(b) $I_{2} (s)$
(c) $C u^{2 +}$
(d) $A g^{+}$

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Using the standard electrode potential , find out the pair between which redox reaction is not feasible .E0 values : Fe3+/Fe2+ = +0.77;I2/I− = + 0.54;Cu2+/Cu = +0.34; Ag+/Ag = +0.80V

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