if the cell potential at 298 k in 0.096 V.if the ph of solution is 1.396, then caculate concentration of sn2+.

Asked by JAY ACHARYA | 25th Oct, 2012, 05:47: PM

Expert Answer:

We know that
[H+]=antilog 1.396
E=Eo-0.0591/2 log [Sn+2]/[H+]
0.096=0.14-0.0591/2 log [Sn+2]/[24.886]2
Solving this equation we can get the value of [Sn+2]

Answered by  | 23rd Apr, 2013, 12:39: PM

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