1e03-tut10-a0

# 1e03-tut10-a0 - CHEM 1E03 TUTORIAL#10 Solutions Page 1 Note...

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Unformatted text preview: CHEM 1E03 TUTORIAL #10 Solutions _____________________________________________________________________ Page 1 Note: the current textbook uses E o cell = E o red (cathode) - E o red (anode) while other books use E o cell = E o red (cathode) + E o ox (anode). The two formulae are equivalent. 1. Use the E o red table (last page) to answer the following problems. a) Calculate the standard cell potential, E o cell , for: Sn(s) + Cu 2+ (aq) Sn 2+ (aq) + Cu(s) Sn 2+ + 2 e- Sn E o red = 0.14 V Cu 2+ + 2 e- Cu E o red = + 0.34 V Sn + Cu 2+ Sn 2+ + Cu E o cell = E o red (Cu) - E o red (Sn) = + 0.48 V b) Decide whether Cl 2 (g) (bubbled through an acid solution) will oxidize Mn 2+ (aq) to MnO 4- (aq) under standard conditions. MnO 4- (aq) + 8 H + (aq) + 5 e- Mn 2+ (aq) + 4 H 2 O E o red = + 1.51 V Cl 2 + 2 e- 2Cl- (aq) E o red = + 1.36 V The E o red values show that MnO 4- is a stronger oxidizing agent than Cl 2 (g). Therefore, Cl 2 (g) will not oxidize Mn 2+ in acid solution under standard conditions. 2. Using the E o red table (last page), select a suitable substance for each of the following transformations (assume that all soluble substances are present in 1.00 M concentrations): a) a reducing agent capable of reducing Ag + to Ag but not Cu 2+ to Cu....
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## This document was uploaded on 03/29/2012.

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1e03-tut10-a0 - CHEM 1E03 TUTORIAL#10 Solutions Page 1 Note...

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