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Al^(3+)(aq)+3e^(-)rarrAl(s) E^(@)=-1.66V...

`Al^(3+)(aq)+3e^(-)rarrAl(s) E^(@)=-1.66V`
`Mn^(2+)(aq)+2e^(-)rarrMn(s) E^(@)=-1.18V`
What process occurs at the anode of a volatic cell utilizing these two half-reactions?

A

`Al(s)rarrAl^(3+)(aq)+3e^(-)`

B

`Al^(3+)(aq)+3e^(-)rarrAl(s)`

C

`Mn(s)rarrMn^(2+)(aq)+2e^(-)`

D

`Mn^(2+)(aq)+2e^(-)rarrMn(s)`

Text Solution

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The correct Answer is:
A
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Al^(3+)(aq)+3e^(-)rarrAl(s) E^(@)=-1.66V Mn^(2+)(aq)+2e^(-)rarrMn(s) E^(@)=-1.18V What is the standard potential of a volaic cell produced by using these two half-reactions?

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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?

Electrode potential data are given below: Fe^(3+)(aq)+e^(-)rarrFe^(2+)(aq):E^(@)=+0.77V Al^(3+)+3e^(-)rarrAl(s):E^(@)=-1.66V Br_(2)(aq)+2e^(-)rarr2Br^(-)(aq):E^(@)=+1.08V , Based on the data, the reducing power of Fe^(2+) Al and Br^(-) will increase in the order

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