Question 18.PS.10: Using the Nernst Equation Consider this electrochemical reac......
Using the Nernst Equation
Consider this electrochemical reaction:
Zn(s) + Ni^{2+}(aq) → Zn^{2+}(aq) + Ni(s)
The standard cell potential E^\circ_{cell} = 0.51 V. Find the voltage if the Ni^{2+} concentration is 5.0 M and the Zn^{2+} concentration is 0.050 M.
0.57 V
Strategy and Explanation We use the Nernst equation to solve the problem. Two moles of electrons are transferred from 1 mol Zn to 1 mol Ni^{2+}, giving n = 2. At 298 K,
E^\circ_{cell} = 0.51 V – \frac{0.0592 V}{2} \log \frac{(conc. Zn^{2+})}{(conc. Ni^{2+})}
E^\circ_{cell} = 0.51 V – \frac{0.0592 V}{2} \log (\frac{0.050}{5.0}) = 0.51 V – \frac{0.0592 V}{2} \log(10^{-2})
= 0.51 V – \frac{0.0592 V}{2} (- 2.00) = 0.57 V
Reasonable Answer Check The reactant concentration of Ni^{2+} is 5.0 M, larger than the standard-state value of 1.0 M, and the product concentration of Zn^{2+} is 0.050 M, smaller than the standard-state value. Each of these departures from standard-state conditions tends to make the voltage under these conditions slightly larger than the standard cell potential (E^\circ_{cell} = 0.51 V), and it is.