# Question #0bb5c

Mar 20, 2017

My lengthy analysis below...

#### Explanation:

The notation in (a) means that your electrodes consist of a silver electrode in a solution of $A {g}^{+}$ ions (like $A g N {O}_{3}$ 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 $A g | | A {g}^{+}$ cell is greater in potential, it serves as the cathode, meaning that the half-reaction is

$A {g}^{+} + {e}^{-}$$\rightarrow A g \left(s\right)$

The hydrogen cell is the anode, so oxidation occurs

${H}_{2} \left(g\right) \rightarrow 2 {H}^{+} + 2 {e}^{-}$

Together, the reaction is

$2 A {g}^{+} + {H}_{2} \rightarrow 2 A g \left(s\right) + 2 {H}^{+}$ cell voltage 0.80 V

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

$B {r}_{2} + 2 {e}^{-}$$\rightarrow 2 B {r}^{-}$ potential= +1.06 V

$C {r}_{2} {O}_{7}^{2 -} + 14 {H}^{+} + 6 {e}^{-}$$\rightarrow 2 C {r}^{2 +} + 7 {H}_{2} O$ 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

$C {r}_{2} {O}_{7}^{2 -} + 14 {H}^{+} + 6 {e}^{-}$$\rightarrow 2 C {r}^{2 +} + 7 {H}_{2} O$

$2 B {r}^{-}$$\rightarrow B {r}_{2} + 2 {e}^{-}$

(We now know that the cell consists of one solution containing $B {r}^{-}$ ions and a second solution of $C {r}_{2} {O}_{7}^{2 -}$ ions into which the electrodes are placed. $B {r}_{2}$ and Cr^(3+) are the products.)

Overall reaction:

$6 B {r}^{-} + C {r}_{2} {O}_{7}^{2 -} + 14 {H}^{+}$$\rightarrow 2 C {r}^{2 +} + 7 {H}_{2} O + 3 B {r}_{2}$

Cell voltage 1.33 V - 1.06 V = 0.27 V