Question

A voltaic cell consists of a Zn/Zn2+ half-cell and a Ni/Ni2+ half-cell at 25 ∘C. The initial concentrations of Ni2+ and Zn2+ are 1.30mol/L and 0.130 mol/L, respectively.

Zn2+(aq)+2e−→Zn(s)E∘=−0.76V
Ni2+(aq)+2e−→Ni(s)E∘=−0.23V

What are the concentrations of Ni2+ and Zn2+ when the cell potential falls to 0.45 V?

Express your answers using two significant figures separated by a comma.

Answer 1

Nerns equation to find Standard cell potential is,

E_cell = E^o_cell - (0.0591/n) log Q. --------------- (1)

Given Half cell reactions:

Zn²⁺(aq) + 2e− ------> Zn (s) E^∘= −0.76V
Ni²⁺ (aq) + 2e− ------> Ni(s) E^∘= −0.23V

As Standard reduction potential value is more -ve for Zn²⁺ it will act as reducing agent and get oxidized itself.

Hence Ni2+ will be reduced.

As oxidation occur at Zn it is an Anode and hence Ni will act as Cathode

Hence a balanced redox reaction is n =2 ie 2 electron transfer reaction,

Zn (s) + Ni²⁺ (aq) ------> Ni (s) + Zn²⁺(aq).

Q = [Zn²⁺]/[Ni²⁺] ---------------- (2)

Standar electrod potential is given as,

E^o_cell = E^o_cathode − E^o_anode. = (-0.23) - (-0.76) = +0.53 V. ------------- (3)

Initially we have, [Ni²⁺] = 1.30 M and [Zn²⁺] = 0.130 M.

Let change in [Zn²⁺] be +X M accompanied by -ve change in [Ni²⁺].

Hence new [Zn²⁺] = (0.130 + X) M and [Ni²⁺] = (1.30 - X) M.

The Cell potential corresponding to these new concentrations is

E_cell = 0.45 V ---------------- (4)

Using new concentrations and eq. (2), (3) and (4) in eq. (1) we get,

0.45 = (+0.53)- (0.0591/2) log [(0.130+X)/(1.30-X)].

0.45 = (+0.53)- (0.02955) log [(0.130+X)/(1.30-X)].

log [(0.130+X)/(1.30-X)]. = 0.53 - 0.45

log [(0.130+X)/(1.30-X)]. = 0.08

(0.130+X)/(1.30-X). = 10^0.08.

(0.130+X)/(1.30-X). = 1.2

(0.130+X)= 1.2 (1.30-X) On cross multiplying,

0.130 + X = 1.56 - 1.2X

2.2 X = 1.56 - 0.130

2.2 X = 1.43

X = 1.43 / 2.2

X = 0.65 M

Hence,

new [Zn²⁺] = (0.130 + X) = 0.130 + 0.65 = 0.780 M

new [Ni²⁺] = (1.30 - 0.65) = (1.30 - 0.65) = 0.65 M.

Answer : (0.78 mol/L, 0.65 mol/L)

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