Yes , Thanks. Then why students are even made to calculate them if that value of emf is not correct in practical circumstances where we geniunely need to use the cell ?
If no opposing voltage is applied , but voltmeter has very high resistance (which generally it has) , then also current would be very less , then do we obtain reading as -ΔG/nF ?
So if we were to connect a voltmeter to two electrodes of a galvanic cell , the reading it show experimentally is different from what we calculate from E = -ΔG/nF ?
been though it carefully several times , understood most of the things . But for galvanic cell part , can you please elaborate how reaction happen reversibly at constant Pressure ? I mean for ideal gas reaction , in reversible pathway we define 3 steps in which step 1 and 3 are responsible for...
Yes right , no problem with that. I just want to ask that when ΔH for a reaction is given, it is equal to heat released/absorbed when reaction happen in a closed container at constant external pressure and temperature , right ?
That's what I am assuming , Is there any other way of maintaining constant pressure irreversibly ? When it is mentioned for example that combustion of methane happens at 1atm , that means external pressure is 1atm throughout (that's how I have understood it till now )
but In actual irreversible path at constant P and T , Q = ΔH ? My reasoning : ΔH = ΔU + ΔPV , applying it for initial and final states (internal pressure is same initially and finally only , while external pressure is constant throughout) , ΔH = pΔV = -W = Q .
Yes got it , as pressure is not constant throughout we can't equate q = ΔH , can you confirm that for actual reaction taking place (not this reversible pathway) ΔS of surroundings = -ΔH/T but ΔS of system not equal to -ΔH/T (indeed equal to what it comes out in reversible pathway)
got it , so we are going from pure reactants (initially in a cylinder containing pure reactants connected to reactor by SPM) to pure products (in cylinders connected to reactor on the other side) ?