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We consider two half cells:

Half cell A comprises a platinum electrode immersed in an aqueous solution containing Mn2+ (10 mM) and MnO4- (4 mM) ions at pH=4.
Half cell B is constructed with a silver electrode immersed in an aqueous potassium chromate solution (8 mM) in the presence of solid Ag2CrO4. The pH of this half cell B is pH=9.
Both half cells are connected with a salt bridge. The voltage of the cell is measured at 25°C: it is 0.573 V. Use the following standard reduction potentials:

E°+0.800 V E0 = +1.491 V E +1.679 V Eo +0.564 V Eo -1.181 V --> Mn2+ + 2 H2O Mno4H+3e MnO, +e Mn2* +2 e -> MnO, + 2 H2O 2- MnO4

a. Calculate the solubility product of silver chromate.
b. Explain, why the given pH values are essential for this problem.
c. Determine – through calculation – the standard reduction potentials for

MnO2 +4H+ +3e- ?? Mn2+ +2H2O

and

MnO42- +4H+ +2e- ?? MnO2 +2H2O

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Answer #1

From the data Mn2+10x103 M = 10.2 M 4x103 M Half cell A Reduction reaction EMno, Mn2+ + 4H2O ˇ 1-4-1.419V Mn MnO4-+8H++5e ___Half cell B Oxidation reaction EAg0.800 v Ag Overall equation Eo cell EO cathode EO Anode -1.419-0.800 -0.619 V

Use Nemst equation 0.059, [products] E cell-E log reactants 2+ 0.573-0.619-log 0.059, AgMnFrom the data 0573-0.619.0059ogAg 102) Ag ](102) (410 101) 0.059 3410)(10) Ag 1-6.42 10 M Ksp [Ag ][CrO4 (6.42x10): (8x10 )

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