(Solved): Please help with Part C MISSED THIS? Read Section 20.61 A voltaic cell consists of a Pb/Pb2+ half-ce ...

MISSED THIS? Read Section 20.61 A voltaic cell consists of a $\mathrm{Pb}/{\mathrm{Pb}}^{2+}$ half-cell and $\mathrm{a}\mathrm{Cu}/{\mathrm{Cu}}^{2+}$ half-cell at 25 What is the initial cell potential? ${}^{?}\mathrm{C}$. The initial concentrations of ${\mathrm{Pb}}^{2+}$ and ${\mathrm{Cu}}^{2+}$ are $0.0530\mathrm{M}$ and $1.60\mathrm{M}$, Express your answer using two significant figures. respectively. $E_{\text{cell }}=E_{\text{cell }}^{?}?\frac{0.0592}{n}Y\log Q$ where $E_{\text{cell }}$ is the cell potential in volts, $E_{\text{cell }}^{?}$ is the standard cell potential in volts, $n$ is the number of moles of electrons transferred in the reaction, and $Q$ is the reaction quotient. Inserting the values into the equation gives $E_{\text{cell }}=0.47\mathrm{V}?\frac{0.0592\mathrm{V}}{2}\log \frac{0.0530M}{1.60M}$ - Part $B$ What is the cell potential when the concentration of ${\mathrm{Cu}}^{2+}$ has fallen to $0.200M$ ? Express your answer using two significant figures. Correct The balanced equation for this redox reaction is $\mathrm{Pb}(s)+{\mathrm{Cu}}^{2+}(aq)?{\mathrm{Pb}}^{2+}(aq)+\mathrm{Cu}(s)$ and since the concentration of ${\mathrm{Cu}}^{2+}$ drops from $1.60M$ to $0.200M$, or a change of $1.40M$, then the concentration of ${\mathrm{Pb}}^{2+}$ will increase proportionally by $1.40M$ as the stoichiometry is one to one. The cell potential can be calculated using the equation $\mathrm{Pb}(s)+{\mathrm{Cu}}^{2+}(aq)?{\mathrm{Pb}}^{2+}(aq)+\mathrm{Cu}(s)$ and since the concentration of ${\mathrm{Cu}}^{2+}$ drops from $1.60M$ to $0.200M$, or a change of $1.40M$, then the concentration of ${\mathrm{Pb}}^{2+}$ will increase proportionally by $1.40M$ as the stoichiometry is one to one. The cell potential can be calculated using the equation $E_{\text{cell }}=E_{\text{cell }}^{?}?\frac{0.0592}{n}\mathrm{V}\log Q$ where $E_{\text{cell }}$ is the cell potential in volts, $E_{\text{cell }}^{?}$ is the standard cell potential in volts, $n$ is the number of moles of electrons transferred in the redox reaction, and $Q$ is the reaction quotient. Inserting the values into the equation then gives $E_{\text{cell }}=0.47\mathrm{V}?\frac{0.0592}{2}\mathrm{V}\log \frac{(0.0530M+1.40M)}{0.200M}$ Part C What is the concentration of ${\mathrm{Pb}}^{2+}$ when the cell potential falls to $0.360\mathrm{V}$ ? Express your answers using three significant figures. $x$ Incorrect; Try Again; 2 attempts remaining