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  • Calculate the value of n if, <img alt="\mathrm{E_{cell}^{0}= 0.23 \ and \ K_{eq}= 6}" src="/latex-image/?%5Cmathrm%7BE_%7Bcell%7D%5E%7B0%7D%20%3D%200.023%20%5C%20and%20%5C%20K_%7Beq%7D%3D%206%

Calculate the value of n if, \mathrm{E_{cell}^{0}= 0.23 \ and \ K_{eq}= 6}

Option: 1

1


Option: 2

2


Option: 3

3


Option: 4

4


Answers (1)

best_answer

\begin{array}{l}{\text {In general, }} \\\\ \mathrm{{E_{\text {(cell ) }}^{\ominus}=\frac{0.0591}{n } \log K_{C}}}\end{array}

Thus, the above equation gives a relationship between the equilibrium constant of the reaction and standard potential of the cell in which that reaction takes place. Thus, equilibrium constants of the reaction, difficult to measure otherwise, can be calculated from the corresponding Eo value of the cell.

 

\\0.023 = \frac{0.0591}{n} log _{10}6 \\\\ \\n = \frac{0.0591}{0.23} log _{10}(2\times3) \\\\ \\n = 2.6 (log _{10}(2) + log _{10}(3)) \\\\ \\n = 2.6 (0.3010+0.4771) \\\\ \\n = 2.6 \times 0.7781 \\\\\Rightarrow n =2

Therefore,option(2) is correct

Posted by

Shailly goel

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