18-13. Calculate the potential of a copper cloctrode immersed in (a) 0.0380 M Cu(NO_(3))_(2) (b) 0.0650 M in NaCl and saturated with CuCl.

To calculate the potential of a copper electrode immersed in different solutions, we can use the Nernst equation. The Nernst equation relates the electrode potential to the concentration of the ions in solution.<br /><br />The general form of the Nernst equation is:<br /><br />\[ E = E^0 - \frac{0.0592}{n} \log Q \]<br /><br />where:<br />- \( E \) is the electrode potential,<br />- \( E^0 \) is the standard electrode potential,<br />- \( n \) is the number of moles of electrons transferred in the half-reaction,<br />- \( Q \) is the reaction quotient.<br /><br />For a copper electrode, the half-reaction is:<br /><br />\[ Cu+} + 2e^- \rightarrow Cu \]<br /><br />So, \( n = 2 \).<br /><br />(a) For a 0.0380 M $Cu(NO_{3})_{2}$ solution:<br /><br />The reaction quotient \( Q \) is given by:<br /><br />\[ Q = [Cu^{2+}] = 0.0380 \, M \]<br /><br />Substituting the values into the Nernst equation:<br /><br />\[ E = E^0 - \frac{0.0592}{2} \log(0.0380) \]<br /><br />Using the standard electrode potential for copper, \( E^0 = +0.34 \, V \):<br /><br />\[ E = 0.34 - \frac{0.0592}{2} \log(0.0380) \]<br /><br />\[ E = 0.34 - 0.0296 \log(0.0380) \]<br /><br />\[ E = 0.34 - 0.0296(-1.42) \]<br /><br />\[ E = 0.34 + 0.0420 \]<br /><br />\[ E = 0.382 \, V \]<br /><br />(b) For a 0.0650 M NaCl solution saturated with CuCl:<br /><br />The reaction quotient \( Q \) is given by:<br /><br />\[ Q = [Cu^{2+}] = 0.0650 \, M \]<br /><br />Substituting the values into the Nernst equation:<br /><br />\[ E = E^0 - \frac{0.0592}{2} \log(0.0650) \]<br /><br />Using the standard electrode potential for copper, \( E^0 = +0.34 \, V \):<br /><br />\[ E = 0.34 - \frac{0.0592}{2} \log(0.0650) \]<br /><br />\[ E = 0.34 - 0.0296 \log(0.0650) \]<br /><br />\[ E = 0.34 - 0.0296(-1.81) \]<br /><br />\[ E = 0.34 + 0.0537 \]<br /><br />\[ E = 0.3937 \, V \]<br /><br />Therefore, the potential of the copper electrode in:<br />(a) 0.0380 M $Cu(NO_{3})_{2}$ is approximately 0.382 V.<br />(b) 0.0650 M NaCl saturated with CuCl is approximately 0.3937 V.