On Spontaneity of the Redox Reactions
Question 1:


You are mixing the sign conventions for work from physics and chemistry. Sadly, they are not the same. In chemistry, the work done by the system on the surroundings is negative to signify energy leaving the system. Any transfer of energy (heat, work, etc.) from the system to the surroundings is denoted with a negative sign to ensure the energy change of the system is negative (energy is lost). Chemistry is concerned ultimately only with the system (usually the chemical reaction under investigation), and not the surroundings (apparatus, equipment, experimenter, laboratory, and the rest of the universe). 

Thus, for pressure-volume work:

$$W=-p\Delta V$$
$$\delta W=-pdV$$

We need this negative sign. The system does pressure-volume work on the surroundings when the system expands (think a piston in an automotive engine) so $dV&gt;0$. Without the negative sign, we would assume that the system gains energy and the surroundings lose energy when the system expands.

Thus, in the chemistry formulation of the first law (assuming only pressure-volume work) is:

$$dU=\delta Q + \delta W =TdS - pdV$$

In physics, some consideration is given to the surroundings (and in some cases the surroundings may be more interesting than the system, such as in the proper definition of entropy). Especially in the early days of physics, work leaving the system was useful in that it could be utilized. Thus, in physics $W=p\Delta V$ means work is being done by the system on the surroundings. Thus, the first law as formulated in physics is:

$$dU=\delta Q - \delta W = TdS -pdV$$

Note however, regardless of the sign convention in the definition of $W$ as a function of $p$ and $\Delta V$, we still end up with $$dU=TdS-pdV$$

The sign of $W_{\text{max}}$ in chemistry needs to be negative to follow the same conditions of chemistry. Thus, the equation in your chemistry textbook is correct. An equivalent definition in physics would be: 

$$W_{\text{max}}=nFE_{\text{cell}}$$
$$\Delta G_\text{cell}=-W_\text{max}=-nFE_\text{cell}$$

In chemistry the negative sign goes in the definition of work. In physics the negative sign goes in front of the work term. Don't mix and match conventions, and you will get consistent answers.


  Question 2:


$\Delta G = 0$ is the definition of equilibrium not the conditions for reversibility. A reaction can be reversible (and most electrochemical reactions are) and still be far from equilibrium at the initial conditions.


  Question 3:


Temperature is not required to be constant here. Also, note that the term $W_\text{max}$ refers to the maximum possible electrochemical work. Usually less electrochemical work is done because heat is also generated (which changes the temperature). Also, most electrochemical systems are not closed. Even a battery is not a closed system (by the strict physics definition). It's hard to maintain constant pressure in a closed system. 


  Question 4:


Technically the formulation of the first law that you used is explicitly for pressure-volume work done by closed gaseous systems. As most redox reactions occur in solution in open systems, that formulation is invalid for a different reason. 


  Question 5: 


Given that your initial equation is correct, then for a spontaneous reaction:

$$\Delta G &lt; 0$$
$$\therefore -nFE_\text{cell} &lt; 0$$
$$nFE_\text{cell} &gt; 0$$
$$E_\text{cell} &gt; 0$$
$$E_\text{red} - E_\text{ox} &gt; 0$$