Question

for a cell involving one electron Ecell=0.059v at 298k, the equilibrium constant for the cell reaction is​

Answer 1

Answer:

1.0x10^10

Explanation:

Ecell=$E^{0}$cell-0.059/nlogQ................(basic formula

now at equillibrium Q=Keq

and

Ecell=0

0=$E^{0}$cell-0.059/1   logKeq

log Keq=$\frac{E^{0} cell}{0.059}$=0.59/0.059=10

Keq=10^10=1X10^10