A voltaic cell is constructed that uses the following reaction and operates at 2

Question

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.

0.48

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Correct

Part B

What is the emf of this cell when [Ni2+]= 2.50 M and [Zn2+]= 0.140 M ?

Express your answer using two significant figures.

Part C

What is the emf of the cell when [Ni2+]= 0.220 M and [Zn2+]= 0.870 M ?

Express your answer using two significant figures.

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 =

0.48

  V  

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Correct

Part B

What is the emf of this cell when [Ni2+]= 2.50 M and [Zn2+]= 0.140 M ?

Express your answer using two significant figures.

E =   V  

Explanation / Answer

   Zn(s) ----------------------> Zn^2+ (aq) + 2e^-       E0 = 0.76v

Ni^2+ (aq) + 2e^- ------------> Ni(s)                     E0 = -0.23v

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Zn(s) + Ni^2+ (aq) ------------------> Zn^2+ (aq) + Ni(s)   E0 emf = 0.53v

    n = 2

Eemf = E0emf-0.0592/n logQ

   Eemfl = E0 emf - 0.0592/n log[Zn^2+]/[Ni^2+]

             = 0.53 - 0.0592/2 log0.14/2.5

             = 0.53 -0.0296*-1.2518 = 0.57V

c.

n(s) + Ni^2+ (aq) ------------------> Zn^2+ (aq) + Ni(s)   E0 emf = 0.53v

    n = 2

Eemf = E0emfl -0.0592/n logQ

   Eemf = E0 emf - 0.0592/n log[Zn^2+]/[Ni^2+]

             = 0.53 - 0.0592/2 log0.87/0.22

             = 0.53 -0.0296*0.597

             = 0.51V >>>answer

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