... in the silver half cell. ==> Repeat this for each of the listed concentrations (0.1 M, 0.01 M, 0.001M, 0.0001 M, 0.003 M, 0.00033 M) of silver nitrate solutions, by diluting each solution accordingly and replacing the salt bridge each time. Calculations: (1) log([Ag+ (aq)]/ M - see table (2) Electrode potential of the silver half cell for each concentration: * 0.1 M silver nitrate solution: Taking E? of copper half cell to be actual value of: + 0.34 V: E? cell = E?positive - E?negative (or E rgt -E lft) + 0.27 = e - (+ 0.34) ? E? of silver electrode = + 0.61 V * 0.01 M silver nitrate solution: Taking E? of copper half cell to be actual value of: + 0.34 V: E? cell = E?positive - E?negative (or E rgt -E lft) + 0.26 = e - (+ 0.34) ? E? of silver electrode = + 0.60 V * 0.0033 M silver nitrate solution: Taking E? of ...

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