What would the standard cell potential in volts be for

Mn(s) + 2Ce4+(aq) -> Mn2+(aq) + 2Ce3+(aq)?

We don't do homework but we can help you understand. What about the problem do you not understand? Do you know to look up the cell potential for the Mn to Mn^+2 and add it to the cell potential for the Ce^+4 ==>Ce^+3?

That should give you the cell potential for the reaction. Post your work if you get stuck.

The two values I got for those cell potentials were 1.61 (Ce^4+ + 2e- -> Ce^3+) and -1.18 for Mn^2+ + 2e^- -> Mn. Which equals 0.43V.

I wasn't sure if I double the value for Ce^4 ... as they have 2's at front of them?

No, you do not double the value for Ce^+4. Sorry it took me so long to get back to you.

To determine the standard cell potential, you can use the Nernst equation, which relates the cell potential to the standard cell potential and the concentrations of the species involved.

The standard cell potential, E°cell, is a measure of the tendency of a redox reaction to occur. It is given by the difference in the standard electrode potentials, E°, for the reduction half-reaction and the oxidation half-reaction.

In this case, the half-reactions involved are:

Reduction half-reaction: Ce4+(aq) + e- -> Ce3+(aq)

The standard electrode potential for this half-reaction can be found in a table of standard reduction potentials. Let's assume that the value is E°reduction = +1.44 V.

Oxidation half-reaction: Mn(s) -> Mn2+(aq) + 2e-

The standard electrode potential for this half-reaction can also be found in a table of standard reduction potentials. Let's assume that the value is E°oxidation = -1.18 V.

To find the standard cell potential, E°cell, you subtract the standard electrode potential of the oxidation half-reaction from the standard electrode potential of the reduction half-reaction:

E°cell = E°reduction - E°oxidation
E°cell = (+1.44 V) - (-1.18 V)
E°cell = +2.62 V

Therefore, the standard cell potential for the given reaction is +2.62 volts.