A voltaic cell is constructed that uses the following reaction and operates at 298 K: Zn(s)+Ni2+(aq)→Zn2+(aq)+Ni(s). Part A What is the emf of this cell under standard conditions? Express your answer using two significant figures. E∘ E ∘ = nothing V Request Answer Part B What is the emf of this cell when [Ni2+]= 2.00 M and [Zn2+]= 0.200 M ? Express your answer using two significant figures. E E = nothing V Request Answer Part C What is the emf of the cell when [Ni2+]= 0.290 M and [Zn2+]= 0.999 M ? Express your answer using two significant figures.

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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). Part A What is the emf of this cell under standard conditions? Express your answer using two significant figures. E∘ E ∘ = nothing V Request Answer Part B What is the emf of this cell when [Ni2+]= 2.00 M and [Zn2+]= 0.200 M ? Express your answer using two significant figures. E E = nothing V Request Answer Part C What is the emf of the cell when [Ni2+]= 0.290 M and [Zn2+]= 0.999 M ? Express your answer using two significant figures.

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Finti Finti · Answer 1 · 2025-03-21T23:15:59-04:00

Answer:

1. 0.559 V

2. 0.514 V

Explanation:

Nernst Equation - Can be used to find the cell potential at any moment in during a reaction or at conditions other than standard-state.

E = E₁ - RT/nF ㏑Q

E = cell potential (V) under specific conditions

E₁= cell potential at standard-state conditions

R = ideal gas constant = 8.314 J/mol-K

T = temperature (kelvin), which is generally 25C (298 K)

n = number of moles of electrons transferred in the balanced equation

F = Faraday's constant, the charge on a mole of electrons = 95,484.56 C/mol

lnQ = the natural log of the reaction quotient at the moment in time

First calculate E₀ for the cell.

Zn(s) ==> Zn²⁺(aq) + 2e⁻ . . .E₀ = +0.76 V

Ni⁺²(aq) + 2e⁻ ==> Ni(s) . . . E₀ = -0.23 V

Zn(s) + Ni⁺²(aq) ==> Zn⁺²(aq) + Ni(s) . .E₀ cell = +0.53 V

Then use the Nernst equation:

E cell = E₀ cell - 0.059/n log Q = 0.53 - 0.059/2 log ([Zn⁺²]/[Ni⁺²]) = 0.53 - 0.0295 log (0.200/2.0) = 0.559 V

For the second set of conditions,

E cell = 0.53 - 0.0295 log (0.999/0.290) = 0.514 V

okpalawalter8 okpalawalter8 · Answer 2 · 2025-03-21T23:15:32-04:00

The emf of the cell when [Ni2+]= 0.290 M and [Zn2+]= 0.999 M is mathematically given as

E cell= 0.514 V

What is the emf of the cell when [Ni2+]= 0.290 M and [Zn2+]= 0.999 M ?

Generally, the equation for the Chemical reaction is mathematically given as

Zn(s)+Ni2+(aq)→Zn2+(aq)+Ni(s).

Therefore

Zn(s) --> Zn2⁺(aq) + 2e- . . .E₀ = +0.76 V

Ni⁺2(aq) + 2e⁻ ---> Ni(s) . . . E₀ = -0.23 V

hence

E cell = E₀ cell - 0.059/n log Q

E cell = 0.53 - 0.059/2 log ([Zn⁺²]/[Ni⁺²])

E cell = 0.53 - 0.0295 log (0.200/2.0)

E cell = 0.559 V

In conclusion

E cell = 0.53 - 0.0295 log (0.999/0.290)

E cell= 0.514 V