1.

Given the following cell at 25^(@) C . What will be the potential of the cell ? Given pK_(a) of CH_(3) COOH = 4.74

Answer»

`-0.42` V
`0.42 V `
`-0.19 V `
`0.19 V `

Solution :It is a concentration cell , THEREFORE `E_("cell")^(@) = 0`
pHof `W_(A) = (1)/(2) ( pk_(a) - log c)`
`= (1)/(2) (4.74 - log 10^(-3)) = 3.87`
pH of NaOH = 14 - 3 = 11
`therefore E = -0.059 (pH_(c)- pH_(a))`
`= -0.059 (11 - 3.87) = -0.42 V `


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