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Cu^(+) + e^(-) rightarrow Cu, E^(@) = x1...

`Cu^(+) + e^(-) rightarrow Cu`, `E^(@) = x1volt`,
`Cu^(2+) + 2e^(-) rightarrow Cu`, `E^(@) = x_(2)` volt, then for
`Cu^(2+) + e^(-) rightarrow Cu^(+), E^(@)` (volt) will be :

A

`x_(1) - 2x_(2)`

B

`x_(1) + 2x_(2)`

C

`x_(1) - x_(2)`

D

`2x_(2) - x_(1)`

Text Solution

Verified by Experts

The correct Answer is:
D
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Cu^(+) + e rarr Cu, E^(@) = X_(1) volt, Cu^(2+) + 2e rarr Cu, E^(@) = X_(2) X_(2) volt For Cu^(2+) + e rarr Cu^(+), E^(@) will be :

Given the standard reduction potentials, which statement is correct? {:(Cu^(2)(aq) + 2e^(-) rightarrow Cu(s), E^(@) = 0.34V), (2H^(+)(aq) + 2e^(-) rightarrow H_(2)(g), E^(@)= 0.0V), (Cr^(3+)(aq) + 2e^(-) rightarrow Cr(s),E^(@) = -0.73V):}

Which of the half reactions, when coupled, will make a galvanic cell that will produce the largest voltage under standard conditons? P. Cu^(2+)(aq) + 2e^(-) rightarrow Cu(s) E^(@) = +0.34V Q. Pb^(2+)(aq) + 2e^(-) rightarrow Pb(s) E^(@) = -0.13V R. Ag^(+)(aq) + 3e^(-) rightarrow Al(s) E^(@) = -1.66V

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

Au^(3+) + 3^(-) rightarrow Au(s) E^(@) = 1.50V Cu^(2+) + 2e^(-) rightarrow Cu(s) E^(@) = 0.34V According to the standard reduction potentials above, a substance that can oxidize only one of these metals must have an E^(@) value:

Goven : (i) CU^(2+) + 2e^- rarr Cu, E^@ = 0.337 V (ii) Cu^(2+) +e^- rarr Cu^(+) , E^2=0.1 5 3 V .

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