Using table of standard reduction potentials, find the ksp of Hg2Br2. 1)Write the balanced half-reactions and their voltages.2) Label the cathode, anode, which cell is the oxidation and reduction and whether it is positive or negative. Indicate direction of electron flow, the concentrations of all ionic species and the composition of the electrodes.3) Calculate the voltage of the cell and then using that voltage, calcuate the Ksp of your compound. Please show all work. Thank you again. This is how the question is worded below, feel free to LMK if there is an issue. **Educational Content Transcription****Overall Voltage:** [No data provided]**Cell Diagram:**The diagram is a schematic representation of a galvanic (voltaic) cell. It features two vertical, cylindrical shapes placed side by side, resembling test tubes or beakers. These are connected by a horizontal bridge, likely representing a salt bridge or a porous barrier.- **Left Vertical Cylinder:** Represents one half-cell of the electrochemical setup. This may contain an electrode submerged in a solution of ions specific to that electrode. - **Right Vertical Cylinder:** Represents the other half-cell, similarly structured with an electrode and a solution of ions.- **Horizontal Bridge:** A connector typically symbolizing a salt bridge. This component allows the flow of ions between the two half-cells to maintain charge balance and complete the circuit, essential for the flow of electrons.This setup illustrates how a chemical reaction can generate an electric current through the transfer of electrons across the electrodes and the conductive pathway. **Determination of Ksp**Using a table of standard reduction potentials, calculate the solubility product (Ksp) of the compound assigned to you. Write down the two balanced half-reactions and their voltages. Label the anode, the cathode, which cell is the oxidation and reduction and whether it is positive or negative. Indicate the direction of electron flow, the concentrations of all ionic species, and the composition of the electrodes. Calculate the voltage of the cell and then, using the voltage, calculate the Ksp of your compound. MUST SHOW ALL WORK!Calculation of the Ksp of __________ (assigned compound)Half reactions:Overall Voltage: (Note: The image does not contain any graphs or diagrams.)

a) The balanced half cell reactions are as follows:' Anode (oxidation) 2Hg(l)→Hg22+(aq)+2e- E°cell=0.796 V Cathode (reduction) Hg2Br2(s)+2e-→2Hg(l)+2Br-(aq) E°cell=1.392 Vb) The diagram representation is as follows: c) Calculation of cell voltage: E°cell=Ecathode-Eanode=1.392 V-0.796 V=0.596 V Thus the voltage of the cell is 0.596 V Calculation of Ksp at 298 K: ∆G°=-RT ln Ksp ...... (1)∆G°=-nFE°cell ...... (2)Putting (1)=(2)-RT ln Ksp=-nFE° cell-8.314×298×ln Ksp=-2×96500×0.596 Vln Ksp=46.43Ksp=1.46×1020 Thus Ksp of Hg2Br2 will be 1.46×1020.