Question

calculate the cell potential of the following cell at 298k
Ag|AgNO3 [0.01M] || AgNO3 [1.0M] | Ag​

Answer 1

Answer:

cell potential will be +0.118v

Explanation:

since in the question it is given all the reaction is taking place in the standard temperature i.e.25°C

so is the formula .

E=E°-0.059/n ×log10(product)/(reactant).

since cathode and anode are same so E° will be zero. i.e.E°=0. same number of electron are transferring so n value will be 1 cause it will be get cancelled .

E=0-0.059/1×log10(10^-2)/1

E= -0.059×-2×log10× 10. (log 10×10 =1)

E = -0.059×-2

E =+0.118 v