c. What would the pH of the cell have to be for the voltage to increase by 20.0%? please answer all parts 8.Answer the questions below based on the following electrochemical cell:Cu(s)|Cu(NO,),(aq, 1.0 m)||Mn(NO,),(aq, 1.0 m), HNO,(aq)|MnO,(s)|Pt(,)a. Write the individual half reactions and the overall cell reaction and determine thestandard cell potential, E (V), using the attached table of reduction potentials.ANS(Fake = 1.00 V)b. Calculate the AGrxn and the equilibrium constant, K, for this reaction.ANSc. What would the pH of the cell have to be for the voltage to increase by20.0%?

Standard electrode potential (Eocell) The electrode potential under standard conditions (at 298 K and 1 atm pressure ) is called standard electrode potential Cell potential (Ecell ) The cell potential of emf is the difference between the electrode potentials of the two electrodes that causes the production of electric current. Ecell= Ecathode- Eanode Nernst equation It gives the relationship between electrode potential and concentration of ionic species. For a reaction, the Nernst equation can be written as; Mn++ne-→M(s)E=Eocell-RTnF ln [M][Mn+]E=electrode potentialEocell=standard reduction potentialR=gas constantT=absolute temperatureF=faraday's constantn=number of electrons involved in the reactions a) Anode reaction; 3Cu(s)+8 HNO3(aq)→3Cu(NO3)2(aq)+2NO(g)+4H2O(l) Eocell=-0.34VCathode reactionMnO2(s)+4H+(aq)+2e-→Mn2+(aq)+2H2O (l) Eocell=1.23 V Eocell=Eocathode-Eoanode=EoMnO2/Mn2+-EoCu/Cu2+=1.23-(-0.34)=1.57 Vb) The net reaction is; 3Cu(s)+MnO2(s)+12H+→3Cu2+(aq) + Mn2+(aq)+6H2O(l)+ 2NO(g) The equilibrium constant KeqEocell=RTnFln KeqEocell=1.57 VConsidering standard conditions,R=8.314 J/K molT=298 Kn=2F=96485 CEocell=RTnFln Keq1.57=8.314×2982×96485 ln Keq ln Keq=122.2Keq=1.27×1053 The equation relating ∆Go and Keq is;∆Go =-RT ln KeqR=8.314 J/K molT=25oC=298 K (room temperature)As Keq=1.27×1053∆Go=-RT ln Keq=-(8.314×298) ln (1.27×1053)=-302.96 kJc) 3Cu(s)+MnO2(s)+12H+→3Cu2+(aq) + Mn2+(aq)+6H2O(l)+ 2NO(g) Eocell=0.059nlog[Cu2+]3 [Mn2+][H+]12Eocell=1.57n=2 (as two electrons are transferred during reaction)[Cu2+] =1.0 m[Mn2+]=1.0 mOn increasing 20% of the volatge,Eocell=1.57+20% of 1.57=1.88 VSo the Nernst equation becomes,1.88 V=0.0592log[1]3 [1][H+]121.88=0.029 log1[H+]12 log1[H+]12=63.7-log [H+] 12=63.7-12 log [H+]=63.7- log [H+]=5.3pH=5.3 Answer a) The standard cell potential is : 1.57 V b) ∆Go rxn and Keq is -302.96 kJ and 1.27×1053 respectively. c) The pH of the cell for the voltage to increase by 20% is : 5.3