E - values of some redox couples are given below- On the basis of these values choose the coorect option- E - values- frac Br - 2 Br -1-90 1 frac Ag - Ag s -0-80 frac Cu -2- Cu s -0-34 frac - I - 2 S I -0-54 -a Cu will reduce Br -b Cu will reduce Agc Cu will reduce I -d Cu will reduce Br 2

Given that E^⊖ values of Br_2/Br^ =+1.90 V Ag^+/Ag(S)=+0.80V Cu^2+/Cu(s)=+0.34V I_2(S)/I^ =+0.54V Br^ /Br_2=+1.90 V The E^⊖ values show that copper will reduce ...

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Standard XII

Chemistry

Electrochemical Series

E ⊖ values of...

Question

$E^{⊖}$ 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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Solution

Given that $E^{⊖}$ values of

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

The $E^{⊖}$ values show that copper will reduce $B r_{2}$ , if the $E^{⊖}$ of the following redox reaction is positive.
Now
$2 C u + B r_{2} \to C u B r_{2} C u \to C u^{2 +} + 2 e^{-}; E^{⊖} = - 0.34 V \frac{B r_{2} + 2 e^{-} \to 2 B r^{-}; E = + 1.09 V}{C u + B r_{2} \to C u B r_{2}; E^{⊖} = + 0.75 V}$
Since $E^{⊖}$ of this reaction is positive. therefore, Cu can reduce $B r_{2}$ while other reaction will give negative value.

Answer: (d). Cu will reduce $B r_{2}$

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Similar questions

E^{⊝}

values of some redox couples are given. On the basis of these values choose the correct option:

E^{⊝}

values:

B r_{2} / B r^{-} = + 1.9; A g^{+} / A g (s) = + 0.8

C u^{2 +} / C u = + 0.34; I_{2} (s) / I^{-} = 0.54

E^{0}

values of some redox couples are given below. On the basis of these values choose the correct option.

E^{0}

values

B r_{2} | B r^{-}

= +1.90V;

A g^{+} | A g_{(} s)

= +0.80V;

C u^{2 +} | C u_{(} s)

= + 0.34V;

I_{2 (s)} | I^{-}

= +0.54V

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) $A g^{+}$

Q. The standard

E^{0}

values of few redox couples are

Z n^{2 +} / Z n = - 0.76 V

A g^{+} / A g = + 0.08 V

C u^{2 +} / C u = 0.34 V

. Choose the correct option?

Q. On the basis of standard electrode potential of redox couples given below, find out which of the following is the strongest oxidising agent.

(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)

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E⊖ values of some redox couples are given below. On the basis of these values choose the coorect option. E⊖ values : Br2Br−=+1.90Ag+Ag(s)=+0.80Cu2+Cu(s)=+0.34I2(S)I−=+0.54; (a) Cu will reduce Br− (b) Cu will reduce Ag (c) Cu will reduce I− (d) Cu will reduce Br2