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1 [2 points) You have constiucted a 1 M quinhydronetquinone concentration cell with buffers of pH = 4.3 and pH = 2,2 senor-ding to u: following…

1 [2 points) You have constiucted a 1 M quinhydronetquinone concentration cell with buffers of pH = 4.3 and pH = 2,2 senor-ding to u: following…

− 7.79:, , *log ([&)] ‘ [&)]’)= − 7.79:, , *log ([&)] ‘ [&)]’)= Which can be simplified using properties of logs: <=>>=− 7.79:, , *log ([&)] ‘ [&)]’)= − 7.79:, , [(2log([H+])-(2log([H+])]Pulling out a -2 we end up with a simplified equation in terms of the pH in each half reaction<=>>= ?7.79:,(?) , [-log([H+])-(-log([H+])]= +0.0592*(pH-pH) =0.0592*DpH = +0.0592*DpHUsing the new Nernst equation as derived here, calculate the cell potential for this electrochemical cell. Attachment 1Attachment 21 [2 points) You have constiucted a 1 M quinhydronetquinone concentration cell with buffersof pH = 4.3 and pH = 2,2 senor-ding to flu: following notation:C(st graphite] l pH 4.3 buffer, quinhydroneillei) || pH 2.2 buffer, quiuhydronetllui) | (Is, gmphite] Where we will have solutions of quinone {Q} and quinhydrone (Hill) in the aqueous phase inboth of out half eells, and graphite as our electrode. This can he thought of as an electrochemical cellwhere quinone is our cathode and quinhydrone is the attach: with booster of It electrons; Q + 214* + rte—i HJQH2Q -. o + 2st + le’Net: Q +2H++ HEQ—r Q +2H++ HQQ’ + ei:ii§::ii:+—i; noting that “13 ”“1033“ “3’“ concentrations in each half cell are different clue to the differences in pH. We are assumingthe same concentration of Q and QH; in both half eelle however, so our reaction quotientbecomes: The reaction quotient for this reaction is Q = H+ 2Q = [ +12[H ]Where [H+] is the concentration of hydroniunt ion in the the anode halfeell and [H+]

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