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An electrochemical cell involves the Au3+/Au and the acidic MnO4-/Mn2+ half-cells. For standard conditions, the anode...

An electrochemical cell involves the Au3+/Au and the acidic MnO4-/Mn2+ half-cells. For standard conditions, the anode is the gold cell. Calculate the pH (to two decimal places) required to make the gold cell the cathode. Assume an Ecell of 0.01 V.

E (Au +) = 1.42V

E (8H+ +MnO4-) = 1.55V

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Answer #1

An electrochemical Cell Au+3) Au Il rinot Irigt2 Cathod . Anod Reduction Oxidation Ecell = 0.01 V Etreet) = 1.42V E (84+ innoE = Ecell + 0.0591 pH 0.01 = -0.13V+0.0591 pH 0.0591. pH = 0.01 +0:13V 0.0591 pH = 0.14 pH = 0.14 000591 pH = 2.3688 s PH = 2

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