Question

25 mL of 0.1M ZnSO4 was placed in a beaker. 25 mL of 0.1 M CuSO4...

25 mL of 0.1M ZnSO4 was placed in a beaker. 25 mL of 0.1 M CuSO4 was placed in a ceramic cup. A strip of Zn was placed in the ZnSO4 and a strip of Cu was placed in the CuSO4 . The ceramic cup with the CuSO4 was placed in the beaker. A voltmeter was connected to the Zn strip and the Cu strip and registered 10.59 volts DC.

What is the Cell Voltage?

What is the Cu half equation?

What is the Zn half equation?

What is the Net cell equation?

What is the Eo of Cu half equation?

What is the Eo of Zn half equation?

What is the Eo of Net Cell equation?

Cu half equation :

Cu+2 + 2e- --------------------> Cu

Zn half equation :

Zn ---------------------> Zn+2 + 2e-

net cell equation :

Zn + Cu+2 -----------------> Zn+2 + Cu

Eo cell of Cu half equation :

Eo = +0.34 V

Eo cell of Zn half equation :

Eo = -0.76 V

What is the Eo of Net Cell equation

Eocell = Eo cathode - Eo anode

= 0.34 - (-0.76)

= 1.10 V

What is the Cell Voltage?

Q = [Zn+2] / [Cu+2]

Q = 0.1 / 0.1 = 1

if Q = 1 ,   Eo cell = Ecell

cell voltage   = 1.10 V

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