For the disproportionation reaction 2Cu-aq- Cus-Cu2-aq at 298 K- In K where K is the equilibrium constant is -10-1Nearest integer-Givrn- ECu2-Cu-o -0-16V- ECu-Cuo-0-52V- RTF-0-025

2Cu^+(aq)⇌ Cu(s)+Cu^2+(aq) Reduction potential of Cu^+/Cu couple is higher, so, nbsp; E_cell^0=E_Cu^+/Cu^0 E_Cu^2+/Cu^+^0 = 0.52 0.16 =0.36 V Δ G = nfE^0 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Effective Equilibrium Constant

For the dispr...

Question

For the disproportionation reaction $2 C u^{+} (a q) ⇌ C u (s) + C u^{2 +} (a q)$ at $298 K,$ In K (where K is the equilibrium constant) is ____ $\times 10^{- 1}$ (Nearest integer).
Givrn: $(E_{C u^{2 +} / C u^{+}}^{o} = 0.16 V; E_{C u^{+} / C u}^{o} = 0.52 V; \frac{R T}{F} = 0.025)$

144.0

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

144

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

144.00

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

$2 C u^{+} (a q) ⇌ C u (s) + C u^{2 +} (a q)$
Reduction potential of $C u^{+} / C u$ couple is higher, so,
$E_{c e l l}^{0} = E_{C u^{+} / C u}^{0} - E_{C u^{2 +} / C u^{+}}^{0}$

$= 0.52 - 0.16$

$= 0.36 V$

$Δ G = - n f E^{0}$
$Δ G = - R T l n K_{e q}$

$E_{c e l l}^{o} = \frac{R T}{n F} ln K_{e q}$

$0.36 = \frac{0.025}{1} ln K_{e q}$

ln $K_{e q} = 14.4$

ln $K_{e q} = 144 \times 10^{- 1}$

Suggest Corrections

Similar questions

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

is unstable in solution and undergoes simultaneous oxidation and reduction, according to the reaction:

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

choose correct

E^{o}

for above reaction if

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

and

E_{{C u}^{2 +} / {C u}^{+}}^{o} = 0.15 V

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. Given the equilibrium constant,

K_{c}

of the reaction :

C u (s) + 2 A g^{+} (a q) \to C u^{2 +} (a q) + 2 A g (s)

10 \times 10^{15}

, calculate the

E_{c e l l}^{o}

of this reaction at

298 K

[2.303 \frac{R T}{F} a t 298 K = 0.059 V]

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:
$2 C u^{+} (a q) \to C u (s) + C u^{2 +} (a q)$

Q. The equilibrium constant for the equation:

E^{0} = 0.46 V

at 298 K is:

C u + 2 A g^{+} (a q) ⟶ {C u}^{2 +} (a q) + 2 A g

Join BYJU'S Learning Program

For the disproportionation reaction 2Cu+(aq)⇌Cu(s)+Cu2+(aq) at 298 K, In K (where K is the equilibrium constant) is ____×10−1(Nearest integer). Givrn: (EoCu2+/Cu+=0.16V; EoCu+/Cu=0.52V;RTF=0.025)

2Cu+(aq)⇌Cu(s)+Cu2+(aq) Reduction potential of Cu+/Cu couple is higher, so, E0cell=E0Cu+/Cu−E0Cu2+/Cu+ =0.52−0.16 =0.36 V ΔG=−nfE0 ΔG=−RT lnKeq Eocell=RTnF lnKeq 0.36=0.0251ln Keq ln Keq=14.4 ln Keq=144×10−1

For the disproportionation reaction $2 C u^{+} (a q) ⇌ C u (s) + C u^{2 +} (a q)$ at $298 K,$ In K (where K is the equilibrium constant) is ____ $\times 10^{- 1}$ (Nearest integer).
Givrn: $(E_{C u^{2 +} / C u^{+}}^{o} = 0.16 V; E_{C u^{+} / C u}^{o} = 0.52 V; \frac{R T}{F} = 0.025)$