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A1^(3+) + 3e^(-) to Al(s), E^(@) = -1.66...

`A1^(3+) + 3e^(-) to Al(s)`, `E^(@) = -1.66V`
`Cu^(2+)+ 2e^(-) rightarrow CU(s)`, `E^(@)=+0.34V`
What voltage is produced under standard conditions by combining the half reactions with these standard electrode potentials?

A

1.32 V

B

2.00 V

C

2.30 V

D

4.34 V

Text Solution

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The correct Answer is:
B
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Standard electrode potential data is given below : Fe^(3+) (aq) + e ^(-) rarr Fe^(2+) (aq) , E^(o) = + 0.77 V Al^(3+) (aq) + 3e^(-) rarr Al (s) , E^(o) = - 1.66 V Br_(2) (aq) + 2e^(-) rarr 2Br^(-) (aq) , E^(o) = +1.08 V Based on the data given above, reducing power of Fe^(2+) Al and Br^(-) will increase in the order :

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Given: (i) Cu^(2+)+2e^(-) rarr Cu, E^(@) = 0.337 V (ii) Cu^(2+)+e^(-) rarr Cu^(+), E^(@) = 0.153 V Electrode potential, E^(@) for the reaction, Cu^(+)+e^(-) rarr Cu , will be

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