consider the cell

Pt(s)I H2 (g. 0.100 bar)I H+ (aq, pH= 2.54)IICl^- ( aq, 0.200 MI Hg2Cl2(S)IHg (l)IPt(s)

a. write a reduction reaction and nernst equation for each half- cell. For the Hg2Cl2 half reaction, E^0 = 0.268 V

b. Find E for the net cell reaction and state whether reduction will occur at the left ot right hand electrodes

I'm sure you know the equation for the H2.

2H^+ + 2e ==> H2(g)
E=Eo-(0.059/n)*log(pH2/H^+) but you need to add the partial pressures and concentrations to that. Here is a site you can read about the SCE.
http://en.wikipedia.org/wiki/Saturated_calomel_electrode

To write reduction reactions and Nernst equations for each half-cell in the given cell, we need to have the half-cell reactions for each electrode. Let's analyze the given information.

The cell notation is: Pt(s)|H2(g, 0.100 bar)|H+(aq, pH = 2.54)||Hg2Cl2(s)|Hg(l)|Pt(s)

a. Half-Cell Reactions and Nernst Equations:

1. Reduction Half-Cell: H+ (aq) + e- → ½ H2(g)
The reduction half-cell equation can be written as: 2H+ (aq) + 2e- → H2(g)
The Nernst equation for this half-cell can be written as: E1 = E1° - (RT/nF) ln([H+]^2)
Here, E1 is the cell potential, E1° is the standard reduction potential, R is the ideal gas constant (8.314 J/(mol·K)), T is the temperature in Kelvin, n is the number of electrons transferred (in this case, n = 2), F is the Faraday constant (96,485 C/mol), and [H+] is the concentration of H+ ions.

2. Reduction Half-Cell: Hg2Cl2(s) + 2e- → 2Hg(l)
The Nernst equation for this half-cell can be written as: E2 = E2° - (RT/nF) ln([Hg2Cl2]/[Hg]^2)
Here, E2 is the cell potential, E2° is the standard reduction potential, [Hg2Cl2] is the concentration of Hg2Cl2, and [Hg] is the concentration of Hg.

b. To find E for the net cell reaction, we can use the Nernst equation for each half-cell and subtract the two equations:

E_net = E2 - E1

Given E2° = 0.268 V and pH = 2.54, we can plug in the values into the equations to calculate the net cell potential.

Now, let's solve for E_net and determine whether reduction will occur at the left or right-hand electrodes based on the sign of E_net.