91Ó°ÊÓ

• The shorthand notation of a cell reaction indicates phase boundary by a single vertical line (|) and salt bridge is denotes by a double vertical line .
• In this notation on the left side ofsalt bridge oxidation reactionwhich occurred at anode and on the right side of salt bridge; reduction reaction which occurred at cathode.
• The electrodes are written on the extreme corners; anode on extreme left and cathode on extreme right. The reactants in each half-cell are always first, followed by the products.
• If the standard cell potential is positive, then the reaction isspontaneousand if the standard cell potential is negative, then the reaction isnon-spontaneous

Given:

In the given reaction hydrogen is oxidized at anode and bromine is reduced at cathode.

Anode reaction:

\({{\rm{H}}_2}(g) \to 2{{\rm{H}}^ + }(aq) + 2{e^ - }\quad {E^o} = 0.00\;{\rm{V}}\)

Cathode reaction:

\(B{r_2}(aq) + 2{e^ - } \to 2B{r^ - }(aq)\quad {E^o} = + 1.0873\;{\rm{V}}\)

The overall or net cell reaction is as follows:

\(\begin{aligned}{l}{H_2}(g) \to 2{H^ + }(aq) + 2{e^ - }\\\frac{{B{r_2}(aq) + 2{e^ - } \to 2B{r^ - }(aq)}}{{{H_2}(g) + B{r_2}(aq) \to 2HBr(aq)}}\end{aligned}\)

Now calculate the standard cell potential at \({25\circ }{\rm{C}}\), using the following expression:

\(\begin{aligned}{l}E_{{\rm{cell }}}^o = E_{{\rm{cathode }}}^o - E_{{\rm{anode }}}^o\\E_{{\rm{cell }}}^o = + 1.0873\;{\rm{V}} - (0.00\;{\rm{V}})\\ = + 1.0873\;{\rm{V}}\end{aligned}\)

For spontaneity:

\(\Delta {\rm{G}} = - {\rm{nFEE}}_{{\rm{cell }}}^{\rm{o}}\)

Where \({\rm{n}}\) is the number of electrons, \({\rm{F}}\) is Faradays constant ( \(96500{\rm{C}}),\Delta {\rm{G}}\) is change in Gibbs energy, and \({\rm{E}}_{{\rm{cell }}}^{\rm{o}}\) is cell potential.

Substituting the values:

\(\begin{aligned}{l}\Delta {\rm{G}} = - 2 \times 96500 \times 1.09\\\Delta {\rm{G}} = - 210370\;{\rm{J}}\\\Delta {\rm{G}} = - 210.370\;{\rm{kJ}}\end{aligned}\)

Since, value of \(\Delta G\) is negative so, the reaction is spontaneous.