Calculate the cell potential for the following reaction as written at 51 degrees C, given that [Zn2+] = 0.837 M and [Sn2+] = 0.0180 M.

Question

Zn(s) + Sn^2+(aq) = Zn^2+(aq)+Sn(s)
E = __ V

Reduction Half-Reaction Standard Potential Ered° (V)
Zn2+(aq) + 2e– → Zn(s) –0.76
Sn2+(aq) + 2e– → Sn(s) –0.14

I was hoping someone could tell me if this is correct:
-0.76 - 0.14 = -0.90 V
(-0.9) - (8.314 * 324 K)/(2 * 96485) * ln[0.837/0.0180]
= (-0.9) - 0.053596262
= -0.95359 V?

Answer 1

It isn't correct although I didn't finish working the problem.
Zn ==> Zn^2+ + 2e = +0.76
Sn^2+ + 2e ==> Sn = -=0.14
Therefore, for the rxn
Zn + Sn^2+ ==> Sn + Zn^2+ E = 0.76+(-0.14) = 0.62 v.
I changed the Zn to an oxidation, left the Sn as a reduction, added the oxidn half to the redn half and added the two corresponding E values to find Ecell.
I think the rest is ok if you change the 0.9 to 0.62. By the way reporting to 5 significant figures is ridiculous.