20.7 Q3 in decimals plz **Electrolysis of CsI Solution**A 1.00 M solution of CsI was electrolyzed using inert electrodes. Assume that all oxidations and reductions occur at their standard potentials (no overvoltage).- **Question**: What is the minimum voltage that must be applied to make the electrolysis occur?- **Instructions**: You will need to write the oxidation half-reaction, the reduction half-reaction, and determine the overall potential for the electrolysis.

Caesium is a highly reactive metal. It reacts with iodine to produce caesium iodide. The standard cell potential for this process is +3.57 V, hence the reaction is spontaneous. Caesium is a metal belonging to s-block elements and 1st group. It contains one electron in its valence shell. It readily loses this electrons to form Cs+ ion. Iodine atom is a nonmetal which has 7 electrons in its valence shell, hence it will accept one electron to form iodide ion. Since iodine exist in its diatomic, hence one iodine molecule I2 will accept two electrons to give 2 iodide ( 2I- ) ions. The final reaction is:- 2Cs + I2 ------> 2CsI E0cell = +3.57 VFor the electrolysis of CsI, the potential of external cell (voltage of the current applied) should be more than the cell potential ( 3.57 V). When the Electrolysis starts, the Cs+ get reduced to caesium metal at cathode where as the iodide ion get oxidized to iodine (I2) at anode. The standard electrode potential for this reaction will be equal to _ 3.57 V. Since the concentration of the solution is 1 M, hence the reaction quotient (Q = concentration of products divided by the concentration of reactants) will be equal to 1. Ecell = E0cell - (2.303RT / nF) log Q log Q = log 1 = 0 Hence , Ecell = E0cell = -3.57 V