Both metals Mg and Fe can reduce copper from a solution having Cu 2- ion- according to equilibria- Mg s - Cu 2- aq - Mg 2- aq -Cu s - K 1 -6- 10 90 Fe s - Cu 2- aq - Fe 2- aq -Cu s - K 1 -3- 10 26Which metal will remove cupric ion from the solution to a greater extent-

The correct option is A Mg will remove more Cu ^ 2+ than Fe .Since, K _ 1 gt; K _ 2 , the product in the first reaction is much more f ...

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Standard XII

Chemistry

Reaction Quotient

Both metals ...

Question

Both metals $M g$ and $F e$ can reduce copper from a solution having ${C u}^{2 +}$ ion, according to equilibria.
$M g (s) + {C u}^{2 +} (a q) ⇌ {M g}^{2 +} (a q) + C u (s); K_{1} = 6 \times 10^{90}$
$F e (s) + {C u}^{2 +} (a q) ⇌ {F e}^{2 +} (a q) + C u (s); K_{1} = 3 \times 10^{26}$
Which metal will remove cupric ion from the solution to a greater extent?

M g

will remove more

{C u}^{2 +}

than

F e

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F e

will remove more

{C u}^{2 +}

than

M g

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F e

will remove equal

{C u}^{2 +}

M g

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None of these

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Solution

The correct option is A $M g$ will remove more ${C u}^{2 +}$ than $F e$ .
Since, $K_{1} > K_{2}$ , the product in the first reaction is much more favoured than in the second one. $M g$ thus, removes more ${C u}^{2 +}$ from solution than $F e$ does.

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Similar questions

Q. Both

M g

and

F e

metal can reduce copper from a solution having

C u^{2 +}

ion, according to equilibria.

M g_{(s)} + C u^{2 +} ⇋ M g^{2 +} + C u_{(s)}; K_{1} = 6 \times 10^{90}

F e_{(s)} + C u^{2 +} ⇋ F e^{2 +} + C u_{(s)}; K_{2} = 3 \times 10^{26}

Choose the correct option regarding above equilibrium.

Q. Which of the following metals or metal ions will be reduced by

F e

A g^{+} (a q) + e^{-} - > A g (s) E^{\circ} = + 0.80 V

C u^{2 +} (a q) + 2 e^{-} - - > C u (s) E^{\circ} = + 0.34 V

P b^{2 +} (a q) + 2 e^{-} - - > P b (s) E^{\circ} = - 0.13 V

F e^{2 +} (a q) + 2 e^{-} - - > F e (s) E^{\circ} = - 0.44 V

A l^{3 +} (a q) + 3 e^{-} - - > A l (s) E^{\circ} = - 1.66 V

Q. Which of the following reactions will not occur? Why not?
(a)

{MgSO}_{4} (aq) + Cu (s) \underset{}{\to} {CuSO}_{4} (aq) + Mg (s)

(b)

{CuSO}_{4} (aq) + Fe (s) \underset{}{\to} {FeSO}_{4} (aq) + Cu (s)

(c)

{MgSO}_{4} (aq) + Fe (s) \underset{}{\to} {FeSO}_{4} (aq) + Mg (s)

Q. State, giving reasons, which of the following reactions will occur and which will not:
(a) MgSO₄ (aq) + Cu(s) → CuSO₄ (aq) + Mg(s)
(b) CuSO₄ (aq) + Fe (s) → FeSO₄(aq) + Cu(s)
(c) MgSO₄ (aq) + Fe (s) → FeSO₄ (aq) + Mg(s)

Q. A test for the complete removal of

C u^{2 +}

ions from a solution of

C u^{2 +} (a q .)

is to add

N H_{3} (a q .)

. A blue colour signifies the formation of complex

[C u (N H_{3})_{4}]^{2 +}

that has

K_{f} = 1.1 \times 10^{13}

and thus confirms the presence of

C u^{2 +}

in the solution. 250 mL of 0.1 M

C u S O_{4} (a q .)

is electrolyzed by passing a current of 3.5 ampere for 1350 seconds. Then, sufficient quantity of

N H_{3} (a q .)

is added to the electrolyzed solution maintaining

[N H_{3}] = 0.1 M

. If

[C u (N H_{3})_{4}]^{2 +}

is detectable upto its concentration as low as

1 \times 10^{- 5}

. Pick the correct statement(s):

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Both metals Mg and Fe can reduce copper from a solution having Cu2+ ion, according to equilibria.Mg(s)+Cu2+(aq)⇌Mg2+(aq)+Cu(s);K1=6×1090Fe(s)+Cu2+(aq)⇌Fe2+(aq)+Cu(s);K1=3×1026Which metal will remove cupric ion from the solution to a greater extent?

The correct option is A Mg will remove more Cu2+ than Fe.Since, K1>K2, the product in the first reaction is much more favoured than in the second one. Mg thus, removes more Cu2+ from solution than Fe does.

The correct option is A $M g$ will remove more ${C u}^{2 +}$ than $F e$ .
Since, $K_{1} > K_{2}$ , the product in the first reaction is much more favoured than in the second one. $M g$ thus, removes more ${C u}^{2 +}$ from solution than $F e$ does.