Question

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 E^o of Cu half equation?

What is the E^o of Zn half equation?

What is the E^o of Net Cell equation?

Answer 1

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 E^o 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