Question

Suppose the 0.1 M Zn 2+ solution has been diluted (instead of the Cu 2+ solution), would the measured cell potentials have increased or decreased? Explain why the change occurred.

Answer 1

Suppose the 0.1 M Zn 2+ solution has been diluted (instead of the Cu 2+ solution), would the measured cell potentials have increased or decreased? Explain why the change occurred.

Electrod potential depends on the concentration of metallic ions in the solution. The variation of electrode potential and hence the cell potential with the concentration of metallic ions in the solution is given by Nernst equation.

In a galvanic cell, Zn(s) | Cu²⁺ (aq.) || Zn²⁺ (aq.) | Cu(s)

Zn(s) + Cu²⁺ (aq.) <------------- > Zn²⁺ (aq.) + Cu(s)……………..(n=2 i.e. 2 electron transfer reaction)

According to the Nernst equation the EMF or cell potential is given as,

E_cell = E⁰_cell – (0.0591/2) log{[Zn²⁺]/[Cu²⁺]}

E_cell = E⁰_cell – (0.02955) log{[Zn²⁺]/[Cu²⁺]}

(0.0591/2) log{[Zn²⁺]/[Cu²⁺]} this term is subtracted from E⁰_cell and hence we can say that value of E_cell i.e. cell potential will increase (in terms of the magnitude) as the value of the term log{[Zn²⁺]/[Cu²⁺]} decreases (or becomes –ve) (0.02995 is constant)…………(statement-1)

Its clear that the value of log{[Zn²⁺]/[Cu²⁺]} will decrease if the ratio [Zn²⁺]/[Cu²⁺] decreases. As we dilute Zn²⁺ solution, [Zn²⁺] concentration will decrease and ultimately the ratio [Zn²⁺]/[Cu²⁺] will decrease (as we are not changing [Cu2+]) and hence the value of log{[Zn²⁺]/[Cu²⁺]} will decrease……………..(statement-2)

With statement-1 and 2 we can say that the cell potential will increase (in terms of magnitude) on dilution of Zn²⁺ solution keeping [Cu²⁺] constant.

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