Question

Consider the reaction occurring at 25∘C.

A(s)+ B2+(aq)→ A   2+(aq)+ B(s) ΔGorxn=−11.0 kJ

Determine the value of E∘cell and K for the reaction and complete the table.

[B2+] (M)	[A2+] (M)	Q	Ecell	ΔGrxn
1.00	1.00
1.00	1.00×10−4
1.00×10−4	1.00
1.18×10−2	1.00

What is the value Ecell and K

Answer 1

R= 8.314 J.\/mol/c
T= 298 K
F= 96485 C
n= number of electrons involved in the oxidation reduction =2**

dG0 = -RTlnK
use this formula to get K
lnK = -{-11 x10^3 J]/8.314 x298 K=

K= 84.76

dG0= -nFE0
E0 = -{-11 x10^3}/96485 C x2 = 0.057 V
**************************************

dG = dG0 + RTlnQ use this to get dG
Ecell = E0cell - 0.0592/nlog Q to get Ecell
1.
Q = [A2+]/[B2+] = 1/1 = 1
lnQ = 0
dG = dG0 = -11 KJ
Ecell = E0cell - 0.0592/nlog Q
Ecell = E0cell =0.057 V
*************
2
same as above solve all
Q=1x 10^-4/1=1x 10^-4
lnQ=ln[ 1x 10^-4 ] = -9.2
dG= -11000 -RTlnQ= -11000-2477.6 [-9.2] =11794J

Ecell = E0cell - 0.0592/nlog Q
Ecell = 0.057 V -0.0296x{-4} =0.1754 V
**********************
3
Q= [1]/[1x10^-4] =10000
ln Q = 9.2
log Q= 4
dG = -11000 -RTlnQ =-33794 J
Ecell = E0cell - 0.0592/nlog Q
=0.057 V -0.0296x{4} =- 0.0614 V
***********************
Q = 1/1.18 x10^-2 =84.75
ln Q=4.44
log q =1.92
dG =-11000 -RTlnQ = -22000 j
Ecell = E0cell - 0.0592/nlog Q
= 0.000168 V
***********************
Thank you