Question

43
The dissociation constant for [Ag(NH3)2]* into Ag* and
NH3 is 10^-13 at 298 K.if E°(Ag*/Ag)=0.8 V, then E° for the half cell [Ag(NH3)2l* + e- =Ag+ 2NH3 will be...???
(A) 0.33 V (B) - 0.33 V (C) -0.033 V (D) 0.033

please give me right answer with understanding solution very fast ..I give you brainlis thank you..​

Answer 1

Given that,

The dissociation constant for $[Ag(NH_{3})_{2}]^{+}$ into Ag⁺ and NH₃ is $6\times10^{-13}$ at 298 K.

$k_{c}=6\times10^{-13}$

Anode reaction is

$E^{o}=0.8\ V$ for $Ag^{+}+e\Rightarrow Ag$

Cathode reaction is

$(Ag(NH_{3})_{2})^{+}+e\Rightarrow Ag+ 2NH_{3}$

We need to calculate the value of $E_{cell}^{o}$

Using formula of $E_{cell}$

$E_{cell}=E_{cell}^{o}+\dfrac{0.0599}{n}\times\log k_{c}$

Put the value into the formula

$0=E_{cell}^{o}+\dfrac{0.059}{1}\times\log(10^{-13})$

$E_{cell}^{o}=-\dfrac{0.059}{1}\times\log(10^{-13})$

$E_{cell}^{o}=-0.767\ V$

We need to calculate the value of E⁰ for the half cell

Using formula of $E^{o}_{cell}$

$E^{o}_{cell}=E^{o}_{c}-E^{o}_{a}$

Put the value into the formula

$-0.767=E^{o}_{c}-0.8$

$E^{o}_{c}=-0.767+0.8$

$E^{o}_{c}=0.033\ V$

Hence, The value of E⁰ for the half cell is 0.033 V.

(D) is correct option.