Question

The correct cell diagram for the following reaction and $E^\circ$ for the cell is:
$2AgBr{}_{(8)} + H{}_2 \to 2Ag{}_{(8)} + 2H{}^ + 2Br{}^ -$
$E{}_{AgBr/Ag/Br{}^ - }^\circ = + 0.10V$
A. $(Pt)H{}_2/H{}^ + //Br{}^ - /AgBr/Br{}_2(Pt);E^\circ = 0.1V$
B. $(Pt)H{}_2/H{}^ + //Br{}^ - /AgBr/Br{}_2(Pt);E^\circ = - 0.10V$
C. $(Pt)Br{}_2/AgBr/Br{}^ - //H{}^ + /H{}_2(Pt);E^\circ = 0.10V$
D. $(Pt)Br{}_2/AgBr/Br{}^ - //H{}^ + /H{}_2(Pt);E^\circ = - 0.10V$

Answer 1

Electrons flow from anode to cathode or from the oxidation half cell to reduction half cell. In terms of E cells of the half reactions, the electrons flow from more negative to more positive half reactions. A cell diagram is a representation of an electrochemical cell.

Complete step by step answer:
$2 \times [AgBr{}_{(s)} + e{}^ - \to Ag{}_{(s)} + Br_{(aq)}^ - ]$ (Reduction potential)
$H{}_2 \to 2H{}^ + + 2e$(Oxidation potential)
$2AgBr + H{}_2 \to 2Ag + 2Br{}^ -$ (Net cell reaction)
Cell reaction: $(Pt)H{}_2/H{}^ + //Br{}^ - /AgBr/Br{}_2(Pt);E^\circ = 0.1V$

Additional information:
When a redox reaction takes place, the electrons will be transferred from one species to another and if the reaction is spontaneous the energy is released which we can use to do work. To get this energy, the reaction must be divided into two separate half reactions known as oxidation and reduction. The reactions are added in two different containers and a wire is used to drive the electrons from one side to the other side. When all this is done a voltaic or galvanic cell is made. A voltaic cell is basically an electrochemical cell that uses spontaneous energy and generates electricity.

Note: when we draw a cell diagram, we have to follow the following; the anode will always be placed on the left side, and the cathode will always be placed on the right side. The salt bridge is represented by double vertical lines. The difference in the phase of an element will be represented by a single vertical line and the change in the oxidation state will be represented by commas.