1.

A cell consists of two hydrogen electrode . The negative electrode is in contact with a solution having pH = 6 . The positive electrode is in contact with a solution of pH = x . Calculate the value of x if the e.m.f. of the cell is found to be 0.118 V at 298 K .

Answer»


Solution :The cell may be represented as `H_2 |H^+ (10^(-6) M) ||M^+ (M_1) |H_2`
`E = E^(@) - (0.059)/(1) "log" ([H^(+)]_("anode"))/([H^(+)]_("cathode"))`
`0.118 = 0 - (0.059)/(1)"log" ((10^(-6)))/(M_(1))`
log `((10^(-6)))/(M_(1))= - (0.118)/(0.059) = -2`
`(10^(-6))/(M_(1)) = 10^(-2)`
`therefore M_(1) = (10^(-6))/(10^(-2)) = 10^(-4)` M
`therefore X = 4`


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