Question

Which of the following is the cell reaction that occurs when the following half cells are combined?

{\text{B}}{{\text{r}}_2}{\text{ + 2 }}{{\text{e}}^ - }{\text{ }} \to {\text{ 2 B}}{{\text{r}}^ - }\left( {1M} \right){\text{; }}{E^o}{\text{ = + 1}}{\text{.09 V}} \\\\$$ $${\text{A) 2 B}}{{\text{r}}^ - }{\text{ + }}{{\text{I}}_2}{\text{ }} \to {\text{ B}}{{\text{r}}_2}{\text{ + 2 }}{{\text{I}}^ - } \\\ {\text{B ) }}{{\text{I}}_2}{\text{ + B}}{{\text{r}}_2}{\text{ }} \to {\text{ 2 }}{{\text{I}}^ - }{\text{ + 2 B}}{{\text{r}}^ - } \\\ {\text{C ) B}}{{\text{r}}_2}{\text{ + 2 }}{{\text{I}}^ - }{\text{ }} \to {\text{ }}{{\text{I}}_2}{\text{ + 2 B}}{{\text{r}}^ - } \\\ {\text{D ) 2 }}{{\text{I}}^ - } + {\text{ 2 B}}{{\text{r}}^ - }{\text{ }} \to {\text{ }}{{\text{I}}_2}{\text{B}}{{\text{r}}_2} \\\\$$

Answer 1

A cell reaction is spontaneous when the cell potential is positive. If the cell potential is negative, then the reaction is non spontaneous.

Complete step by step answer:
In a galvanic cell, two electrodes are dipped in respective electrolyte solutions. Each electrode represents a half cell. Two half cells together represent a redox couple. In the redox couple, one half cell undergoes reduction whereas the other half cell undergoes oxidation.
To start an electrochemical reaction, two redox couples need to be connected with a salt bridge and a wire.
A galvanic cell or simple battery is made of two electrodes. Each of the electrodes of a galvanic cell is known as a half cell. In a battery, the two half cells form an oxidizing-reducing couple. When two half cells are connected via an electric conductor and salt bridge, an electrochemical reaction is started.
Bromine has higher reduction potential than iodine.
The electrode having higher reduction potential value undergoes reduction. Hence, bromine will be reduced to bromide ion and iodide ion will be oxidized to iodine.
Write down the given half reactions

{\text{B}}{{\text{r}}_2}{\text{ + 2 }}{{\text{e}}^ - }{\text{ }} \to {\text{ 2 B}}{{\text{r}}^ - }\left( {1M} \right){\text{; }}{E^o}{\text{ = + 1}}{\text{.09 V}} \\\\$$ Subtract the half reaction for iodine from the half reaction for bromine to obtain overall cell reaction. $${\text{ B}}{{\text{r}}_2}{\text{ + 2 }}{{\text{e}}^ - }{\text{ }} \to {\text{ 2 B}}{{\text{r}}^ - }\left( {1M} \right){\text{; }}{E^o}{\text{ = + 1}}{\text{.09 V}} \\\ \left[ {{{\text{I}}_2}{\text{ + 2 }}{{\text{e}}^ - }{\text{ }} \to {\text{ 2 }}{{\text{I}}^ - }\left( {1M} \right){\text{; }}{E^o}{\text{ = + 0}}{\text{.54 V }}} \right] \\\ \- - - - - - - - - - - - - - - - - - - - - - - \\\ {\text{B}}{{\text{r}}_2}{\text{ + 2 }}{{\text{I}}^ - }\left( {1M} \right){\text{ }} \to {\text{ 2 B}}{{\text{r}}^ - }{\text{ + }}{{\text{I}}_2}\left( {1M} \right) \\\ {\text{ }}{E^o}{\text{ = + 1}}{\text{.09 V}} - {\text{0}}{\text{.54 V}} = {\text{0}}{\text{.55 V}} > 0 \\\ $$ Positive value of standard cell potential suggests that the reaction is spontaneous as written. _**Hence, the option C ) is the correct option.**_ **Note:** A reaction is spontaneous if it has negative value of standard Gibbs free energy change. The standard Gibbs free energy change will be negative when the standard cell potential will be positive.