Question

Determine the overall reaction and its standard cell potential at $25\circ \mathrm{C}$ for these reactions. Is the reaction spontaneous at standard conditions? Assume the standard reduction for $\mathrm{B}{\mathrm{r}}_2(l)$ is the same as for $\mathrm{B}{\mathrm{r}}_2(aq)$.

Short answer

The overall reaction is $H_2(g)+B{r}_2(aq)\to 2HBr(aq)$

The standard cell potential at $25\circ \mathrm{C}$ is $+1.0873\mathrm{V}$

The value of $\mathrm{\Delta }G$ is negative, and then the reaction is spontaneous.

Step-by-step solution

Define standard cell potential

• 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

Determine the standard cell potential

Given:

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

Anode reaction:

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

Cathode reaction:

$B{r}_2(aq)+2e^{-}\to 2B{r}^{-}(aq)E^o=+1.0873\mathrm{V}$

The overall or net cell reaction is as follows:

$\begin{array}{r}l{H}_2(g)\to 2H^{+}(aq)+2e^{-}\\ \frac{B{r}_2(aq)+2e^{-}\to 2B{r}^{-}(aq)}{H_2(g)+B{r}_2(aq)\to 2HBr(aq)}\end{array}$

Now calculate the standard cell potential at $25\circ \mathrm{C}$, using the following expression:

$\begin{array}{r}l{E}_{\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{l}}^o=E_{\mathrm{c}\mathrm{a}\mathrm{t}\mathrm{h}\mathrm{o}\mathrm{d}\mathrm{e}}^o-E_{\mathrm{a}\mathrm{n}\mathrm{o}\mathrm{d}\mathrm{e}}^o\\ E_{\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{l}}^o=+1.0873\mathrm{V}-(0.00\mathrm{V})\\ =+1.0873\mathrm{V}\end{array}$

For spontaneity:

$\mathrm{\Delta }\mathrm{G}=-{\mathrm{n}\mathrm{F}\mathrm{E}\mathrm{E}}_{\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{l}}^{\mathrm{o}}$

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

Substituting the values:

$\begin{array}{r}l\mathrm{\Delta }\mathrm{G}=-2\times 96500\times 1.09\\ \mathrm{\Delta }\mathrm{G}=-210370\mathrm{J}\\ \mathrm{\Delta }\mathrm{G}=-210.370\mathrm{k}\mathrm{J}\end{array}$

Since, value of $\mathrm{\Delta }G$ is negative so, the reaction is spontaneous.