calculate the equilibrium constant for the reaction :- Mg(s) | Mg2+(0.0001M) || Cu2+(0.0001M) | Cu(s) Given E^° (Mg2+/Mg)= -2.37 V, E^°=+0.34 V.

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M g_{(s)} + C u^{2 +} (0.0001 M) \to M g^{2 +} (0.001 M) + C u_{(s)}

E_{M g^{2 +} / M g}^{o} = - 2.37

E_{C u^{2 +} / C u}^{o} = + 0.34

298 K

{M g}_{(s)} + {C u}^{2 +} (0.0001 M) \to {M g}^{2 +} (0.001 M) + {C u}_{(s)}

(E^{o})

2.71 V

M g_{(} s) + 2 A g^{+} (0.0001 M) ⇌ M g^{2 +} (0.130 M) + 2 A g_{(s)}

E_{(c e l l)} E_{(c e l l)}^{0} = 3.17 v o l t

M g^{2 +} + 2 e^{-} \to M g; E_{c e l l}^{\circ} (V) = - 2.37

C u^{2 +} + 2 e^{-} \to C u; E_{c e l l}^{\circ} = 0.34

M g_{(s)} + 2 A g_{(a q)}^{+} \to M g_{(a q)}^{2 +} + 2 A g_{(s)}

E_{M g^{2 +} / M g}^{0} = - 2.37 V

E_{A g^{+} / A g}^{0} = 0.80 V

[M g^{2 +}] = 0.001 M; [A g^{+}] = 0.0001 M

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