The equilibrium constant for the reaction will be- Cus - Cu2- aq- 2Cu-aq- ECu2-Cu- - 0-34 V ECu2- Cu- - 0-15 VGiven- log 3-72 - 0-571

The correct option is A 3.72× 10^ 6 lnK = nE0.0591 n= 2 E = 0.19 lnK = 2 × 0.190.0591 K = 3.72 × 10^ 6 #160 ...

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Byju's Answer

Standard XII

Chemistry

Equilibrium Constant from Nernst Equation

The equilibri...

Question

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: $log 3.72 = 0.571$ )

3.72 \times 10^{- 6}

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3.72 \times 10^{- 5}

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3.72 \times 10^{- 7}

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3.72 \times 10^{- 8}

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Solution

The correct option is A $3.72 \times 10^{- 6}$
$l n K$ = $\frac{n E}{0.0591}$
$n = 2$
$E$ = $- 0.19$
$l n K$ = $\frac{2 \times - 0.19}{0.0591}$
$K = 3.72 \times 10^{- 6}$
A is correct answer

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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

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