Question

Question #0bb5c

Answer 1

My lengthy analysis below...

Explanation:

The notation in (a) means that your electrodes consist of a silver electrode in a solution of `Ag^+` ions (like `AgNO_3` dissolved in water). The standard electrode potential for this is +0.80 V.

The other electrode is hydrogen gas bubbled over an inert platinum in an acidic solution (such as 1.0 M HCl). The standard potential is 0.00 V

The double vertical line represents the barrier between the two half-cells.

Since the `Ag||Ag^+` cell is greater in potential, it serves as the cathode, meaning that the half-reaction is

`Ag^+ + e^-` `rarr Ag (s)`

The hydrogen cell is the anode, so oxidation occurs

`H_2(g) rarr 2H^+ + 2e^-`

Together, the reaction is

`2Ag^+ + H_2 rarr 2Ag (s)+2H^+` cell voltage 0.80 V

In (b), we can start with two reduction half-reactions, and their potentials:

`Br_2 + 2 e^-` `rarr 2Br^-` potential= +1.06 V

`Cr_2O_7^(2-) + 14H^+ + 6e^-` `rarr2Cr^(2+)+7H_2O` potential =+1.33V

Both of these occur by passing the chemicals over an inert platinum electrode.

Since the second half-reaction has the greater potential, it is the reduction (and the cathode). The bromine process is the oxidation (and the anode).

So the half-reactions are

`Cr_2O_7^(2-) + 14H^+ + 6e^-` `rarr2Cr^(2+)+7H_2O`

`2Br^-` `rarr Br_2+2e^-`

(We now know that the cell consists of one solution containing `Br^-` ions and a second solution of `Cr_2O_7^(2-)` ions into which the electrodes are placed. `Br_2` and Cr^(3+) are the products.)

Overall reaction:

`6Br^-+Cr_2O_7^(2-) + 14H^+` `rarr2Cr^(2+)+7H_2O+3Br_2`

Cell voltage 1.33 V - 1.06 V = 0.27 V