we constructed galvanic cells with the following solutions (with a their respect

Question

we constructed galvanic cells with the following solutions (with a their respective metals used as cathodes): KNO3 was used to construct a salt bridge to all the following solutions: ZnSO4, CuSO4, FeSO4, NI(NO3)2, KI/I2. we then used a multimeter to test voltaages, this is what we came up with for results: Cell Anode Cathod Voltage Assuming that the reduction potential for the most active metal is

Cell Anode Cathode Voltage

Zn+ Cu Zn Cu 1.01

Zn+ Fe Zn Fe 0.523

Zn+ Ni Zn Ni 0.971

Zn+ I2/I- Zn I2/I- 1.25

Cu+ Fe Fe Cu 0.489

Cu+ Ni Cu Ni 0.015

Cu+ I2/I- Cu I2/I- 0.252

Fe+ Ni Fe Ni 0.516

Fe+ I2/I- Fe I2/I- 0.733

Ni+ I2/I- Ni I2/I- 0.523

1) what is the highest cell voltage that could be constructed from 1/2 cells used in this experiement?

2) suppose that you had used Fe/Fe2+ as the reference reaction instead of the one that include the most active metal. How would your answer to question 1 have changed?

3) Determine value for delta G0 for the following systems using your value for E0

Zn + Fe2+ ==> Zn2+ + Fe

Cu + Zn2+ = =>Cu2+ + Zn

Explanation / Answer

delta Go = - nFEo

   F = Faraday constant = 96485 C/mol

n = no of electrons transferred

3) a) Zn + Fe2+ ==> Zn2+ + Fe Eo = 0.523 V

delta Go = - nFEo

= - 2 x 96465 C/mol x 0.523 V

= -100923 J

b) Cu + Zn2+ = =>Cu2+ + Zn   Eo = 1.01 V

   delta Go = - nFEo

= - 2 x 96465 C/mol x 1.01 V

= -194899 J

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