Hi-Please provide a detailed explantation and a step by step instructions to sol

Question

Hi-Please provide a detailed explantation and a step by step instructions to solve this problem. I have tried looking at other solutions of this problem and not getting it. If you provide step by step instructions, will give a good rating for your answer. Thanks.

Consider the following electrochemical cell at 298 K. One compartment contains a 1.00-M solution of silver nitrate. The other compartment contains a saturated solution of silver iodide. Both electrodes are silver and the measured potential is 0.476 V. From this information, determine the molar solubility of silver iodide at 298 K. Answer is 9.11*10^-9.

Explanation / Answer

Ecell = -0.0592 log Ksp

log Ksp = -Ecell /0.0592

cathode: Ag+ + e Ag(s)

anode: Ag(s) Ag+ + e

n = 1

log Ksp = -Ecell /0.0592

= -0.476 V /0.0592 ]= -8.04

Ksp = 9.12 x 10-9

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