Question #0bb5c

1 Answer
Mar 20, 2017

Answer:

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