Question

25 mL of 0.1M ZnSO_{4} was placed in a beaker. 25 mL of
0.1 M CuSO_{4} was placed in a ceramic cup. A strip of Zn
was placed in the ZnSO_{4} and a strip of Cu was placed in
the CuSO_{4} . The ceramic cup with the CuSO_{4}
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**

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