Calculate the emf of the following cell at 250CZn/Zn 2+(0.04)//Cd 2+ (

1.	Calculate the emf of the following cell at 250CZn/Zn 2+(0.04)//Cd 2+ (0.2) /CdGiven E0 Zn=-0.76Vand E0 Cd=-0.403V
Answer» *Correct* *option* (c) E=+0.36 - (0.059/2)log 2×102 Explanation: E=E - 0.059/n *log* [oxdised *state]/[REDUCED* *state]* cell n= number of electrons taken PART in the reaction is equal to 2 E=[E 2+E *zn/zn cd/cd] -0.059log 0.004/0.2 = [ 0.763 - 0.403] - 0.059/2 log* 1/50 =0.36 - 0.059/2 *log* 2× -2 10 *please* *mark* me *BRAINLIEST*

Calculate the emf of the following cell at 250CZn/Zn 2+(0.04)//Cd 2+ (0.2) /CdGiven E0 Zn=-0.76Vand E0 Cd=-0.403V

Answer»

Correct option (c) E=+0.36 - (0.059/2)log 2×102

Explanation:

E=E - 0.059/n log [oxdised state]/[REDUCED state]

cell

n= number of electrons taken PART in the reaction is equal to 2

E=[E 2+E

zn/zn cd/cd] -0.059log 0.004/0.2

= [ 0.763 - 0.403] - 0.059/2 log 1/50

=0.36 - 0.059/2 log 2× -2