)|Microsoft Office Homex: HW13 . Secton 049 X C ezto.mheducation.com/hm.tpx Ques

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)|Microsoft Office Homex: HW13 . Secton 049 X C ezto.mheducation.com/hm.tpx Question 11 (of 12) value 0.58 points Ch. Ex. 73-Calculate Ecell Values and Determine Cathode 1 out of 3 attempts Be sure to answer all parts. As A concentration cell consists of two SnSn.. half-cells. The electrolyte in compartment A is 0.16 M Sa(NO).. The electrolyte in B is 0.87 M Sa(NO Which half-cell houses the cathode? What is the voltage of the cell? Cathode: half-cell A half-cell B Voltage of cell: References eBook & Resources 00 TOSHIBA

Explanation / Answer

for concentration cell Eocell = 0

Ecell = - 0.05916 / n log [Sn+2]anode / [Sn+2]cathode

         = - 0.05916 / 2 log (0.16 / 0.87)

Ecell = 0.022 V.

here Ecell become positive when 0.16 M Sn(NO3)2 as anode and 0.87 M Sn(NO3)2 as cathode.

so here Electrolyte B is a cathode

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