InterviewSolution
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InterviewSolution
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Carefully observe the given figure and using data provided find the `EMF` of shown Galvenic cell in volt: Solution A is `0.1M` each in `NH_(4)OH` and `NH_(4)CI` and solution B is `0.1M` each in `CH_(3)COOH` and `CH_(3)COONa^(+)`. [Given: `K_(a)(CH_(3)COOH) = 10^(-5), K_(b) (NH_(4)OH) = 10^(-5)` and `(2.303RT)/(F) = 0.06` volt] |
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Answer» Correct Answer - `0.24` `pH` of solution (A) : Solution A is a basic Buffer `pOH = pK_(b) +log.(["Salt"])/([Base]) rArr pOH = 5` `rArr pH = 9 rArr [H^(+)]_(A) = 10^(-9)M` `pH` of solution (B): Solution B is Acidic Buffer `pH = pK_(a) + (["Salt"])/([Base])` `rArr pH = 5 +log.((0.1))/((0.1)) rArr pH = 5` `rArr [H^(+)]_(B) = 10^(-5)` Now, The cell is a concentration cell Cell reaction: `2H_(B)^(+)(aq) +2e hArr H_(2,B)(g)` `H_(2,A) (g) hArr 2H_(A)^(+) +2e` `bar(2H_(B)^(+)(aq)+H_(2,A)(g)overset(n=2)hArr)2H_(A)^(+)(aq)+H_(2,B)(g)` Nernst Equation for cell `E_(cell) = 0-(0.06)/(2) log.([H^(+)]_(A)^(2))/([H^(+)]_(B)^(2))` `=0.06 log ((10^(-4))/(1))rArr +0.24` volt. |
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