# Question e6bb7

Jun 2, 2016

Bromide ion will not be able to spontaneously reduce silver ion.

#### Explanation:

In order to answer this question, we will need to write the reduction and oxidation half equations and to look up their corresponding standard reduction potentials:

Reduction reaction: $A {g}^{+} + 1 {e}^{-} \to A g \text{ " " } {\xi}^{\circ} = 0.80 V$

Oxidation reaction: $2 B {r}^{-} \to B {r}_{2} + 2 {e}^{-} \text{ " } - {\xi}^{\circ} = - 1.09 V$

The sum of these two equations would give the redox reaction between bromide and silver ions.

Reduction: $\textcolor{red}{\times 2} \left(A {g}^{+} + \cancel{1 {e}^{-}} \to A g\right)$
Oxidation: $\underline{\text{ " } 2 B {r}^{-} \to B {r}_{2} + \cancel{2 {e}^{-}}}$
Redox: " "2Ag^(+)+2Br^(-)->2Ag+Br_2" "xi_("redox")^@=-0.29V#

Note that the redox potential is the sum of both half equations potentials:

${\xi}_{\text{redox")^@=xi_("red")^@+xi_("ox}}^{\circ} = 0.80 V + \left(- 1.09 V\right) = - 0.29 V$

Since the overall redox potential of this redox reaction is negative, therefore, the reaction is not spontaneous in the forward direction and thus, bromide ion will not be able to spontaneously reduce silver ion.

Electrochemistry | The Galvanic Cell.