1. What is the value of the equilibrium constant for the reaction occurring in the cell shown below at 298 K?
Ag(s) | Ag+(aq, 1 M) || Cu2+(aq, 1 M) | Cu(s)
Report your answer to four significant figures in scientific notation (1.234E-3 or 1.234E3).
Can anyone explain the steps for me and work me through it i have done it 5 times already and i just cannot get it.
Look up Ag reduction potential. Since the concn is 1M, the terms -(0.0592/n)log(1/Ag^+) is zero and Ehalfcell is just Eo.
The same applies to the Cu half cell. Add the Ag==>Ag^+ half cell which in my text is -0.799 to the Cu^+2 ==> Cu half cell (which in my text is +0.337) to obtain the Ecell.
nEF = -RTlnK
n = 2, E is about -0.46 but you need to look up the numbers and use those from your text or notes, F is 96,485, T is 298 and solve for K. Since the cell voltage is -, it isn't spontaneous and we would expect k to be small number.