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.
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.
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