consider the galvanic cell constructed with the following materials $\mathrm{Zn}$ electrode, $\mathrm{Zn}$ (NO
3)2, Fe electrode, $\mathrm{Fe}(\mathrm{NO} 3) 3 \mathrm{a}$. sketch a diagram of the galvanic cell identifying each material and labeling the cathode and anode $b$ determine the overall cell reaction and calculate $E$ cell c calculate delta $G$ and $K$ for the cell reaction at 25 degrees celsius $\mathrm{d}$. calculate Ecell at 25 degrees celcius when $[\mathrm{Zn} 2+] 1.0$ times 10 to the $-2 \mathrm{M}$ and [Fe $2+$ ] 1.0 times 10 to the -4 th $M$

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∻Solution∻ ‖ 1 ⋮ **Sketch the diagram**: A galvanic cell consists of two half-cells, each containing an electrode and an electrolyte. The Zn electrode is the anode where oxidation occurs, and the Fe electrode is the cathode where reduction occurs. The anode and cathode are connected by a salt bridge, and the external circuit connects the electrodes to allow for electron flow. ‖ ‖ 2 ⋮ **Determine the overall cell reaction**: The anode half-reaction is $\mathrm{Zn} ightarrow \mathrm{Zn}^{2+} + 2e^-$, and the cathode half-reaction is $\mathrm{Fe}^{3+} + e^- ightarrow \mathrm{Fe}^{2+}$. The overall cell reaction is $\mathrm{Zn} + \mathrm{Fe}^{3+} ightarrow \mathrm{Zn}^{2+} + \mathrm{Fe}^{2+}$. ‖ ‖ 3 ⋮ **Calculate $E_{ ext{cell}}$**: The standard electrode potentials ($E^\circ$) for the half-reactions are $E^\circ_{ ext{Zn}^{2+}/ ext{Zn}} = -0.76\ ext{V}$ and $E^\circ_{ ext{Fe}^{3+}/ ext{Fe}^{2+}} = 0.77\ ext{V}$. The standard cell potential is $E^\circ_{ ext{cell}} = E^\circ_{ ext{cathode}} - E^\circ_{ ext{anode}} = 0.77\ ext{V} - (-0.76\ ext{V}) = 1.53\ ext{V}$. ‖ ‖ 4 ⋮ **Calculate $\Delta G$ and $K$**: The change in free energy $\Delta G^\circ$ is calculated using the equation $\Delta G^\circ = -nFE^\circ_{ ext{cell}}$, where $n$ is the number of moles of electrons transferred (2 moles), $F$ is Faraday's constant ($96485\ ext{C/mol}$), and $E^\circ_{ ext{cell}}$ is the standard cell potential. The equilibrium constant $K$ is calculated using the equation $\Delta G^\circ = -RT\ln K$, where $R$ is the gas constant ($8.314\ ext{J/(mol·K)}$) and $T$ is the temperature in Kelvin (298 K). ‖ ‖ 5 ⋮ **Calculate $E_{ ext{cell}}$ at given concentrations**: Use the Nernst equation $E_{ ext{cell}} = E^\circ_{ ext{cell}} - \frac{RT}{nF}\ln Q$, where $Q$ is the reaction quotient. Given the concentrations, $Q = \frac{[\mathrm{Zn}^{2+}]}{[\mathrm{Fe}^{3+}]} = \frac{1.0 imes 10^{-2}}{1.0 imes 10^{-4}} = 100$. ‖ ∻1 Answer∻ ⚹ Diagram not provided, but the description of the galvanic cell components and their arrangement is given. ⚹ ∻2 Answer∻ ⚹ The overall cell reaction is $\mathrm{Zn} + \mathrm{Fe}^{3+} ightarrow \mathrm{Zn}^{2+} + \mathrm{Fe}^{2+}$. ⚹ ∻3 Answer∻ ⚹ $E^\circ_{ ext{cell}} = 1.53\ ext{V}$. ⚹ ∻4 Answer∻ ⚹ $\Delta G^\circ = -2 imes 96485\ ext{C/mol} imes 1.53\ ext{V} = -295,082\ ext{J/mol}$ and $K$ can be calculated from $\Delta G^\circ = -RT\ln K$. ⚹ ∻5 Answer∻ ⚹ $E_{ ext{cell}}$ at the given concentrations can be calculated using the Nernst equation. ⚹ ∻Key Concept∻ ⚹ Galvanic cells convert chemical energy into electrical energy through redox reactions, and the cell potential can be calculated using standard potentials and the Nernst equation. ⚹ ∻Explanation∻ ⚹ The galvanic cell operates on the principle of spontaneous redox reactions, with the anode experiencing oxidation and the cathode experiencing reduction. The standard cell potential gives the voltage under standard conditions, while the Nernst equation allows for the calculation of cell potential under non-standard conditions. ⚹◊What is the overall cell reaction for the galvanic cell described in the student's query?⍭ Generate me a similar question◊

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