Both Mg and Fe metal can reduce copper from a solution having Cu2- ion- according to equilibria- Mgs - Cu2- Mg2- - Cus- K1 - 6 - 1090 Fes - Cu2- Fe2- - Cus- K2 - 3 - 1026Choose the correct option regarding above equilibrium-

$E_{F e^{2 +} / F e}^{\circ} = - 0.44 V, E_{C u^{2 +} / C u}^{\circ} = 0.34 V, E_{A g^{2 +} / g}^{\circ} = 0.80 V$ . Will Fe displace Cu from $C u S O_{4}$ solution? Will Ag displace Cu from $C u S O_{4}$ solution?

1. Yes, Yes

2. No, No

3. Yes, No

4. No, Yes

Q. The standard reduction potential for

M g^{2 +} / M g

- 2.37 V

and for

C u^{2 +} / C u

0.337

. The

E_{c e l l}^{o}

for the following reaction is

M g + C u^{2 +} \to M g^{2 +} + C u

Q. Based on the following reactions, arrange the metals in increasing order of their reduction potentials.

Z n + C u^{2 +} \to Z n^{2 +} + C u

M g + Z n^{2 +} \to M g^{2 +} + Z n

C u + 2 A g^{+} \to C u^{2 +} + 2 A g

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Both Mg and Fe metal can reduce copper from a solution having Cu2+ ion, according to equilibria.Mg(s)+Cu2+⇋Mg2++Cu(s);K1=6×1090Fe(s)+Cu2+⇋Fe2++Cu(s);K2=3×1026Choose the correct option regarding above equilibrium.

The correct option is A Mg removes more Cu2+ from solutionSince, 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 $M g$ removes more $C u^{2 +}$ from solution
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.