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Electrochem help (1 Viewer)
- Thread starter bengosha60
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faisalabdul16
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- 2014
The answer is C, exam choice paper, has weird questions in it
Negative ions flow to the anode, positive ions flow to the cathode so for picture 1 , y = anode and z = cathode. Y has reduction potential of -0.24 however it is being oxidised so to work out x you do 0.43-0.24 = 0.19
z = 0.19 = reduction potential.
In second diagram, z is being oxidised and x is being reduced. so -0.19 + x = 1.15V
x = 1.34
Negative ions flow to the anode, positive ions flow to the cathode so for picture 1 , y = anode and z = cathode. Y has reduction potential of -0.24 however it is being oxidised so to work out x you do 0.43-0.24 = 0.19
z = 0.19 = reduction potential.
In second diagram, z is being oxidised and x is being reduced. so -0.19 + x = 1.15V
x = 1.34
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