Ecell=Ecathode-Eanode
=-0.4-(-0.9)
Ecell=0.5
nernst equation:
=ecell-0.059/n log product/react.
=0.5-0.059/2 log 0.1/0.1
=0.5-0.0295×0
=0.5
Calculate the EMF of the cell CR|Cr3+(0.1M)||Fe2+(0.1M)|Fe
Cell reaction- 2Cr(s)+3Fe2+(aq)->2Cr3+(aq)+3Fe(s)
3 answers
How fool completely wrong answer you confused more than before you idiot
If u don't know how to solve don't post the solution bloody hell
If u don't know how to solve don't post the solution bloody hell
E°cell= E°cathode-E°anode
E°cell=-0.44-(-0.74) = 0.30V
Ecell= E°cell - 0.0591/n log [Cr3+]^2/[Fe2+]^3
Ecell= 0.30- 0.0591/6 log[0.1]^2/[0.1]^3
Ecell= 0.30 - 0.0591/6 log 10
Ecell= 0.30 - 0.0591= 0.2409V
E°cell=-0.44-(-0.74) = 0.30V
Ecell= E°cell - 0.0591/n log [Cr3+]^2/[Fe2+]^3
Ecell= 0.30- 0.0591/6 log[0.1]^2/[0.1]^3
Ecell= 0.30 - 0.0591/6 log 10
Ecell= 0.30 - 0.0591= 0.2409V