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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Solution

Given that $E^{⊖}$ values of

$\frac{F e^{3 +} (a q)}{F e^{2 +} (a q)} = + 0.77 \frac{I_{2} (s)}{I^{-} (a q)} = + 0.54, \frac{C u^{2 +} (a q)}{C u (s)} = + 0.34, \frac{A g^{+} (a q)}{A g (s)} = + 0.80 V .$
Since, $E^{⊖}$ of the redox couple $\frac{A g^{+}}{A g}$ is the most positive, i.e., 0.80 V, therefore $A g^{+}$ is the strongest oxidising agent.

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E^{\circ}

E^{\circ}

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

(E^{0} values : F e^{3 +} ∣ F e^{2 +} = + 0.77 V; I_{2} (s) ∣ I^{-} = + 0.54 V; C u^{2 +} | C u = + 0.34 V; A g^{+} ∣ A g = + 0.80 V)

E^{0}

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

I_{2} / I^{-}

C u^{2 +} / C u

A g^{+} / A g

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 $F e^{3 +}$

$E^{⊖}$ values of some redox couples are given below. On the basis of these values choose the coorect option.
$E^{⊖}$ values :

$\frac{B r_{2}}{B r^{-}} = + 1.90 \frac{A g^{+}}{A g (s)} = + 0.80 \frac{C u^{2 +}}{C u (s)} = + 0.34 \frac{I_{2} (S)}{I^{-}} = + 0.54;$

(a) Cu will reduce $B r^{-}$
(b) Cu will reduce $A g$
(c) Cu will reduce $I^{-}$
(d) Cu will reduce $B r_{2}$

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