`Cu^(+) + e^(-) rightarrow Cu`, `E^(@) = x_1volt`, `Cu^(2+) + 2e^(-) rightarrow Cu`, `E^(@) = x_(2)` volt, then for `Cu^(2+) + e^(-) rightarrow Cu^(+)

Question

`Cu^(+) + e^(-) rightarrow Cu`, `E^(@) = x_1volt`, `Cu^(2+) + 2e^(-) rightarrow Cu`, `E^(@) = x_(2)` volt, then for `Cu^(2+) + e^(-) rightarrow Cu^(+)

`Cu^(+) + e^(-) rightarrow Cu`, `E^(@) = x_1volt`,
`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)`

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - D
`Cu^(+)+e^(-)toCu,E^(+)+x_(1)"volt"`
`Cu^(2+)+2e^(-)toCu,x_(2)"volt"`
`CutoCu^(+)+e^(-)-x_(1)"volt"`
`Cu^(2+)+e^(-)toCu^(+)`
`-2xx x_(2) xx f+1xx x_(1)xx f=-1xxE^(@)xxf`
`E^(@)=2x_(2)-x_(1)`

4 views Answered 2023-02-05 15:29