Question

Question: An equilibrium is established according to the following equation

\({K_c} = 4.6\)

What will happen in a solution that is 0.20 M in each \({\rm{Hg}}_2^{2 + },{\rm{NO}}_3^ - ,{{\rm{H}}^ + },{\rm{H}}{{\rm{g}}^{2 + }}\)and \({\rm{HN}}{{\rm{O}}_2}{\rm{\;?\;}}\)

a)\({\rm{H}}{{\rm{g}}_2}^{2 + }\) will be oxidized and \({\rm{N}}{{\rm{O}}_3}\)reduced,

b) \({\rm{H}}{{\rm{g}}_2}^{2 + }\)will be oxidized and \({\rm{N}}{{\rm{O}}_3}\) oxidized.

c)\({\rm{H}}{{\rm{g}}_2}^{2 + }\)will be oxidized and \({\rm{HN}}{{\rm{O}}_2}\)reduced.

d) \({\rm{H}}{{\rm{g}}_2}^{2 + }\)will be oxidized and \({\rm{HN}}{{\rm{O}}_2}\)oxidized.

(e) There will be no change because all reactants and products have an activity of 1.

Short answer

The correct option is option (d)

Step-by-step solution

 Find reaction quotient.

Given information is

-The given equilibrium constant is \({K_c} = 4.6\)

-The concentration of \({\rm{H}}{{\rm{g}}_2}^{2 + },{\rm{N}}{{\rm{O}}_3} - ,{{\rm{H}}^ + },{\rm{H}}{{\rm{g}}^{2 + }}\),and \({\rm{HN}}{{\rm{O}}_2}\) are 0.20M each

We have to see what will happen in a solution in terms of oxidation and reduction.

First we will find the reaction quotient, to determine the direction in which the reaction will go.

\(Q = \frac{{{{\left( {{\rm{H}}{{\rm{g}}^{2 + }}} \right)}^2} \times \left( {{\rm{HN}}{{\rm{O}}_2}} \right)}}{{\left( {{\rm{H}}{{\rm{g}}_2}^{2 + }} \right) \times \left( {{\rm{N}}{{\rm{O}}_3} - } \right) \times {{\left( {{{\rm{H}}^ + }} \right)}^3}}}\)

\(\begin{array}{l} = \frac{{{{0.20}^2} \times 0.20}}{{0.20 \times 0.20 \times {{0.20}^3}}}\\ = 25\end{array}\)

Since \({\rm{Q}} > {{\rm{K}}_c}\) the reaction will move to the left (because we have more product, it will shift toward the reactant side, and favor formation of \({\rm{H}}{{\rm{g}}_2}^{2 + },{\rm{N}}{{\rm{O}}_3}^ - ,{\rm{and}}\,{\rm{\;}}3{{\rm{H}}^ + }\)

Determine the correct option

The oxidation number of Hg on the right side is +2, and on the left side is +1 (since reaction goes from right to left) \({\rm{H}}{{\rm{g}}^{2 + }}\)will be reduced.

The oxidation number of N in \({\rm{HN}}{{\rm{O}}_2}\) is +3 and oxidation number of \({\rm{NO}}_3^ - \)is +5. Therefore \({\rm{HN}}{{\rm{O}}_2}\)is oxidized.

So, the correct option is (d).