Chp9_PExp9pt5

# Chp9_PExp9pt5 - Practice Example 9.5 2 2 1 The cell Zn Zn...

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Practice Example 9.5 The cell: Pt M Cu M Cu M Zn Zn 084 . 0 , 188 . 0 066 . 0 1 2 2 Pt is an inert electrode NO copper metal involved The Zn couple is the anode The Cu ion couple is the cathode 1. The Cell Reaction & Standard State Cell Potential 76 . 0 2 16 . 0 1 0 1 2 0 1 1 2 E e Zn Zn E Cu e Cu Note that the E 0 value is the reduction potential even though we note here that the Zn couple is undergoing oxidation. The net reaction is thus (note that we must balance the number of electrons): 2 1 2 2 2 Zn Cu Zn Cu n = 2 92 . 0 76 . 0 16 . 0 0 0 0 Anode Cath E E E 2. The Reaction Quotient     2 2 2 2 2 2 1 2 10 32 . 1 188 . 0 084 . 0 066 . 0 Cu Cu Zn Q note carefully the distinction of charges and exponents. 3. The Cell Potential. We use the Nernst Equation since we require the cell potential at the specific concentrations specified above.

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