Question

Consider the cell described below at 263 K:

Fe | Fe²⁺ (0.865 M) || Cd²⁺ (0.941 M) | Cd

Given EoCd2+→Cd = -0.403 V, EoFe2+→Fe = -0.441 V. Calculate the cell potential after the reaction has operated long enough for the Fe2+ to have changed by 0.391 mol/L.

Answer 1

you can solve this using Ecell = Enaught - (RT/nF)ln(Q)

Where:

R=8.314

T=Temp in Kelvin 263

n= number of moles of electrons (2 for you)

F= faradays constant 96485 C/mole electrons

Q= reaction quotient ie. Q= [products]/[reactants]

Enaught= (-.403) + (.441)= .038 V

If Fe2+ changes by .391mol/L then you will set up your Q like this and solve for Ecell

X= .393

Ecell= Enaught - (RT/nF)ln ([.941-x]] / [.865+x])

where ln is the natural log.

Pretty sure that will work if it doesnt you may need to flip the signs from .941-x to .865+x but that should give you the answer