Question

using standard potential tables, are the following reactions thermodynamically favorable in the forward direction?

a. Ca^2+(aq) + Zn(s)----->Ca(s) + Zn^2+(aq)

b.2Ag^+(aq) + Ni(s) ------> 2Ag(s) + Ni^2+(aq)

c. Fe(s) + Mn^2+ ------> Fe^2+(aq) + Mn(s)

d. 2Al(s) +3Pb^2+(aq) -----> 2 Al^3+ (aq) + 3Pb(s)

Answer 1

In general, metals tend to make good reducing agents because they can only be oxidized.  The reducing ability of the metal is given by the activity series from their standard potentials.  The more active metal is able to reduce the less active metal cation.  This activity series is:

Li>K>Ca>Na>Mg>Al>Mn >Zn>Cr>Fe>Ni>Sn>Pb>H₂>Cu>Hg>Ag>Pt>Au

Most active Least active

In a reaction more negative E⁰ value will move to the left and less negative or more positive E⁰ value will move to right.

a) Ca²⁺(aq) + Zn(s) -----> Ca(s) + Zn²⁺(aq)

Here Ca is more active metal than Zn, so Zn cannot displace Ca²⁺ from the solution. So the reaction is not thermodynamically favorable in the forward direction.

b) 2Ag⁺(aq) + Ni(s) -----> 2Ag(s) + Ni²⁺(aq)

Here Ni is more active metal than Ag, so Ni can displace Ag⁺ from the solution. So the reaction is thermodynamically favorable in the forward direction.

c) Fe(s) + Mn²⁺(aq)------> Fe²⁺(aq) + Mn(s)

Here Mn is more active metal than Fe, so Fe cannot displace Mn²⁺ from the solution. So the reaction is not thermodynamically favorable in the forward direction.

d) 2Al(s) +3Pb²⁺(aq) -----> 2 Al³⁺(aq) + 3Pb(s)

Here Al is more active metal than Pb, so Al can displace Pb²⁺ from the solution. So the reaction is thermodynamically favorable in the forward direction.