A solution containing `H^(+)` and `D^(+)` ions is in equilibrium with a mixture of `H_(2)` and `D_(2)` gases at `25^(@)C` . If the partial pressure of both the gases are `1.0` atm. The ration `[H^(+)]/[D^(+)]`.
Given, `E_(D^(+)//D)^(@) = -0.003 C, E_("cell") = 0.006 V` :

Options

A. `1.2`
B. `1.1`
C. `0.11`
D. `1.0`

Answer

B

Explanation

`D_(2) + 2H^(+) rarr H_(2) + 2D^(+)`
`E = E_(OP_(D))^(@) + E_(RP_(H))^(@) + (0.0591)/(2)"log"([H^(+)]^(2)xxP_(D_(2)))/(P_(H_(2)) xx [D^(+)]^(2))`
`0.006 = 0.003 + 0 + (0.0591)/(1)"log"([H^(+)])/([D^(+)])`
`(0.0591)/(1)"log"([H^(+)])/([D^(+)]) = 0.0006 - 0.003`
`:. "log"([H^(+)])/([D^(+)]) = 0.0507`
`:. ([H^(+)])/([D^(+)]) = 1.12`