Question

n = 6 and E° = 0.011v for a redox reaction with Q = 7.03 x 10-1. Calculate E. Assume T = 298.15K.

So I did E=E°-(RT/nF)lnQ
Since the temp is 298.15, I substituted0.02569 for (RT/F) and got
0.011v-(0.0259/6)lnQ
and got -.002355

Is that correct?

Answer 1

What did you do with the 2.303? That's
2.303*R*T/F. I suppose you don't need that IF you used ln and not log. But isn't log (or ln) of 0.703 a negative number and that times a negative is a positive.
RT/F IS 0.0259 but you must multiply by 2.303 and divide by 6.
I have 0.011 + 0.00151 = ??

Also, look at your next to last line where you have 0.011 - (0.0259/6)lnQ, that should be 0.02569.