Question

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A voltaic cell is constructed that uses the following reaction and operates at 298 K:
Zn(s)+Ni2+(aq)→Zn2+(aq)+Ni(s).

1.) What is the emf of this cell under standard conditions?

2.) What is the emf of this cell when [Ni2+]= 2.80 M and [Zn2+]= 0.130 M ?

3.) What is the emf of the cell when [Ni2+]= 0.150 M and [Zn2+]= 0.999 M ?

Express your answer using two significant figures.

Answer 1

1.

Anode reaction

Zn -2e  $\rightarrow$ Zn²⁺ , E^o(Zn²⁺/Zn ) = - 0.76 V

Cathode reaction

Ni²⁺ +2e $\rightarrow$ Ni , E^o(Ni²⁺/Ni) = - 0.25 V

So emf at standard condition

E^o_cell = E^o_cathode - E^o_anode = - 0.25 - (-0.76) = 0.51 V

The cell reaction is

Zn(s) + Ni²⁺(aq) $\to$ Ni_{​​​​​​} (s) + Zn²⁺(aq)

Number of electron transferred (n) = 2

Now, nernst equation for the cell reaction at 298 K is

For, Solids, [Zn] =[Ni] = 1

E_cell = E^o_cell - $\frac{0.059}{2}$ log [Zn²⁺]/[Ni²⁺] (1)

Or, E_cell = 0.51 - 0.0295 * log[Zn²⁺]/[Ni²⁺] (2)

2)

When, [Zn²⁺] = 0.130 M, [Ni²⁺]= 2.80 M

Or, E_cell = 0.51 - $\frac{0.059}{2}$ log (0.130/2.80)

Or, E_cell = 0.51 - 0.0295*(-1.33)

Or, E_{cell =} 0.51 + 0.04 = 0.55 V.( In two significant figure)

3)

[Ni²⁺] = 0.150 M, [Zn²⁺]= 0.999 M

Then, using Eq.2

E_cell = 0.51 - 0.0295* log ( 0.999/0.150)

E_cell = 0 51 - 0.0295 * (0.823)

Or, E_{cell =} 0.51 - 0.024 = 0.486 V = 0.49 V ( in two significant figure)