posted by Anon .
Calculate the cell potential for the following reaction as written at 59 ¡ãC, given that [Cr2+] = 0.763 M and [Sn2+] = 0.0170 M. Standard reduction potentials:
Sn2+(aq) + 2e¨C ¡ú Sn(s) E=¨C0.14
Cr2+(aq) + 2e¨C ¡ú Cr(s) E=¨C0.91
Your numbers aren't right and there is nothing to show "as written".
Sn^2+ + 2e ==> Sn Eo = -0.14
Cr^2+ + 2e ==> Cr Eo = -0.91
Use E = Eo - (2.303*RT/n)log(Sn/Sn^2+), plug in R and T (59+273), Sn is 1 and Sn^2+ is 0.0170. Solve for E.
Do the same for Cr^2+. Then reverse the Cr^2+ rxn, change the sign for Cr and add E for Sn. That gives you Ecell