Use tabulated half-cell potentials to calculate δg∘rxn for each of the following reactions at 25 ∘c. 2fe3+(aq)+3sn(s)→2fe(s)+3sn2+(aq)

- for the reaction is -58 kJ/mol.

Solution:- Oxidation half reaction taking place at the anode is--

= 0.14 V

Reduction half equation taking place at the cathode is---

= -0.04V

for the cell =

for the cell = -0.04V + 0.14V = 0.10V

Now we could easily calculate by using the below formula--

= -nF

where n is the number of electrons transferred in the over all reaction. Looking at two half equations the value of n is 6.

F is Farady constant and it's value is 96500 C/mol.

Plugging in the values in the formula...

= -(6)(96500)(0.10)

= -57900 J

Since, 1000 J = 1 kJ

So, = -58 kJ/mol

the ideal gas law equation is as follows

pv = nrt

where p - pressure

v - volume

n - number of moles

r - universal gas constant

t - temperature

so if the pressure, volume and temperature are already known

we are left with n and r

since r is the universal gas constant that has a known fixed value then r too is known

so we are left with 'n'

once we know temperature volume and pressure

we can find the number of moles of gas present using the ideal gas law equation

this is not my own answer, this is an answer for the same question by the user blahblahmali. i do not claim ownership of this account.