Answer to Question #217942 in General Chemistry for Meep

Question

1. Write the half-reactions occurring at the anode and the cathode for the following reactions in a voltaic cell and calculate the standard cell voltage. a. Pb_(s)+2Ag⁺_(aq)->Pb²⁺_(aq)+2Ag_(s) b. MnO_2(aq)+4H⁺_(aq)+2I^-_(aq)->Mn²⁺_(aq)+2H₂O_(l)+I_2(s)
2. Write the cell notation for (a) and (b) in #1.
3. Is the reaction below spontaneous? Prove your answer by showing equations. I_2(s)+2Br^-_(aq)->2I^-_(aq)+Br_2(l)

Answer 1

Pb_(s)+2Ag⁺_(aq)->Pb²⁺_(aq)+2Ag_(s) b. MnO_2(aq)+4H⁺_(aq)+2I^-_(aq)->Mn²⁺_(aq)+2H₂O_(l)+I_2(s)

3.

A chemical reaction is said to be non-feasible or non-spontaneous when the change in Gibbs free energy is positive. It is calculated with the help of the electrode potential of the overall redox cell reaction.

By looking at the standard reduction potentials, we can say that bromine will always get reduced in the following reaction.

Explanation:

For the given chemical reaction:

I_2(s)+2Br^-_(aq)->2I^-_(aq)+Br_2(l) ∆G=1.1×10⁵ j

Here, iodine is getting reduced because it is gaining electrons and bromine is getting oxidized because it is loosing electrons.

We know that:

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the  $E°_{cell}$ of the reaction, we use the equation:

$E°_{cell}=E°_{cathode}-E°_{anode}$

$E°_{cell}=$ $0.53-1.07=-0.54V$

Relationship between standard Gibbs free energy and standard electrode potential follows:

∆G°$=-nFE°_{cell}$

As, the standard electrode potential of the cell is coming out to be negative for the above cell. Thus, the standard Gibbs free energy change of the reaction will become positive making the reaction non-spontaneous.