An electrochemical cell is based on the following two half-reactions:

Ox: Sn(s)--> Sn^2+(aq, 1.70 M)+ 2e
Red: ClO2(g, 0.220 atm)+e --> ClO2^-(aq, 1.80 M)

Compute the cell potential at 25 C.

I think I'm supposed to use the Nernst Equation for this, but I do not know how to set up the equation for Q.

To compute the cell potential at 25°C using the Nernst Equation, you need to determine the reaction quotient (Q). The reaction quotient is similar to the equilibrium constant (K), but it is calculated using the concentrations (or activities) of the species involved in the electrochemical cell reaction.

In this case, the Nernst Equation is given as:

E = E° - (RT/nF) * ln(Q)

where:
E is the cell potential at 25°C,
E° is the standard cell potential at 25°C,
R is the ideal gas constant (8.314 J/(mol·K)),
T is the temperature in Kelvin (25 + 273 = 298 K for 25°C),
n is the number of electrons transferred in the balanced equation, and
F is Faraday's constant (96485 C/mol).

To set up the equation for Q, you need to write the balanced equation for the overall cell reaction. The given half-reactions are:

Ox: Sn(s) → Sn2+(aq, 1.70 M) + 2e
Red: ClO2(g, 0.220 atm) + e → ClO2-(aq, 1.80 M)

To balance the overall reaction, you need to multiply the half-reactions by appropriate coefficients to ensure the same number of electrons are transferred in each half-reaction. In this case, you need to multiply the oxidation half-reaction by 1 and the reduction half-reaction by 2 to balance the electrons:

Ox: Sn(s) → Sn2+(aq, 1.70 M) + 2e
Red: 2ClO2(g, 0.220 atm) + 2e → 2ClO2-(aq, 1.80 M)

Now, add the balanced half-reactions together to obtain the overall reaction:

Sn(s) + 2ClO2(g, 0.220 atm) → Sn2+(aq, 1.70 M) + 2ClO2-(aq, 1.80 M)

The Q for this reaction will be the product of the concentrations (or activities) of the products divided by the product of the concentrations (or activities) of the reactants, each raised to the power equal to their stoichiometric coefficient.

So, for the overall reaction:

Q = [Sn2+(aq)] * [ClO2-(aq)]^2 / [Sn(s)] * [ClO2(g)]^2

Substitute the given concentrations and partial pressure into the equation above. Note that the concentration of Sn(s) is not included because its activity is considered as 1. Also, partial pressure is used for ClO2(g) as it is a gas:

Q = (1.70 M) * (1.80 M)^2 / (0.220 atm)^2

Now, you can proceed to substitute the calculated Q value into the Nernst Equation along with the other given values to compute the cell potential (E).

Calculate Eo based on the reduction potential.

E = Eo-(0.0592/n)log (red form/ox form)
For Sn, Eo (written in reduced form)is
Sn^+2(1.70M) + 2e = Sn Eo = whatever. you look it up.
Substitute into the Nernst equation I have above. n = 2 in the equation and the log Q part is (Sn)/(Sn^+2). (Sn) = 1 (that is the Sn in the pure state) and (Sn^+2) is 1.70 in the problem. Calculate E for Sn that way, THEN TURN the equation around to put it in the oxidized form as it appears in your post, and reverse the sign.
Do the same for ClO2 + e ==> ClO2^- BUT don't reverse the reaction and don't reverse the sign. Then add the Eox + Ered to obtain Ecell.