18. Find the equilibrium constant for the reaction, Cu(s)+ Cu2+(aq) ➡ 2Cu+(aq). E Cu2+/Cu=0.34, E Cu2+/Cu+=0.15 (Given log 3.72=5.71)

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Q. The equilibrium constant for the reaction will be:

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

E_{C u^{2 +} / C u}^{\circ} = 0.34 V E_{C u^{2 +} / C u^{+}}^{\circ} = 0.15 V

(Given:

log 3.72 = 0.571

)

Q. Consider the following half cells

C u_{(s)} \to C u_{(a q)}^{2 +} + 2 e^{-}; E^{o} = - 0.325 V

C u_{(a q)}^{+} + e^{-} \to C u_{(s)}; E^{o} = 0.50 V

The

E^{o}

for the reaction

2 C u_{(a q)}^{+} \to C u_{(a q)}^{2 +} + C u_{(s)}

Q. Find

K_{c}

for the reaction:

C u^{2 +} (a q) + S n^{2 +} (a q) ⟶ S n^{4 +} (a q) + C u (s)

25^{\circ} C

(Given:

E_{C u^{2 +} | C u}^{\circ} = 0.34 V; E_{S n^{4 +} | S n^{2 +}}^{\circ} = 0.15 V

)

Q. The electrode potentials for

C u^{2 +} (a q) + e^{-} \to C u^{+} (a q) and C u^{2 +} (a q) + e^{-} \to C u (s) are + 0.15 V and + 0.50 V

respectively. The value of

E_{C u^{2 +} / C u}^{0}

will be:

Given the half cell reactions:
$C u^{+} (a q) + e^{-} \to C u (s); E^{0} = + 0.52 V C u^{2 +} (a q) + e^{-} \to C u^{+} (a q); E^{0} = + 0.165 V$
The equilibrium constant for the disproportionation reaction at 298 K is:
$2 C u^{+} (a q) \to C u (s) + C u^{2 +} (a q)$

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