QJEEM19Che70b: JEE Main 2019 Chemistry - Class 12 JEEM19Che70b Solution: Given the equilibrium constant: K C of the reaction : Cu(s) +…

Question

Verified Solution • 2025–26 Session moderate

Given the equilibrium constant:

K_C of the reaction :

Cu(s) + 2Ag⁺ (aq)

\to

Cu²⁺ (aq) + 2Ag(s) is

10

\times

10¹⁵, calculate the E

_{cell}^0

of this reaciton at 298 K

[2.303

{{RT} \over F}

at 298 K = 0.059V]

Accepted Answer

We know,

\Delta

G^o = -RTln(K_C) ....(1)

Also

\Delta

G^o = -nF

E_{cell}^o

....(2)

\therefore

-nF

E_{cell}^o

= -RTln(K_C)

\Rightarrow

E_{cell}^o

=

{{RT} \over {nF}}\ln \left( {{K_C}} \right)

=

2.303{{RT} \over {nF}}\log \left( {{K_C}} \right)

=

{{0.059} \over 2}\log \left( {10 \times {{10}^{15}}} \right)

( n = no of electron transferred = 2 )

= 0.059

\times

{{16} \over 2}

= 0.059

\times

8

= 0.472 V

Step-by-Step Verified Solution