# ΔG of an Electrochemical Cell The change in Gibbs Energy ΔG is the maximum non-PV work* that can be obtained from a chemical reaction at constant T and.

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ΔG of an Electrochemical Cell The change in Gibbs Energy ΔG is the maximum non-PV work* that can be obtained from a chemical reaction at constant T and P: ΔG = w max For an electrochemical cell, w max = -nFE ΔG = -nFE E = E° - RT ln Q nF *or minimum energy needed for the nonspontaneous reaction

The Nernst Equation – Example 1 E = E° - 0.0592 V log Q n What is the emf of the following cell at 298 K? This is called a concentration cell. Zn 2+ (aq, 0.500 M)  Zn 2+ (aq, 0.150 M) Q = 0.150 0.500 E = 0 - 0.0592 log.300 = 0.015 V 2 = 0.300 M

The Nernst Equation – Example 2 Find the solubility product for silver thiocyanate at 25°C, given AgSCN(s) + e -  Ag(s) + SCN - (aq) E°(25°C) = 0.0895 V and using the standard reduction potential for Ag + found in Appendix E. Answer: K SP = 1.0 x 10 -12

The Nernst Equation – Example 3 The concentration of potassium ions in the intracellular fluid (ICF) is 135 mM. In the extracellular fluid (ECF), it is 4 mM. What potential must be applied to the cell to keep the K + from diffusing out of the cell? Answer: The ECF must have a potential of 94 mV relative to the ICF in order to keep the K + from diffusing out of the cell.

The Nernst Equation – Example 4 In a galvanic cell with two hydrogen electrodes, both at 298 K, the first electrode has P H2 = 1.00 bar and an unknown concentration of H +. The second hydrogen electrode is the SHE. E cell is found to be 0.211 V and the electron flow is from the first electrode to the second. What is the pH of the solution in the first hydrogen electrode? Answer: pH = 3.56

The Nernst Equation – Example 5 In a galvanic cell at 25°C, the cathode half- reaction is Ag + (aq, 1M) + e -  Ag(s). The anode is a hydrogen electrode with P H2 = 1.00 bar. It is in a buffer containing 0.10 M benzoic acid and 0.050 M sodium benzoate. E cell (25°C) = 1.030 V. What is the pK a of the benzoic acid? Answer: pK a = 4.20

The Nernst Equation – Example 6 4Fe 2+ (aq) + O 2 (g)+ 4H + (aq)  4Fe 3+ (aq) +2H 2 O(l) What is the cell emf at 25°C when [Fe 2+ ]=1.3M, [Fe 3+ ]=0.010M, P O2 = 0.5 bar, and pH=3.50? E = Eº -.0592 log Q …need n, Q, and Eº n Answer: E = 0.37V

The Nernst Equation – Example 7 2Fe 3+ (aq) + H 2 (g)  2Fe 2+ (aq) + 2H + (aq) What is the cell emf at 25°C when [Fe 2+ ]=0.0010M, [Fe 3+ ]=2.50M, P H2 = 0.85 bar, and pH=5.00? E = Eº -.0592 log Q …need n, Q, and Eº n Answer: E = 1.27V

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