Question

3. Let's say that we didn't have access to a standard H , electrode. Instead, we made our half-cell potential measurements relative to the tin half-cell. (a) What would the E be for the half-reaction: Sn(s) Sn?(aq) + 2e"? Explain (b) What would the I", be for the oxidation half-reactions of (1) Cu(s). (2) Ni(s), and (3) Ag(s)? (s) What would the E be for the Cu-Ag electrochemical cell relative to the tin electrode? Provide both half-cell reactions, potentials relative to tin, and the net cell reaction. (d) We found a standard H/H, electrode and redid the calculations necessary for part of this question. What would be now? (e) Compare the answers you got for parts c and d. What can you conclude about Ell?

Answer 1

Answes ) az khat would the Eo be the half reaction Sals) - snt (aq) + 28? here from reduction potentials then here we get. :. Soltre 2 sn (s) - 0:13 By inverting this, then the potentials becomes Positive . Enew = -013-013 6) oxidation half neactions of ci) culs), W Nils) & (3) Agls) . from the given Reactions. Eux font here cox -) Ag+ + e 2 Ag(s) +0.1996 - inverted diodation Potential a -0.1996 > Cu2+ + 2e e culs) + 0.387 inrated Coundations . Potential = -0,387 -) Nidttae - 2 Nils) -0.25 inverted Oxidation potential = -0.25 = 0.25v - Cuett

(c) so be for the cu-Ag electrochemical cell relative to the tin election. Ecell Cag) = 0.7996 +0.13 S = 0.9296v for the Agt here given in as reference. E cell cou) -0.337+0:13. = 0.467 v cu given as reference. both half cell reactions, potentials. relative to tin, and the net cell reaction. d) standard H+ (He Electrode and nedid the calculation necessary for parte here i, e gocell will be the same, since their reference will have been cancelled e) By comparing the conclusion about the Poell. here from the previous answers and al are Similar, but since their reference will be Cancelled here total difference lies b/w cung Rotential difference hence solved