if zn2+/zn electrode is diluted 1000 times then the change in electrode potential is Share with your friends Share 1 Ravjot Kaur answered this Dear studentE =Eo-0.0592log[Zn2+]E=Eo-0.0592log[Zn2+]1000E-E0 =0.0059 Regards -13 View Full Answer Shubham Raj answered this E=E• - 0.059/2 log [zn2+] Then, .Change in e, Eʿ=E• - 0.059/2 log [zn2+/1000] -9
