The ionic product of water at 298 K is `10^(-14) M^(2)` . The standard emf of the cell producing the reaction `H^(+)(aq)+OH^(-)(aq)rarr H_(2)O(l)` will be

Options

A. 0.723V
B. `-0.723V`
C. `0.82V`
D. `-0.82V`

Answer

B

Explanation

The equilibrium constant of the reaction `H^(+) (aq) + OH^(-) (aq) hArr H_(2)O (I)`
`K_(eq)= ([H_(2)O)])/([H^(+)] [OH^(-)])= (55.56M)/(10^(-14) M^(2)= 55.56 xx 10^(14 ) M^(-1)`
Since `Delta G^(@)= -nFE^(@)= -RT ln K^(@)`, we have
`E^(@)= (RT ln K^(@))/(nF)= ((2.303RT)/(nF)) log K^(@)= (0.059V) log (55.56 xx 10^(-14)) = -0.723V`