In the concentration cell shown, Ecell = +0.0296 V at 298 K. What is the concentration at the cathode?
The concentration of the anode is .01M Fe2+
In the electrochemical cell shown in the question above, what concentration of Fe2+ at the cathode would be required to reach a cell potential of 0.05 V at 298 K?
Any help greatly appreciated
- hcbiochemLv 72 months ago
The voltage of a concentration cell is given as:
E = -0.0592/n log [Fe2+]anode/[Fe2+]cathode
0.0296 V = -0.0592 / 2 log 0.01/[Fe2+]
[Fe2+] = 0.10 M
Use the same equation,
0.050 V = -0.0592/2 log (0.01/[Fe2+]
[Fe2+] = 0.50 M