In a cell reaction, `Cu_((s))+ 2Ag_((aq))^(+) to Cu_((aq))^(2+) + 2Ag_((s)) E_"cell"^@`=+0.46 V . If the concentration of `Cu^(2+)` ions is doubled then `E_"cell"^@` will be

Options

A. Decrease by small fraction
B. Doubled
C. Unchanged
D. Decrease by small fraction

Answer

A

Explanation

Nernst equation: `E_("cell")= E_("cell")^(@)- RT//nF ln Q` ...(i)
`E_("cell") rarr` Emf of cell.
`E_("cell") rarr` Standard reduction potential.
` R rarr` Universal gas constant.
`T rarr` Temperature in Kelvin.
`n rarr` moles of electrons.
`F rarr` Faraday.s constant.
`Q rarr` Reaction Quotient.
`2Ag^(+) (aq) + Cu(s) rarr Cu^(2+) (aq) + 2Ag(s)` ....(ii)
Q = `["Products"]^(a)//["Reactants"]^(b)`
a and b `rarr` stoichiometric co-efficient of product and reactant respectively. For pure state concentration is taken as unity
`rArr E_("cell") = E_("cell") rarr RT/nF In [Cu_(aq)^(2+)]//[Ag_(aq)^(+)]^(2)` .....(iii)
`rArr` From Equation 4, it is observed that `E_(("cell"))` depends on the concentration of both `Cu^(2+) and Ag^(+)`. It increases with increase in concentration of `Ag^(+)`