Question

Which of the following reactions should have the larger emf under standard conditions? Why?
CuSO4(aq)+Pb(s) ⇌ PbSO4(s)+Cu(s)
Cu(NO3)2(aq)+Pb(s) ⇌ Pb(NO3)2(aq)+Cu(s)

Answer 1

Answer:

Cu(NO3)2(aq)+Pb(s) ⇌ Pb(NO3)2(aq)+Cu(s)

Explanation:

If we look at the both reactions closely, we will quickly discover that the reaction CuSO4(aq)+Pb(s) ⇌ PbSO4(s)+Cu(s) involves PbSO4.

The compound PbSO4 is insoluble in water and sinks to the bottom of the reaction vessel. When this occurs, the concentration of Pb^2+ becomes low. This will bring about a low voltage in the cell.

On the other hand, Pb(NO3)2 is soluble in water hence the cell voltage in this case is higher than the former.

Answer 2

the following reaction should have the larger emf should be Cu(NO3)2(aq)+Pb(s) ⇌ Pb(NO3)2(aq)+Cu(s)

How to find out the larger emf ?

In the case when we look at both reactions so here we discovered that the reaction of

CuSO4(aq)+Pb(s) ⇌ PbSO4(s)+Cu(s) includes PbSO4.

The compound PbSO4 should be insoluble in water and sinks to the bottom of the reaction vessel. At the time When this occurs, the concentration of Pb^2+ becomes low. This will bring about a low voltage in the cell.

On the other hand, Pb(NO3)2 is soluble in water so the cell voltage in this case is more than the former.

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