Question #e6bb7

1 Answer
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: Ag^(+)+1e^(-)->Ag" " " "xi^@=0.80VAg++1eAg ξ=0.80V

Oxidation reaction: 2Br^(-)->Br_2+2e^(-)" " ""-xi^@=-1.09V2BrBr2+2e ξ=1.09V

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

Reduction: color(red)(xx2)(Ag^(+)+cancel(1e^(-))->Ag)
Oxidation: ul(" " "2Br^(-)->Br_2+cancel(2e^(-)))
Redox: " "2Ag^(+)+2Br^(-)->2Ag+Br_2" "xi_("redox")^@=-0.29V

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

xi_("redox")^@=xi_("red")^@+xi_("ox")^@=0.80V+(-1.09V)=-0.29V

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.